https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Algebra+star1234&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T10:13:03ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_17&diff=769072016 AMC 10B Problems/Problem 172016-02-22T01:59:29Z<p>Algebra star1234: /* Solution 2= */</p>
<hr />
<div>==Problem==<br />
<br />
All the numbers <math>2, 3, 4, 5, 6, 7</math> are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br />
<br />
<math>\textbf{(A)}\ 312 \qquad<br />
\textbf{(B)}\ 343 \qquad<br />
\textbf{(C)}\ 625 \qquad<br />
\textbf{(D)}\ 729 \qquad<br />
\textbf{(E)}\ 1680</math><br />
<br />
==Solution 1==<br />
Let us call the six sides of our cube <math>a,b,c,d,e,</math> and <math>f</math> (where <math>a</math> is opposite <math>d</math>, <math>c</math> is opposite <math>e</math>, and <math>b</math> is opposite <math>f</math>.<br />
Thus, for the eight vertices, we have the following products: <math>abc</math>,<math>abe</math>,<math>bcd</math>,<math>bde</math>,<math>acf</math>,<math>cdf</math>,<math>cef</math>, and <math>def</math>.<br />
Let us find the sum of these products:<br />
<math>abc+abe+bcd+bde+acf+cdf+aef+def</math><br />
We notice <math>b</math> is a factor of the first four terms, and <math>f</math> is factor is the last four terms.<br />
<math>b(ac+ae+cd+de)+f(ac+ae+cd+de)</math><br />
Now, we can factor even more:<br />
<math>(b+f)(ac+ae+cd+de)</math><br />
<math>(b+f)(a(c+e)+d(c+e))</math><br />
<math>(b+f)(a+d)(c+e)</math><br />
We have the product. Notice how the factors are sums of opposite faces. The best sum for this is to make <math>(7+2)</math>,<math>(6+3)</math>, and <math>(5+4)</math> all factors.<br />
<math> (7+2)(6+3)(5+4)</math><br />
<math>9</math> <math> *</math> <math> 9</math> <math> *</math> <math>9</math><br />
<math> 729 </math><br />
Thus our answer is <math>\textbf{(D)}\ 729</math>.<br />
<br />
<br />
==Solution 2==<br />
We first find the factorization <math>(b+f)(a+d)(c+e)</math> using the method in Solution 1. By using AM-GM, we get, <math>(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3</math>. To maximize, the factorization, we get the answer is <math>\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_17&diff=769062016 AMC 10B Problems/Problem 172016-02-22T01:58:51Z<p>Algebra star1234: </p>
<hr />
<div>==Problem==<br />
<br />
All the numbers <math>2, 3, 4, 5, 6, 7</math> are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br />
<br />
<math>\textbf{(A)}\ 312 \qquad<br />
\textbf{(B)}\ 343 \qquad<br />
\textbf{(C)}\ 625 \qquad<br />
\textbf{(D)}\ 729 \qquad<br />
\textbf{(E)}\ 1680</math><br />
<br />
==Solution 1==<br />
Let us call the six sides of our cube <math>a,b,c,d,e,</math> and <math>f</math> (where <math>a</math> is opposite <math>d</math>, <math>c</math> is opposite <math>e</math>, and <math>b</math> is opposite <math>f</math>.<br />
Thus, for the eight vertices, we have the following products: <math>abc</math>,<math>abe</math>,<math>bcd</math>,<math>bde</math>,<math>acf</math>,<math>cdf</math>,<math>cef</math>, and <math>def</math>.<br />
Let us find the sum of these products:<br />
<math>abc+abe+bcd+bde+acf+cdf+aef+def</math><br />
We notice <math>b</math> is a factor of the first four terms, and <math>f</math> is factor is the last four terms.<br />
<math>b(ac+ae+cd+de)+f(ac+ae+cd+de)</math><br />
Now, we can factor even more:<br />
<math>(b+f)(ac+ae+cd+de)</math><br />
<math>(b+f)(a(c+e)+d(c+e))</math><br />
<math>(b+f)(a+d)(c+e)</math><br />
We have the product. Notice how the factors are sums of opposite faces. The best sum for this is to make <math>(7+2)</math>,<math>(6+3)</math>, and <math>(5+4)</math> all factors.<br />
<math> (7+2)(6+3)(5+4)</math><br />
<math>9</math> <math> *</math> <math> 9</math> <math> *</math> <math>9</math><br />
<math> 729 </math><br />
Thus our answer is <math>\textbf{(D)}\ 729</math>.<br />
<br />
<br />
==Solution 2===<br />
We first find the factorization <math>(b+f)(a+d)(c+e)</math>. By using AM-GM, we get, <math>(b+f)(a+d)(c+e) \le \left( \frac{a+b+c+d+e+f}{3} \right)^3</math>. To maximize, the factorization, we get the answer is <math>\left( \frac{27}{3} \right)^3 = \boxed{\textbf{(D)}\ 729}</math><br />
==See Also==<br />
{{AMC10 box|year=2016|ab=B|num-b=16|num-a=18}}<br />
{{MAA Notice}}</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=76516AMC historical results2016-02-20T18:08:48Z<p>Algebra star1234: </p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor:<br />
<br />
===AMC 12A===<br />
*Average score:<br />
*AIME floor: 92<br />
<br />
===AMC 12B===<br />
*Average score:<br />
*AIME floor:<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff:<br />
*USAJMO cutoff:<br />
<br />
===AIME II===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff:<br />
*USAJMO cutoff:<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: <br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: <br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: <br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
==1959==<br />
==1958==<br />
==1957==<br />
==1956==<br />
==1955==<br />
==1954==<br />
==1953==<br />
==1952==<br />
==1951==<br />
==1950==<br />
<br />
[[Category:Historical results]]</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=76515AMC historical results2016-02-20T17:57:34Z<p>Algebra star1234: </p>
<hr />
<div><!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --><br />
This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br />
<br />
==2016==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor:<br />
<br />
===AMC 12A===<br />
*Average score:<br />
*AIME floor: 92 (subject to change)<br />
<br />
===AMC 12B===<br />
*Average score:<br />
*AIME floor:<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff:<br />
*USAJMO cutoff:<br />
<br />
===AIME II===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff:<br />
*USAJMO cutoff:<br />
<br />
==2015==<br />
===AMC 10A===<br />
*Average score: 73.39<br />
*AIME floor: 106.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.10<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 69.90<br />
*AIME floor: 99<br />
<br />
===AMC 12B===<br />
*Average score: 66.92<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: 219.0<br />
*USAJMO cutoff: 213.0<br />
<br />
===AIME II===<br />
*Average score: <br />
*Median score: <br />
*USAMO cutoff: 229.0<br />
*USAJMO cutoff: 223.5<br />
<br />
==2014==<br />
===AMC 10A===<br />
*Average score: 63.83<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 71.44<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.01<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 68.11<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 4.88<br />
*Median score: <br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
===AIME II===<br />
*Average score: 5.49<br />
*Median score: <br />
*USAMO cutoff: 211.5<br />
*USAJMO cutoff: 211<br />
<br />
==2013==<br />
===AMC 10A===<br />
*Average score: 72.50<br />
*AIME floor: 108<br />
<br />
===AMC 10B===<br />
*Average score: 72.62<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 65.06<br />
*AIME floor:88.5<br />
<br />
===AMC 12B===<br />
*Average score: 64.21<br />
*AIME floor: 93<br />
<br />
===AIME I===<br />
*Average score: 4.69<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
===AIME II===<br />
*Average score: 6.56<br />
*Median score: <br />
*USAMO cutoff: 209<br />
*USAJMO cutoff: 210.5<br />
<br />
==2012==<br />
===AMC 10A===<br />
*Average score: 72.51<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 76.59<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 64.62<br />
*AIME floor: 94.5<br />
<br />
===AMC 12B===<br />
*Average score: 70.08<br />
*AIME floor: 99<br />
<br />
===AIME I===<br />
*Average score: 5.13<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
===AIME II===<br />
*Average score: 4.94<br />
*Median score: <br />
*USAMO cutoff: 204.5<br />
*USAJMO cutoff: 204<br />
<br />
==2011==<br />
===AMC 10A===<br />
*Average score: 67.61<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 71.78<br />
*AIME floor: 117<br />
<br />
===AMC 12A===<br />
*Average score: 66.77<br />
*AIME floor: 93<br />
<br />
===AMC 12B===<br />
*Average score: 64.71<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: 5.47<br />
*Median score: <br />
*USAMO cutoff: 188<br />
*USAJMO cutoff: 179<br />
<br />
===AIME II===<br />
*Average score: 2.23<br />
*Median score: <br />
*USAMO cutoff: 215.5<br />
*USAJMO cutoff: 196.5<br />
<br />
==2010==<br />
===AMC 10A===<br />
*Average score: 68.11<br />
*AIME floor: 115.5<br />
<br />
===AMC 10B===<br />
*Average score: 68.57<br />
*AIME floor: 118.5<br />
<br />
===AMC 12A===<br />
*Average score: 61.02<br />
*AIME floor: 88.5<br />
<br />
===AMC 12B===<br />
*Average score: 59.58<br />
*AIME floor: 88.5<br />
<br />
===AIME I===<br />
*Average score: 5.90<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
===AIME II===<br />
*Average score: 3.39<br />
*Median score: <br />
*USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br />
*USAJMO cutoff: 188.5<br />
<br />
==2009==<br />
===AMC 10A===<br />
*Average score: 67.41<br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: 74.73<br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 66.37<br />
*AIME floor: <br />
<br />
===AMC 12B===<br />
*Average score: 71.88<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2008==<br />
===AMC 10A===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 10B===<br />
*Average score: <br />
*AIME floor: <br />
<br />
===AMC 12A===<br />
*Average score: 65.6<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 68.9<br />
*AIME floor: 97.5<br />
<br />
===AIME I===<br />
*Average score: <br />
*Median score: <br />
*USAMO floor: <br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2007==<br />
<br />
===AMC 10A===<br />
*Average score: 67.9<br />
*AIME floor: 117<br />
<br />
===AMC 10B===<br />
*Average score: 61.5<br />
*AIME floor: 115.5<br />
<br />
===AMC 12A===<br />
*Average score: 66.8<br />
*AIME floor: 97.5<br />
<br />
===AMC 12B===<br />
*Average score: 73.1<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score: 5<br />
*Median score: 3<br />
*USAMO floor: 6<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2006==<br />
===AMC 10A===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 68.5<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 85.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 85.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2005==<br />
===AMC 10A===<br />
*Average score: 74.0<br />
*AIME floor: 120<br />
<br />
===AMC 10B===<br />
*Average score: 79.0<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 78.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 83.4<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2004==<br />
===AMC 10A===<br />
*Average score: 69.1<br />
*AIME floor: 110<br />
<br />
===AMC 10B===<br />
*Average score: 80.4<br />
*AIME floor: 120<br />
<br />
===AMC 12A===<br />
*Average score: 73.9<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 84.5<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2003==<br />
===AMC 10A===<br />
*Average score: 74.4<br />
*AIME floor: 119<br />
<br />
===AMC 10B===<br />
*Average score: 79.6<br />
*AIME floor: 121<br />
<br />
===AMC 12A===<br />
*Average score: 77.8<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 76.6<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2002==<br />
===AMC 10A===<br />
*Average score: 68.5<br />
*AIME floor: 115<br />
<br />
===AMC 10B===<br />
*Average score: 74.9<br />
*AIME floor: 118<br />
<br />
===AMC 12A===<br />
*Average score: 72.7<br />
*AIME floor: 100<br />
<br />
===AMC 12B===<br />
*Average score: 80.8<br />
*AIME floor: 100<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2001==<br />
===AMC 10===<br />
*Average score: 67.8<br />
*AIME floor: 116<br />
<br />
===AMC 12===<br />
*Average score: 56.6<br />
*AIME floor: 84<br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==2000==<br />
===AMC 10===<br />
*Average score: 64.2<br />
*AIME floor: <br />
<br />
===AMC 12===<br />
*Average score: 64.9<br />
*AIME floor: <br />
<br />
===AIME I===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
===AIME II===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1999==<br />
===AHSME===<br />
*Average score: 68.8<br />
*AIME floor:<br />
<br />
===AIME===<br />
*Average score:<br />
*Median score:<br />
*USAMO floor:<br />
<br />
==1998==<br />
==1997==<br />
==1996==<br />
==1995==<br />
==1994==<br />
==1993==<br />
==1992==<br />
==1991==<br />
==1990==<br />
==1989==<br />
==1988==<br />
==1987==<br />
==1986==<br />
==1985==<br />
==1984==<br />
==1983==<br />
==1982==<br />
==1981==<br />
==1980==<br />
==1979==<br />
==1978==<br />
==1977==<br />
==1976==<br />
==1975==<br />
==1974==<br />
==1973==<br />
==1972==<br />
==1971==<br />
==1970==<br />
==1969==<br />
==1968==<br />
==1967==<br />
==1966==<br />
==1965==<br />
==1964==<br />
==1963==<br />
==1962==<br />
==1961==<br />
==1960==<br />
==1959==<br />
==1958==<br />
==1957==<br />
==1956==<br />
==1955==<br />
==1954==<br />
==1953==<br />
==1952==<br />
==1951==<br />
==1950==<br />
<br />
[[Category:Historical results]]</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=750812016 AMC 10A Problems2016-02-03T22:54:09Z<p>Algebra star1234: /* Problem 4 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <math>\dfrac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
[[2016 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
[[2016 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
For every dollar Ben spent on bagels, David spent <math>25</math> cents less. Ben paid <math>\$12.50</math> more than David. How much did they spend in the bagel store together?<br />
<br />
<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math> <br />
<br />
[[2016 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
<br />
[[2016 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math><br />
==Problem 6==<br />
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br />
<br />
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math><br />
==Problem 7==<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math><br />
==Problem 8==<br />
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays <math>40</math> coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math><br />
==Problem 9==<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
==Problem 10==<br />
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle?<br />
<asy><br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
path rectangle(pair X, pair Y){<br />
return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br />
}<br />
filldraw(rectangle((0,0),(7,5)),gray(0.5));<br />
filldraw(rectangle((1,1),(6,4)),gray(0.75));<br />
filldraw(rectangle((2,2),(5,3)),white);<br />
<br />
label("$1$",(0.5,2.5));<br />
draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br />
draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(1.5,2.5));<br />
draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br />
draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.5,2.5));<br />
draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br />
draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.1,1.5));<br />
draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br />
draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br />
<br />
label("$1$",(3.7,0.5));<br />
draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br />
draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br />
</asy><br />
<br />
<cmath>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</cmath><br />
==Problem 11==<br />
What is the area of the shaded region of the given <math>8 \times 5</math> rectangle?<br />
<br />
<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
<br />
label("$1$",(8,1/2),dir(0));<br />
label("$4$",(8,3),dir(0));<br />
<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
<br />
label("$1$",(0,9/2),dir(180));<br />
label("$4$",(0,2),dir(180));<br />
<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math><br />
==Problem 12==<br />
Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correct statement about the probability <math>p</math> that the product of the three integers is odd?<br />
<br />
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math><br />
==Problem 13==<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
<br />
<math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math><br />
==Problem 14==<br />
How many ways are there to write <math>2016</math> as the sum of twos and threes, ignoring order? (For example, <math>1008\cdot 2 + 0\cdot 3</math> and <math>402\cdot 2 + 404\cdot 3</math> are two such ways.)<br />
<br />
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math><br />
<br />
[[2016 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Seven cookies of radius <math>1</math> inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((0,0),1));<br />
draw(circle((1,sqrt(3)),1));<br />
draw(circle((-1,sqrt(3)),1)); <br />
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br />
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi</math><br />
<br />
==Problem 16==<br />
A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}</math> counterclockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(B)}</math> clockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(C)}</math> reflection about the <math>x</math>-axis <br />
<br />
<math>\textbf{(D)}</math> reflection about the line <math>y = x</math> <br />
<br />
<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.<br />
<br />
==Problem 17==<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math><br />
<br />
==Problem 18==<br />
Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math><br />
==Problem 19==<br />
In rectangle ABCD, <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ration <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?<br />
==Problem 20==<br />
For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>?<br />
<br />
<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math><br />
<br />
==Problem 21==<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
==Problem 22==<br />
For some positive integer <math>n</math>, the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3</math>. How many positive integer divisors does the number <math>81n^4</math> have?<br />
<br />
<math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math><br />
==Problem 23==<br />
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
<br />
<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
<br />
==Problem 24==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
==Problem 25==<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math></div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=750802016 AMC 10A Problems2016-02-03T22:53:44Z<p>Algebra star1234: /* Problem 2 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <math>\dfrac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
[[2016 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
[[2016 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
For every dollar Ben spent on bagels, David spent <math>25</math> cents less. Ben paid <math>\$12.50</math> more than David. How much did they spend in the bagel store together?<br />
<br />
<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math> <br />
<br />
[[2016 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
[[2016 AMC 10A Problems/Problem 4|Solution]]<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
==Problem 5==<br />
A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math><br />
==Problem 6==<br />
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br />
<br />
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math><br />
==Problem 7==<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math><br />
==Problem 8==<br />
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays <math>40</math> coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math><br />
==Problem 9==<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
==Problem 10==<br />
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle?<br />
<asy><br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
path rectangle(pair X, pair Y){<br />
return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br />
}<br />
filldraw(rectangle((0,0),(7,5)),gray(0.5));<br />
filldraw(rectangle((1,1),(6,4)),gray(0.75));<br />
filldraw(rectangle((2,2),(5,3)),white);<br />
<br />
label("$1$",(0.5,2.5));<br />
draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br />
draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(1.5,2.5));<br />
draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br />
draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.5,2.5));<br />
draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br />
draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.1,1.5));<br />
draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br />
draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br />
<br />
label("$1$",(3.7,0.5));<br />
draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br />
draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br />
</asy><br />
<br />
<cmath>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</cmath><br />
==Problem 11==<br />
What is the area of the shaded region of the given <math>8 \times 5</math> rectangle?<br />
<br />
<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
<br />
label("$1$",(8,1/2),dir(0));<br />
label("$4$",(8,3),dir(0));<br />
<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
<br />
label("$1$",(0,9/2),dir(180));<br />
label("$4$",(0,2),dir(180));<br />
<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math><br />
==Problem 12==<br />
Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correct statement about the probability <math>p</math> that the product of the three integers is odd?<br />
<br />
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math><br />
==Problem 13==<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
<br />
<math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math><br />
==Problem 14==<br />
How many ways are there to write <math>2016</math> as the sum of twos and threes, ignoring order? (For example, <math>1008\cdot 2 + 0\cdot 3</math> and <math>402\cdot 2 + 404\cdot 3</math> are two such ways.)<br />
<br />
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math><br />
<br />
[[2016 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Seven cookies of radius <math>1</math> inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((0,0),1));<br />
draw(circle((1,sqrt(3)),1));<br />
draw(circle((-1,sqrt(3)),1)); <br />
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br />
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi</math><br />
<br />
==Problem 16==<br />
A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}</math> counterclockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(B)}</math> clockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(C)}</math> reflection about the <math>x</math>-axis <br />
<br />
<math>\textbf{(D)}</math> reflection about the line <math>y = x</math> <br />
<br />
<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.<br />
<br />
==Problem 17==<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math><br />
<br />
==Problem 18==<br />
Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math><br />
==Problem 19==<br />
In rectangle ABCD, <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ration <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?<br />
==Problem 20==<br />
For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>?<br />
<br />
<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math><br />
<br />
==Problem 21==<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
==Problem 22==<br />
For some positive integer <math>n</math>, the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3</math>. How many positive integer divisors does the number <math>81n^4</math> have?<br />
<br />
<math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math><br />
==Problem 23==<br />
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
<br />
<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
<br />
==Problem 24==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
==Problem 25==<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math></div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=750792016 AMC 10A Problems2016-02-03T22:53:32Z<p>Algebra star1234: /* Problem 3 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <math>\dfrac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
[[2016 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
[[2016 AMC 10A Problems/Problem 2|Solution]]<br />
==Problem 3==<br />
For every dollar Ben spent on bagels, David spent <math>25</math> cents less. Ben paid <math>\$12.50</math> more than David. How much did they spend in the bagel store together?<br />
<br />
<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math> <br />
<br />
[[2016 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
[[2016 AMC 10A Problems/Problem 4|Solution]]<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
==Problem 5==<br />
A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math><br />
==Problem 6==<br />
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br />
<br />
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math><br />
==Problem 7==<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math><br />
==Problem 8==<br />
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays <math>40</math> coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math><br />
==Problem 9==<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
==Problem 10==<br />
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle?<br />
<asy><br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
path rectangle(pair X, pair Y){<br />
return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br />
}<br />
filldraw(rectangle((0,0),(7,5)),gray(0.5));<br />
filldraw(rectangle((1,1),(6,4)),gray(0.75));<br />
filldraw(rectangle((2,2),(5,3)),white);<br />
<br />
label("$1$",(0.5,2.5));<br />
draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br />
draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(1.5,2.5));<br />
draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br />
draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.5,2.5));<br />
draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br />
draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.1,1.5));<br />
draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br />
draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br />
<br />
label("$1$",(3.7,0.5));<br />
draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br />
draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br />
</asy><br />
<br />
<cmath>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</cmath><br />
==Problem 11==<br />
What is the area of the shaded region of the given <math>8 \times 5</math> rectangle?<br />
<br />
<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
<br />
label("$1$",(8,1/2),dir(0));<br />
label("$4$",(8,3),dir(0));<br />
<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
<br />
label("$1$",(0,9/2),dir(180));<br />
label("$4$",(0,2),dir(180));<br />
<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math><br />
==Problem 12==<br />
Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correct statement about the probability <math>p</math> that the product of the three integers is odd?<br />
<br />
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math><br />
==Problem 13==<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
<br />
<math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math><br />
==Problem 14==<br />
How many ways are there to write <math>2016</math> as the sum of twos and threes, ignoring order? (For example, <math>1008\cdot 2 + 0\cdot 3</math> and <math>402\cdot 2 + 404\cdot 3</math> are two such ways.)<br />
<br />
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math><br />
<br />
[[2016 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Seven cookies of radius <math>1</math> inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((0,0),1));<br />
draw(circle((1,sqrt(3)),1));<br />
draw(circle((-1,sqrt(3)),1)); <br />
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br />
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi</math><br />
<br />
==Problem 16==<br />
A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}</math> counterclockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(B)}</math> clockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(C)}</math> reflection about the <math>x</math>-axis <br />
<br />
<math>\textbf{(D)}</math> reflection about the line <math>y = x</math> <br />
<br />
<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.<br />
<br />
==Problem 17==<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math><br />
<br />
==Problem 18==<br />
Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math><br />
==Problem 19==<br />
In rectangle ABCD, <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ration <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?<br />
==Problem 20==<br />
For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>?<br />
<br />
<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math><br />
<br />
==Problem 21==<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
==Problem 22==<br />
For some positive integer <math>n</math>, the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3</math>. How many positive integer divisors does the number <math>81n^4</math> have?<br />
<br />
<math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math><br />
==Problem 23==<br />
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
<br />
<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
<br />
==Problem 24==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
==Problem 25==<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math></div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=750782016 AMC 10A Problems2016-02-03T22:53:15Z<p>Algebra star1234: /* Problem 3 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <math>\dfrac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
[[2016 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
[[2016 AMC 10A Problems/Problem 2|Solution]]<br />
==Problem 3==<br />
For every dollar Ben spent on bagels, David spent <math>25</math> cents less. Ben paid <math>\$12.50</math> more than David. How much did they spend in the bagel store together?<br />
<br />
<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math> \\<br />
[[2016 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
[[2016 AMC 10A Problems/Problem 4|Solution]]<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
==Problem 5==<br />
A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math><br />
==Problem 6==<br />
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br />
<br />
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math><br />
==Problem 7==<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math><br />
==Problem 8==<br />
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays <math>40</math> coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math><br />
==Problem 9==<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
==Problem 10==<br />
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle?<br />
<asy><br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
path rectangle(pair X, pair Y){<br />
return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br />
}<br />
filldraw(rectangle((0,0),(7,5)),gray(0.5));<br />
filldraw(rectangle((1,1),(6,4)),gray(0.75));<br />
filldraw(rectangle((2,2),(5,3)),white);<br />
<br />
label("$1$",(0.5,2.5));<br />
draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br />
draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(1.5,2.5));<br />
draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br />
draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.5,2.5));<br />
draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br />
draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.1,1.5));<br />
draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br />
draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br />
<br />
label("$1$",(3.7,0.5));<br />
draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br />
draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br />
</asy><br />
<br />
<cmath>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</cmath><br />
==Problem 11==<br />
What is the area of the shaded region of the given <math>8 \times 5</math> rectangle?<br />
<br />
<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
<br />
label("$1$",(8,1/2),dir(0));<br />
label("$4$",(8,3),dir(0));<br />
<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
<br />
label("$1$",(0,9/2),dir(180));<br />
label("$4$",(0,2),dir(180));<br />
<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math><br />
==Problem 12==<br />
Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correct statement about the probability <math>p</math> that the product of the three integers is odd?<br />
<br />
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math><br />
==Problem 13==<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
<br />
<math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math><br />
==Problem 14==<br />
How many ways are there to write <math>2016</math> as the sum of twos and threes, ignoring order? (For example, <math>1008\cdot 2 + 0\cdot 3</math> and <math>402\cdot 2 + 404\cdot 3</math> are two such ways.)<br />
<br />
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math><br />
<br />
[[2016 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Seven cookies of radius <math>1</math> inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((0,0),1));<br />
draw(circle((1,sqrt(3)),1));<br />
draw(circle((-1,sqrt(3)),1)); <br />
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br />
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi</math><br />
<br />
==Problem 16==<br />
A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}</math> counterclockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(B)}</math> clockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(C)}</math> reflection about the <math>x</math>-axis <br />
<br />
<math>\textbf{(D)}</math> reflection about the line <math>y = x</math> <br />
<br />
<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.<br />
<br />
==Problem 17==<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math><br />
<br />
==Problem 18==<br />
Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math><br />
==Problem 19==<br />
In rectangle ABCD, <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ration <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?<br />
==Problem 20==<br />
For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>?<br />
<br />
<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math><br />
<br />
==Problem 21==<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
==Problem 22==<br />
For some positive integer <math>n</math>, the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3</math>. How many positive integer divisors does the number <math>81n^4</math> have?<br />
<br />
<math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math><br />
==Problem 23==<br />
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
<br />
<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
<br />
==Problem 24==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
==Problem 25==<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math></div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=750772016 AMC 10A Problems2016-02-03T22:52:48Z<p>Algebra star1234: </p>
<hr />
<div>==Problem 1==<br />
What is the value of <math>\dfrac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
[[2016 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
[[2016 AMC 10A Problems/Problem 2|Solution]]<br />
==Problem 3==<br />
For every dollar Ben spent on bagels, David spent <math>25</math> cents less. Ben paid <math>\$12.50</math> more than David. How much did they spend in the bagel store together?<br />
<br />
<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math><br />
[[2016 AMC 10A Problems/Problem 3|Solution]]<br />
==Problem 4==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
[[2016 AMC 10A Problems/Problem 4|Solution]]<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
==Problem 5==<br />
A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math><br />
==Problem 6==<br />
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br />
<br />
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math><br />
==Problem 7==<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math><br />
==Problem 8==<br />
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays <math>40</math> coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math><br />
==Problem 9==<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
==Problem 10==<br />
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle?<br />
<asy><br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
path rectangle(pair X, pair Y){<br />
return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br />
}<br />
filldraw(rectangle((0,0),(7,5)),gray(0.5));<br />
filldraw(rectangle((1,1),(6,4)),gray(0.75));<br />
filldraw(rectangle((2,2),(5,3)),white);<br />
<br />
label("$1$",(0.5,2.5));<br />
draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br />
draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(1.5,2.5));<br />
draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br />
draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.5,2.5));<br />
draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br />
draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.1,1.5));<br />
draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br />
draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br />
<br />
label("$1$",(3.7,0.5));<br />
draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br />
draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br />
</asy><br />
<br />
<cmath>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</cmath><br />
==Problem 11==<br />
What is the area of the shaded region of the given <math>8 \times 5</math> rectangle?<br />
<br />
<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
<br />
label("$1$",(8,1/2),dir(0));<br />
label("$4$",(8,3),dir(0));<br />
<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
<br />
label("$1$",(0,9/2),dir(180));<br />
label("$4$",(0,2),dir(180));<br />
<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math><br />
==Problem 12==<br />
Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correct statement about the probability <math>p</math> that the product of the three integers is odd?<br />
<br />
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math><br />
==Problem 13==<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
<br />
<math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math><br />
==Problem 14==<br />
How many ways are there to write <math>2016</math> as the sum of twos and threes, ignoring order? (For example, <math>1008\cdot 2 + 0\cdot 3</math> and <math>402\cdot 2 + 404\cdot 3</math> are two such ways.)<br />
<br />
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math><br />
<br />
[[2016 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Seven cookies of radius <math>1</math> inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((0,0),1));<br />
draw(circle((1,sqrt(3)),1));<br />
draw(circle((-1,sqrt(3)),1)); <br />
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br />
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi</math><br />
<br />
==Problem 16==<br />
A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}</math> counterclockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(B)}</math> clockwise rotation about the origin by <math>90^{\circ}</math>. <br />
<br />
<math>\textbf{(C)}</math> reflection about the <math>x</math>-axis <br />
<br />
<math>\textbf{(D)}</math> reflection about the line <math>y = x</math> <br />
<br />
<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.<br />
<br />
==Problem 17==<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math><br />
<br />
==Problem 18==<br />
Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math><br />
==Problem 19==<br />
In rectangle ABCD, <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ration <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?<br />
==Problem 20==<br />
For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>?<br />
<br />
<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math><br />
<br />
==Problem 21==<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
==Problem 22==<br />
For some positive integer <math>n</math>, the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3</math>. How many positive integer divisors does the number <math>81n^4</math> have?<br />
<br />
<math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math><br />
==Problem 23==<br />
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
<br />
<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
<br />
==Problem 24==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
==Problem 25==<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math></div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_1&diff=750732016 AMC 10A Problems/Problem 12016-02-03T22:51:31Z<p>Algebra star1234: </p>
<hr />
<div>What is the value of <math>\dfrac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
===Solution===<br />
<br />
<math>\frac{11!-10!}{9!} = \frac{9!(10 \cdot 11 -10)}{9!} = 10 \cdot 11 - 10 =\boxed{\textbf{(B)} 100}</math></div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=750682016 AMC 10A Problems2016-02-03T22:49:06Z<p>Algebra star1234: /* Problem 1 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <math>\dfrac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
[[2016 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
==Problem 3==<br />
For every dollar Ben spent on bagels, David spent <math>25</math> cents less. Ben paid <math>\$12.50</math> more than David. How much did they spend in the bagel store together?<br />
<br />
<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math><br />
==Problem 4==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
==Problem 5==<br />
A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math><br />
==Problem 6==<br />
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br />
<br />
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math><br />
==Problem 7==<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math><br />
==Problem 8==<br />
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays <math>40</math> coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math><br />
==Problem 9==<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
==Problem 10==<br />
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle?<br />
<asy><br />
size(6cm);<br />
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}<br />
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<br />
label("$1$",(0.5,2.5));<br />
draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br />
draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(1.5,2.5));<br />
draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br />
draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.5,2.5));<br />
draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br />
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<br />
label("$1$",(4.1,1.5));<br />
draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br />
draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br />
<br />
label("$1$",(3.7,0.5));<br />
draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br />
draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br />
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<br />
<cmath>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</cmath><br />
==Problem 11==<br />
What is the area of the shaded region of the given <math>8 \times 5</math> rectangle?<br />
<br />
<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
<br />
label("$1$",(8,1/2),dir(0));<br />
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<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
<br />
label("$1$",(0,9/2),dir(180));<br />
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<br />
<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math><br />
==Problem 12==<br />
Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correct statement about the probability <math>p</math> that the product of the three integers is odd?<br />
<br />
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math><br />
==Problem 13==<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
<br />
<math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math><br />
==Problem 14==<br />
How many ways are there to write <math>2016</math> as the sum of twos and threes, ignoring order? (For example, <math>1008\cdot 2 + 0\cdot 3</math> and <math>402\cdot 2 + 404\cdot 3</math> are two such ways.)<br />
<br />
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math><br />
==Problem 15==<br />
Seven cookies of radius <math>1</math> inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((0,0),1));<br />
draw(circle((1,sqrt(3)),1));<br />
draw(circle((-1,sqrt(3)),1)); <br />
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br />
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi</math><br />
<br />
==Problem 16==<br />
A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?<br />
<br />
<math>\textbf{(A)}</math> counterclockwise rotation about the origin by <math>90^{\circ}</math>.<br />
<math>\textbf{(B)}</math> clockwise rotation about the origin by <math>90^{\circ}</math>.<br />
<math>\textbf{(C)}</math> reflection about the <math>x</math>-axis<br />
<math>\textbf{(D)}</math> reflection about the line <math>y = x</math><br />
<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.<br />
<br />
==Problem 17==<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math><br />
<br />
==Problem 18==<br />
Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math><br />
==Problem 19==<br />
In rectangle ABCD, <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ration <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?<br />
==Problem 20==<br />
For some particular value of <math>N</math>, when <math>(a+b+c+d+1)^N</math> is expanded and like terms are combined, the resulting expression contains exactly <math>1001</math> terms that include all four variables <math>a, b,c,</math> and <math>d</math>, each to some positive power. What is <math>N</math>?<br />
<br />
<math>\textbf{(A) }9 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19</math><br />
<br />
==Problem 21==<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
==Problem 22==<br />
For some positive integer <math>n</math>, the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3</math>. How many positive integer divisors does the number <math>81n^4</math> have?<br />
<br />
<math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math><br />
==Problem 23==<br />
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
<br />
<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
<br />
==Problem 24==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
==Problem 25==<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math></div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems&diff=750652016 AMC 10A Problems2016-02-03T22:47:35Z<p>Algebra star1234: /* Problem 2 */</p>
<hr />
<div>==Problem 1==<br />
What is the value of <math>\dfrac{11!-10!}{9!}</math>?<br />
<br />
<math>\textbf{(A)}\ 99\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 121\qquad\textbf{(E)}\ 132</math><br />
<br />
==Problem 2==<br />
For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>?<br />
<br />
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math><br />
<br />
==Problem 3==<br />
For every dollar Ben spent on bagels, David spent <math>25</math> cents less. Ben paid <math>\$12.50</math> more than David. How much did they spend in the bagel store together?<br />
<br />
<math>\textbf{(A)}\ \$37.50 \qquad\textbf{(B)}\ \$50.00\qquad\textbf{(C)}\ \$87.50\qquad\textbf{(D)}\ \$90.00\qquad\textbf{(E)}\ \$92.50</math><br />
==Problem 4==<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
==Problem 5==<br />
A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box?<br />
<br />
<math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math><br />
==Problem 6==<br />
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?<br />
<br />
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math><br />
==Problem 7==<br />
The mean, median, and mode of the <math>7</math> data values <math>60, 100, x, 40, 50, 200, 90</math> are all equal to <math>x</math>. What is the value of <math>x</math>?<br />
<br />
<br />
<math>\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100</math><br />
==Problem 8==<br />
Trickster Rabbit agrees with Foolish Fox to double Fox's money every time Fox crosses the bridge by Rabbit's house, as long as Fox pays <math>40</math> coins in toll to Rabbit after each crossing. The payment is made after the doubling, Fox is excited about his good fortune until he discovers that all his money is gone after crossing the bridge three times. How many coins did Fox have at the beginning?<br />
<br />
<math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 45</math><br />
==Problem 9==<br />
A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and so on up to <math>N</math> coins in the <math>N</math>th row. What is the sum of the digits of <math>N</math>?<br />
<br />
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math><br />
==Problem 10==<br />
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner rectangle?<br />
<asy><br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
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return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle;<br />
}<br />
filldraw(rectangle((0,0),(7,5)),gray(0.5));<br />
filldraw(rectangle((1,1),(6,4)),gray(0.75));<br />
filldraw(rectangle((2,2),(5,3)),white);<br />
<br />
label("$1$",(0.5,2.5));<br />
draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead));<br />
draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(1.5,2.5));<br />
draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead));<br />
draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.5,2.5));<br />
draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead));<br />
draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));<br />
<br />
label("$1$",(4.1,1.5));<br />
draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead));<br />
draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));<br />
<br />
label("$1$",(3.7,0.5));<br />
draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead));<br />
draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead));<br />
</asy><br />
<br />
<cmath>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</cmath><br />
==Problem 11==<br />
What is the area of the shaded region of the given <math>8 \times 5</math> rectangle?<br />
<br />
<asy><br />
<br />
size(6cm);<br />
defaultpen(fontsize(9pt));<br />
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);<br />
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));<br />
<br />
label("$1$",(1/2,5),dir(90));<br />
label("$7$",(9/2,5),dir(90));<br />
<br />
label("$1$",(8,1/2),dir(0));<br />
label("$4$",(8,3),dir(0));<br />
<br />
label("$1$",(15/2,0),dir(270));<br />
label("$7$",(7/2,0),dir(270));<br />
<br />
label("$1$",(0,9/2),dir(180));<br />
label("$4$",(0,2),dir(180));<br />
<br />
</asy><br />
<br />
<math>\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math><br />
==Problem 12==<br />
Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correct statement about the probability <math>p</math> that the product of the three integers is odd?<br />
<br />
<math>\textbf{(A)}\ p<\dfrac{1}{8}\qquad\textbf{(B)}\ p=\dfrac{1}{8}\qquad\textbf{(C)}\ \dfrac{1}{8}<p<\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math><br />
==Problem 13==<br />
Five friends sat in a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?<br />
<br />
<math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math><br />
==Problem 14==<br />
How many ways are there to write <math>2016</math> as the sum of twos and threes, ignoring order? (For example, <math>1008\cdot 2 + 0\cdot 3</math> and <math>402\cdot 2 + 404\cdot 3</math> are two such ways.)<br />
<br />
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math><br />
==Problem 15==<br />
Seven cookies of radius <math>1</math> inch are cut from a circle of cookie dough, as shown. Neighboring cookies are tangent, and all except the center cookie are tangent to the edge of the dough. The leftover scrap is reshaped to form another cookie of the same thickness. What is the radius in inches of the scrap cookie?<br />
<br />
<asy><br />
draw(circle((0,0),3));<br />
draw(circle((0,0),1));<br />
draw(circle((1,sqrt(3)),1));<br />
draw(circle((-1,sqrt(3)),1)); <br />
draw(circle((-1,-sqrt(3)),1)); draw(circle((1,-sqrt(3)),1)); <br />
draw(circle((2,0),1)); draw(circle((-2,0),1)); </asy><br />
<br />
<math>\textbf{(A) } \sqrt{2} \qquad \textbf{(B) } 1.5 \qquad \textbf{(C) } \sqrt{\pi} \qquad \textbf{(D) } \sqrt{2\pi} \qquad \textbf{(E) } \pi</math><br />
<br />
==Problem 16==<br />
<br />
==Problem 17==<br />
Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the probability that at least <math>\tfrac{3}{5}</math> of the green balls are on the same side of the red ball. Observe that <math>P(5)=1</math> and that <math>P(N)</math> approaches <math>\tfrac{4}{5}</math> as <math>N</math> grows large. What is the sum of the digits of the least value of <math>N</math> such that <math>P(N) < \tfrac{321}{400}</math>?<br />
<br />
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math><br />
<br />
==Problem 18==<br />
Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math><br />
==Problem 19==<br />
In rectangle ABCD, <math>AB=6</math> and <math>BC=3</math>. Point <math>E</math> between <math>B</math> and <math>C</math>, and point <math>F</math> between <math>E</math> and <math>C</math> are such that <math>BE=EF=FC</math>. Segments <math>\overline{AE}</math> and <math>\overline{AF}</math> intersect <math>\overline{BD}</math> at <math>P</math> and <math>Q</math>, respectively. The ration <math>BP:PQ:QD</math> can be written as <math>r:s:t</math> where the greatest common factor of <math>r,s</math> and <math>t</math> is 1. What is <math>r+s+t</math>?<br />
==Problem 20==<br />
<br />
==Problem 21==<br />
Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>?<br />
<br />
<math>\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}</math><br />
==Problem 22==<br />
For some positive integer <math>n</math>, the number <math>110n^3</math> has <math>110</math> positive integer divisors, including <math>1</math> and the number <math>110n^3</math>. How many positive integer divisors does the number <math>81n^4</math> have?<br />
<br />
<math>\textbf{(A) }110\qquad\textbf{(B) }191\qquad\textbf{(C) }261\qquad\textbf{(D) }325\qquad\textbf{(E) }425</math><br />
==Problem 23==<br />
A binary operation <math>\diamondsuit</math> has the properties that <math>a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c</math> and that <math>a\,\diamondsuit \,a=1</math> for all nonzero real numbers <math>a, b,</math> and <math>c</math>. (Here <math>\cdot</math> represents multiplication). The solution to the equation <math>2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p+q?</math><br />
<br />
<math>\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601</math><br />
<br />
==Problem 24==<br />
A quadrilateral is inscribed in a circle of radius <math>200\sqrt{2}</math>. Three of the sides of this quadrilateral have length <math>200</math>. What is the length of the fourth side?<br />
<br />
<math>\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500</math><br />
==Problem 25==<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
<br />
<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math></div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_24&diff=733182015 AMC 10B Problems/Problem 242015-11-29T20:57:19Z<p>Algebra star1234: /* Solution */</p>
<hr />
<div>==Problem==<br />
Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin <math>p_0=(0,0)</math> facing to the east and walks one unit, arriving at <math>p_1=(1,0)</math>. For <math>n=1,2,3,\dots</math>, right after arriving at the point <math>p_n</math>, if Aaron can turn <math>90^\circ</math> left and walk one unit to an unvisited point <math>p_{n+1}</math>, he does that. Otherwise, he walks one unit straight ahead to reach <math>p_{n+1}</math>. Thus the sequence of points continues <math>p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)</math>, and so on in a counterclockwise spiral pattern. What is <math>p_{2015}</math>?<br />
<br />
<math> \textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13) </math><br />
<br />
==Solution==<br />
(Used from 2015 AMC 10/12 B Math Jam)<br />
<br />
The first thing we would do is track Aaron's footsteps:<br />
<br />
He starts by taking <math>1</math> step East and <math>1</math> step North, ending at <math>(1,1)</math> after <math>2</math> steps and about to head West.<br />
<br />
Then he takes <math>2</math> steps West and <math>2</math> steps South, ending at <math>(-1,-1</math>) after <math>2+4</math> steps, and about to head East.<br />
<br />
Then he takes <math>3</math> steps East and <math>3</math> steps North, ending at <math>(2,2)</math> after <math>2+4+6</math> steps, and about to head West.<br />
<br />
Then he takes <math>4</math> steps West and <math>4</math> steps South, ending at <math>(-2,-2)</math> after <math>2+4+6+8</math> steps, and about to head East.<br />
<br />
From this pattern, we can notice that for any integer <math>k \ge 1</math> he's at <math>(-k, -k)</math> after <math>2 + 4 + 6 + ... + 4k</math> steps, and about to head East. There are <math>2k</math> terms in the sum, with an average value of <math>(2 + 4k)/2 = 2k + 1</math>, so:<br />
<br />
<cmath>2 + 4 + 6 + ... + 4k = 2k(2k + 1)</cmath><br />
<br />
If we substitute <math>k = 22</math> into the equation: <math>44(45) = 1980 < 2015</math>. So he has <math>35</math> moves to go. This makes him end up at <math>(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_22&diff=733172015 AMC 10B Problems/Problem 222015-11-29T20:46:42Z<p>Algebra star1234: /* Problem */</p>
<hr />
<div>==Problem==<br />
In the figure shown below, <math>ABCDE</math> is a regular pentagon and <math>AG=1</math>. What is <math>FG + JH + CD</math>?<br />
<asy><br />
pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));<br />
//(0,0) is a convenient point<br />
//E1 to prevent conflict with direction E(ast)<br />
pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0];<br />
draw(A--B--C--D--E1--A);<br />
draw(A--D--B--E1--C--A);<br />
draw(F--I--G--J--H--F);<br />
label("$A$",A,N);<br />
label("$B$",B,E);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$E$",E1,W);<br />
label("$F$",F,NW);<br />
label("$G$",G,NE);<br />
label("$H$",H,E);<br />
label("$I$",I,S);<br />
label("$J$",J,W);<br />
</asy><br />
<math>\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10} </math><br />
<br />
==Solution==<br />
<br />
Triangle <math>AFG</math> is isosceles, so <math>AG=AF=1</math>. Using the symmetry of pentagon <math>FGHIJ</math>, notice that <math>\triangle IGF \cong \triangle JHG \cong \triangle DIJ \cong AFG</math>. Therefore, <math>JH=AF=1</math>.<br />
<br />
Since <math>\triangle AJH \sim \triangle AFG</math>, <math>\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1</math>. From this, we get <math>FG=\frac{\sqrt{5} -1}{2}</math>.<br />
<br />
Since <math>\triangle DIJ \cong \triangle AFG</math>, <math>DJ=DI=AF=1</math>. Since <math>\triangle AFG \sim ADC</math>, <math> \frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac{\frac{\sqrt{5} - 1}{2}}{CD}</math>. Thus, <math>CD = (\sqrt{5} - 1) + (\frac{\sqrt{5} - 1}{2})^2 = \frac{\sqrt{5} +1}{2}</math><br />
<br />
Therefore, <math>FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Geometry Problems]]</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=AMC_8_Problems_and_Solutions&diff=73252AMC 8 Problems and Solutions2015-11-27T14:54:59Z<p>Algebra star1234: </p>
<hr />
<div>The 2015 AMC 8 was held on November 17th, 2015.<br />
<br />
[[AMC 8]] / [[AJHSME]] problems and solutions.<br />
* [[2016 AMC 8]]<br />
* [[2015 AMC 8]]<br />
* [[2014 AMC 8]]<br />
* [[2013 AMC 8]]<br />
* [[2012 AMC 8]]<br />
* [[2011 AMC 8]]<br />
* [[2010 AMC 8]]<br />
* [[2009 AMC 8]]<br />
* [[2008 AMC 8]]<br />
* [[2007 AMC 8]]<br />
* [[2006 AMC 8]]<br />
* [[2005 AMC 8]]<br />
* [[2004 AMC 8]]<br />
* [[2003 AMC 8]]<br />
* [[2002 AMC 8]]<br />
* [[2001 AMC 8]]<br />
* [[2000 AMC 8]]<br />
* [[1999 AMC 8]]<br />
* [[1998 AJHSME]]<br />
* [[1997 AJHSME]]<br />
* [[1996 AJHSME]]<br />
* [[1995 AJHSME]]<br />
* [[1994 AJHSME]]<br />
* [[1993 AJHSME]]<br />
* [[1992 AJHSME]]<br />
* [[1991 AJHSME]]<br />
* [[1990 AJHSME]]<br />
* [[1989 AJHSME]]<br />
* [[1988 AJHSME]]<br />
* [[1987 AJHSME]]<br />
* [[1986 AJHSME]]<br />
* [[1985 AJHSME]]<br />
<br />
== Resources ==<br />
* [[American Mathematics Competitions]]<br />
* [[AMC Problems and Solutions]]<br />
* [[Mathematics competition resources]]</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_2&diff=732022015 AMC 8 Problems/Problem 22015-11-26T14:44:49Z<p>Algebra star1234: /* Solution 3 */</p>
<hr />
<div>Point <math>O</math> is the center of the regular octagon <math>ABCDEFGH</math>, and <math>X</math> is the midpoint of the side <math>\overline{AB}.</math> What fraction of the area of the octagon is shaded?<br />
<br />
<math>\textbf{(A) }\frac{11}{32} \quad\textbf{(B) }\frac{3}{8} \quad\textbf{(C) }\frac{13}{32} \quad\textbf{(D) }\frac{7}{16}\quad \textbf{(E) }\frac{15}{32}</math><br />
<br />
<asy><br />
pair A,B,C,D,E,F,G,H,O,X;<br />
A=dir(45);<br />
B=dir(90);<br />
C=dir(135);<br />
D=dir(180);<br />
E=dir(-135);<br />
F=dir(-90);<br />
G=dir(-45);<br />
H=dir(0);<br />
O=(0,0);<br />
X=midpoint(A--B);<br />
<br />
fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br />
draw(A--B--C--D--E--F--G--H--cycle);<br />
<br />
dot("$A$",A,dir(45));<br />
dot("$B$",B,dir(90));<br />
dot("$C$",C,dir(135));<br />
dot("$D$",D,dir(180));<br />
dot("$E$",E,dir(-135));<br />
dot("$F$",F,dir(-90));<br />
dot("$G$",G,dir(-45));<br />
dot("$H$",H,dir(0));<br />
dot("$X$",X,dir(135/2));<br />
dot("$O$",O,dir(0));<br />
draw(E--O--X);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
Since octagon <math>ABCDEFGH</math> is a regular octagon, it is split into 8 equal parts, such as triangles <math>\bigtriangleup ABO, \bigtriangleup BCO, \bigtriangleup CDO</math>, etc. These parts, since they are all equal, are <math>\frac{1}{8}</math> of the octagon each. The shaded region consists of 3 of these equal parts plus half of another, so the fraction of the octagon that is shaded is <math>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\boxed{\text{(D) }\dfrac{7}{16}}.</math><br />
<br />
==Solution 2==<br />
<asy><br />
pair A,B,C,D,E,F,G,H,O,X,a,b,c,d,e,f,g;<br />
A=dir(45);<br />
B=dir(90);<br />
C=dir(135);<br />
D=dir(180);<br />
E=dir(-135);<br />
F=dir(-90);<br />
G=dir(-45);<br />
H=dir(0);<br />
O=(0,0);<br />
X=midpoint(A--B);<br />
a=midpoint(B--C);<br />
b=midpoint(C--D);<br />
c=midpoint(D--E);<br />
d=midpoint(E--F);<br />
e=midpoint(F--G);<br />
f=midpoint(G--H);<br />
g=midpoint(H--A);<br />
<br />
fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br />
draw(A--B--C--D--E--F--G--H--cycle);<br />
<br />
dot("$A$",A,dir(45));<br />
dot("$B$",B,dir(90));<br />
dot("$C$",C,dir(135));<br />
dot("$D$",D,dir(180));<br />
dot("$E$",E,dir(-135));<br />
dot("$F$",F,dir(-90));<br />
dot("$G$",G,dir(-45));<br />
dot("$H$",H,dir(0));<br />
dot("$X$",X,dir(135/2));<br />
dot("$O$",O,dir(0));<br />
draw(E--O--X);<br />
draw(B--F);<br />
draw(A--O);<br />
draw(D--H);<br />
draw(C--G);<br />
draw(a--e);<br />
draw(b--f);<br />
draw(c--g);<br />
draw(d--O);<br />
</asy><br />
<br />
The octagon has been divided up into 16 identical triangles (and thus they each have equal area). Since the shaded region occupies 7 out of the 16 total triangles, the answer is <math>\boxed{\textbf{(D)}~\dfrac{7}{16}}</math>.<br />
<br />
==Solution 3==<br />
<br />
For starters what I find helpful is to divide the whole octagon up into triangles as shown here:<br />
<asy><br />
pair A,B,C,D,E,F,G,H,O,X;<br />
A=dir(45);<br />
B=dir(90);<br />
C=dir(135);<br />
D=dir(180);<br />
E=dir(-135);<br />
F=dir(-90);<br />
G=dir(-45);<br />
H=dir(0);<br />
O=(0,0);<br />
X=midpoint(A--B);<br />
<br />
fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75));<br />
draw(A--B--C--D--E--F--G--H--cycle);<br />
<br />
dot("$A$",A,dir(45));<br />
dot("$B$",B,dir(90));<br />
dot("$C$",C,dir(135));<br />
dot("$D$",D,dir(180));<br />
dot("$E$",E,dir(-135));<br />
dot("$F$",F,dir(-90));<br />
dot("$G$",G,dir(-45));<br />
dot("$H$",H,dir(0));<br />
dot("$X$",X,dir(135/2));<br />
dot("$O$",O,dir(0));<br />
draw(E--O--X);<br />
draw(C--O--B);<br />
draw(B--O--A);<br />
draw(A--O--H);<br />
draw(H--O--G);<br />
draw(G--O--F);<br />
draw(F--O--E);<br />
draw(E--O--D);<br />
draw(D--O--C);<br />
</asy><br />
<br />
Now it is just a matter of counting the larger triangles remember that <math>\triangle BOX</math> and <math>\triangle XOA</math> are not full triangles and are only half for these purposes. We count it up and we get a total of <math>\frac{3.5}{8}</math> of the shape shaded. We then simplify it to get our answer: <math>\frac{7}{16}</math> or <math>\textbf{(D)}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=1|num-a=3}}<br />
{{MAA Notice}}</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_22&diff=723602015 AMC 10B Problems/Problem 222015-10-04T13:37:34Z<p>Algebra star1234: /* Solution */</p>
<hr />
<div>==Problem==<br />
In the figure shown below, <math>ABCDE</math> is a regular pentagon and <math>AG=1</math>. What is <math>FG + JH + CD</math>?<br />
<asy><br />
pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));<br />
//(0,0) is a convenient point<br />
//E1 to prevent conflict with direction E(ast)<br />
pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0];<br />
draw(A--B--C--D--E1--A);<br />
draw(A--D--B--E1--C--A);<br />
draw(F--I--G--J--H--F);<br />
label("$A$",A,N);<br />
label("$B$",B,E);<br />
label("$C$",C,SE);<br />
label("$D$",D,SW);<br />
label("$E$",E1,W);<br />
label("$F$",F,NW);<br />
label("$G$",G,NE);<br />
label("$H$",H,E);<br />
label("$I$",I,S);<br />
label("$J$",J,W);<br />
</asy><br />
<br />
==Solution==<br />
<br />
Triangle <math>AFG</math> is isosceles, so <math>AG=AF=1</math>. Using the symmetry of pentagon <math>FGHIJ</math>, notice that <math>\triangle IGF \cong \triangle JHG \cong \triangle DIJ \cong AFG</math>. Therefore, <math>JH=AF=1</math>.<br />
<br />
Since <math>\triangle AJH \sim \triangle AFG</math>, <math>\frac{JH}{AF+FJ}=\frac{1}{1+FG}=\frac{FG}{FI}=\frac{FG}1</math>. From this, we get <math>FG=\frac{\sqrt{5} -1}{2}</math>.<br />
<br />
Since <math>\triangle DIJ \cong \triangle AFG</math>, <math>DJ=DI=AF=1</math>. Since <math>\triangle AFG \sim ADC</math>, <math> \frac{AF}{AF+FJ+JD}=\frac1{2+FG} = \frac{FG}{CD}=\frac{\frac{\sqrt{5} - 1}{2}}{CD}</math>. Thus, <math>CD = (\sqrt{5} - 1) + (\frac{\sqrt{5} - 1}{2})^2 = \frac{\sqrt{5} +1}{2}</math><br />
<br />
Therefore, <math>FG+JH+CD=\frac{\sqrt5-1}2+1+\frac{\sqrt5+1}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }</math><br />
<br />
==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=21|num-a=23}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Introductory Geometry Problems]]</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_12&diff=721572012 AIME I Problems/Problem 122015-09-21T17:35:02Z<p>Algebra star1234: /* Solution 1 */</p>
<hr />
<div>==Problem 12==<br />
Let <math>\triangle ABC</math> be a right triangle with right angle at <math>C.</math> Let <math>D</math> and <math>E</math> be points on <math>\overline{AB}</math> with <math>D</math> between <math>A</math> and <math>E</math> such that <math>\overline{CD}</math> and <math>\overline{CE}</math> trisect <math>\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, and <math>p</math> is a positive integer not divisible by the square of any prime. Find <math>m+n+p.</math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
Let <math>CD = 2a</math>. Using angle bisector theorem on <math>\triangle CDB</math>, <math>\frac{2a}{8}=\frac{CB}{15}</math>, so <math>CB = \frac{15a}{4}</math>. Then, drop the altitude from <math>D</math> to <math>CB</math> and call the foot <math>F</math>. Thus, <math>CF = a</math>, <math>CD = a\sqrt{3}</math>, and <math>FB = \frac{11a}{4}</math>. Finally, <math>\tan{B} = \frac{a\sqrt{3}}{\frac{11a}{4}} = \frac{4\sqrt{3}}{11}</math>. Our answer is <math>\boxed{018}</math>.<br />
<br />
=== Solution 2 ===<br />
Without loss of generality, set <math>CB = 1</math>. Then, by the Angle Bisector Theorem on triangle <math>DCB</math>, we have <math>CD = \frac{8}{15}</math>. We apply the Law of Cosines to triangle <math>DCB</math> to get <math>1 + \frac{64}{225} - \frac{8}{15} = BD^{2}</math>, which we can simplify to get <math>BD = \frac{13}{15}</math>. <br />
<br />
Now, we have <math>\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}</math> by another application of the Law of Cosines to triangle <math>DCB</math>, so <math>\cos \angle B = \frac{11}{13}</math>. In addition, <math>\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}</math>, so <math>\tan \angle B = \frac{4\sqrt{3}}{11}</math>. <br />
<br />
Our final answer is <math>4+3+11 = \boxed{018}</math>.<br />
<br />
=== Solution 3 ===<br />
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)<br />
<br />
Find values for all angles in terms of <math>\angle B</math>. <math>\angle CEB = 150-B</math>, <math>\angle CED = 30+B</math>, <math>\angle CDE = 120-B</math>, <math>\angle CDA = 60+B</math>, and <math>\angle A = 90-B</math>.<br />
<br />
Use the law of sines on <math>\triangle CED</math> and <math>\triangle CEB</math>:<br />
<br />
In <math>\triangle CED</math>, <math>\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}</math>. This simplifies to <math>16 = \frac{CE}{\sin (120-B)}</math>.<br />
<br />
In <math>\triangle CEB</math>, <math>\frac{15}{\sin 30} = \frac{CE}{\sin B}</math>. This simplifies to <math>30 = \frac{CE}{\sin B}</math>.<br />
<br />
Solve for <math>CE</math> and equate them so that you get <math>16\sin (120-B) = 30\sin B</math>. <br />
<br />
From this, <math>\frac{8}{15} = \frac{\sin B}{\sin (120-B)}</math>. <br />
<br />
Use a trig identity on the denominator on the right to obtain: <math>\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}</math><br />
<br />
This simplifies to <math>\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cos B + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}</math><br />
<br />
This gives <math>8\sqrt{3}\cos B+8\sin B=30\sin B</math><br />
Dividing by <math>\cos B</math>, we have <math>{8\sqrt{3}}= 22\tan B</math><br />
<br />
<math>\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}</math>. Our final answer is <math>4 + 3 + 11 = \boxed{018}</math>.<br />
<br />
=== Solution 4 ===<br />
(This solution avoids advanced trigonometry)<br />
<br />
Let <math>X</math> be the foot of the perpendicular from <math>D</math> to <math>\overline{BC}</math>, and let <math>Y</math> be the foot of the perpendicular from <math>E</math> to <math>\overline{BC}</math>.<br />
<br />
Now let <math>EY=x</math>. Clearly, triangles <math>EYB</math> and <math>DXB</math> are similar with <math>\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}</math>, so <math>DX=\frac{23}{15}x</math>.<br />
<br />
Since triangles <math>CDX</math> and <math>CEY</math> are 30-60-90 right triangles, we can easily find other lengths in terms of <math>x</math>. For example, we see that <math>CY=x\sqrt{3}</math> and <math>CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x</math>. Therefore <math>XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x</math>.<br />
<br />
Again using the fact that triangles <math>EYB</math> and <math>DXB</math> are similar, we see that <math>\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}</math>, so <math>BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}</math>.<br />
<br />
Thus <math>\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}</math>, and our answer is <math>4+3+11=\boxed{018}</math>.<br />
<br />
=== Solution 5 ===<br />
(Another solution without trigonometry)<br />
<br />
Extend <math>CD</math> to point <math>F</math> such that <math>\overline{AF} \parallel \overline{CB}</math>. It is then clear that <math>\triangle AFD</math> is similar to <math>\triangle BCD</math>.<br />
<br />
Let <math>AC=p</math>, <math>BC=q</math>. Then <math>\tan \angle B = p/q</math>.<br />
<br />
With the Angle Bisector Theorem, we get that <math>CD=\frac{8}{15}q</math>. From 30-60-90 <math>\triangle CAF</math>, we get that <math>AF=\frac{1}{\sqrt{3}}p</math> and <math>FD=FC-CD=\frac{2}{\sqrt{3}}p-\frac{8}{15}q</math>.<br />
<br />
From <math>\triangle AFD \sim \triangle BCD</math>, we have that <math>\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}</math>. Simplifying yields <math>\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1</math>, and <math>\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}</math>, so our answer is <math>4+3+11=\boxed{018}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_12&diff=721562012 AIME I Problems/Problem 122015-09-21T17:34:42Z<p>Algebra star1234: /* Solution 1 */</p>
<hr />
<div>==Problem 12==<br />
Let <math>\triangle ABC</math> be a right triangle with right angle at <math>C.</math> Let <math>D</math> and <math>E</math> be points on <math>\overline{AB}</math> with <math>D</math> between <math>A</math> and <math>E</math> such that <math>\overline{CD}</math> and <math>\overline{CE}</math> trisect <math>\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, and <math>p</math> is a positive integer not divisible by the square of any prime. Find <math>m+n+p.</math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
(This solution uses the angle bisector theorm.)<br />
\newline<br />
Let <math>CD = 2a</math>. Using angle bisector theorem on <math>\triangle CDB</math>, <math>\frac{2a}{8}=\frac{CB}{15}</math>, so <math>CB = \frac{15a}{4}</math>. Then, drop the altitude from <math>D</math> to <math>CB</math> and call the foot <math>F</math>. Thus, <math>CF = a</math>, <math>CD = a\sqrt{3}</math>, and <math>FB = \frac{11a}{4}</math>. Finally, <math>\tan{B} = \frac{a\sqrt{3}}{\frac{11a}{4}} = \frac{4\sqrt{3}}{11}</math>. Our answer is <math>\boxed{018}</math>.<br />
<br />
=== Solution 2 ===<br />
Without loss of generality, set <math>CB = 1</math>. Then, by the Angle Bisector Theorem on triangle <math>DCB</math>, we have <math>CD = \frac{8}{15}</math>. We apply the Law of Cosines to triangle <math>DCB</math> to get <math>1 + \frac{64}{225} - \frac{8}{15} = BD^{2}</math>, which we can simplify to get <math>BD = \frac{13}{15}</math>. <br />
<br />
Now, we have <math>\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}</math> by another application of the Law of Cosines to triangle <math>DCB</math>, so <math>\cos \angle B = \frac{11}{13}</math>. In addition, <math>\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}</math>, so <math>\tan \angle B = \frac{4\sqrt{3}}{11}</math>. <br />
<br />
Our final answer is <math>4+3+11 = \boxed{018}</math>.<br />
<br />
=== Solution 3 ===<br />
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)<br />
<br />
Find values for all angles in terms of <math>\angle B</math>. <math>\angle CEB = 150-B</math>, <math>\angle CED = 30+B</math>, <math>\angle CDE = 120-B</math>, <math>\angle CDA = 60+B</math>, and <math>\angle A = 90-B</math>.<br />
<br />
Use the law of sines on <math>\triangle CED</math> and <math>\triangle CEB</math>:<br />
<br />
In <math>\triangle CED</math>, <math>\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}</math>. This simplifies to <math>16 = \frac{CE}{\sin (120-B)}</math>.<br />
<br />
In <math>\triangle CEB</math>, <math>\frac{15}{\sin 30} = \frac{CE}{\sin B}</math>. This simplifies to <math>30 = \frac{CE}{\sin B}</math>.<br />
<br />
Solve for <math>CE</math> and equate them so that you get <math>16\sin (120-B) = 30\sin B</math>. <br />
<br />
From this, <math>\frac{8}{15} = \frac{\sin B}{\sin (120-B)}</math>. <br />
<br />
Use a trig identity on the denominator on the right to obtain: <math>\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}</math><br />
<br />
This simplifies to <math>\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cos B + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}</math><br />
<br />
This gives <math>8\sqrt{3}\cos B+8\sin B=30\sin B</math><br />
Dividing by <math>\cos B</math>, we have <math>{8\sqrt{3}}= 22\tan B</math><br />
<br />
<math>\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}</math>. Our final answer is <math>4 + 3 + 11 = \boxed{018}</math>.<br />
<br />
=== Solution 4 ===<br />
(This solution avoids advanced trigonometry)<br />
<br />
Let <math>X</math> be the foot of the perpendicular from <math>D</math> to <math>\overline{BC}</math>, and let <math>Y</math> be the foot of the perpendicular from <math>E</math> to <math>\overline{BC}</math>.<br />
<br />
Now let <math>EY=x</math>. Clearly, triangles <math>EYB</math> and <math>DXB</math> are similar with <math>\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}</math>, so <math>DX=\frac{23}{15}x</math>.<br />
<br />
Since triangles <math>CDX</math> and <math>CEY</math> are 30-60-90 right triangles, we can easily find other lengths in terms of <math>x</math>. For example, we see that <math>CY=x\sqrt{3}</math> and <math>CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x</math>. Therefore <math>XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x</math>.<br />
<br />
Again using the fact that triangles <math>EYB</math> and <math>DXB</math> are similar, we see that <math>\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}</math>, so <math>BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}</math>.<br />
<br />
Thus <math>\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}</math>, and our answer is <math>4+3+11=\boxed{018}</math>.<br />
<br />
=== Solution 5 ===<br />
(Another solution without trigonometry)<br />
<br />
Extend <math>CD</math> to point <math>F</math> such that <math>\overline{AF} \parallel \overline{CB}</math>. It is then clear that <math>\triangle AFD</math> is similar to <math>\triangle BCD</math>.<br />
<br />
Let <math>AC=p</math>, <math>BC=q</math>. Then <math>\tan \angle B = p/q</math>.<br />
<br />
With the Angle Bisector Theorem, we get that <math>CD=\frac{8}{15}q</math>. From 30-60-90 <math>\triangle CAF</math>, we get that <math>AF=\frac{1}{\sqrt{3}}p</math> and <math>FD=FC-CD=\frac{2}{\sqrt{3}}p-\frac{8}{15}q</math>.<br />
<br />
From <math>\triangle AFD \sim \triangle BCD</math>, we have that <math>\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}</math>. Simplifying yields <math>\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1</math>, and <math>\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}</math>, so our answer is <math>4+3+11=\boxed{018}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Algebra star1234https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_12&diff=721552012 AIME I Problems/Problem 122015-09-21T17:34:14Z<p>Algebra star1234: /* Solution 1 */</p>
<hr />
<div>==Problem 12==<br />
Let <math>\triangle ABC</math> be a right triangle with right angle at <math>C.</math> Let <math>D</math> and <math>E</math> be points on <math>\overline{AB}</math> with <math>D</math> between <math>A</math> and <math>E</math> such that <math>\overline{CD}</math> and <math>\overline{CE}</math> trisect <math>\angle C.</math> If <math>\frac{DE}{BE} = \frac{8}{15},</math> then <math>\tan B</math> can be written as <math>\frac{m \sqrt{p}}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, and <math>p</math> is a positive integer not divisible by the square of any prime. Find <math>m+n+p.</math><br />
<br />
== Solutions ==<br />
=== Solution 1 ===<br />
(This solution uses the angle bisector theorm.)<br />
Let <math>CD = 2a</math>. Using angle bisector theorem on <math>\triangle CDB</math>, <math>\frac{2a}{8}=\frac{CB}{15}</math>, so <math>CB = \frac{15a}{4}</math>. Then, drop the altitude from <math>D</math> to <math>CB</math> and call the foot <math>F</math>. Thus, <math>CF = a</math>, <math>CD = a\sqrt{3}</math>, and <math>FB = \frac{11a}{4}</math>. Finally, <math>\tan{B} = \frac{a\sqrt{3}}{\frac{11a}{4}} = \frac{4\sqrt{3}}{11}</math>. Our answer is <math>\boxed{018}</math>.<br />
<br />
=== Solution 2 ===<br />
Without loss of generality, set <math>CB = 1</math>. Then, by the Angle Bisector Theorem on triangle <math>DCB</math>, we have <math>CD = \frac{8}{15}</math>. We apply the Law of Cosines to triangle <math>DCB</math> to get <math>1 + \frac{64}{225} - \frac{8}{15} = BD^{2}</math>, which we can simplify to get <math>BD = \frac{13}{15}</math>. <br />
<br />
Now, we have <math>\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}</math> by another application of the Law of Cosines to triangle <math>DCB</math>, so <math>\cos \angle B = \frac{11}{13}</math>. In addition, <math>\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}</math>, so <math>\tan \angle B = \frac{4\sqrt{3}}{11}</math>. <br />
<br />
Our final answer is <math>4+3+11 = \boxed{018}</math>.<br />
<br />
=== Solution 3 ===<br />
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)<br />
<br />
Find values for all angles in terms of <math>\angle B</math>. <math>\angle CEB = 150-B</math>, <math>\angle CED = 30+B</math>, <math>\angle CDE = 120-B</math>, <math>\angle CDA = 60+B</math>, and <math>\angle A = 90-B</math>.<br />
<br />
Use the law of sines on <math>\triangle CED</math> and <math>\triangle CEB</math>:<br />
<br />
In <math>\triangle CED</math>, <math>\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}</math>. This simplifies to <math>16 = \frac{CE}{\sin (120-B)}</math>.<br />
<br />
In <math>\triangle CEB</math>, <math>\frac{15}{\sin 30} = \frac{CE}{\sin B}</math>. This simplifies to <math>30 = \frac{CE}{\sin B}</math>.<br />
<br />
Solve for <math>CE</math> and equate them so that you get <math>16\sin (120-B) = 30\sin B</math>. <br />
<br />
From this, <math>\frac{8}{15} = \frac{\sin B}{\sin (120-B)}</math>. <br />
<br />
Use a trig identity on the denominator on the right to obtain: <math>\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}</math><br />
<br />
This simplifies to <math>\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cos B + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}</math><br />
<br />
This gives <math>8\sqrt{3}\cos B+8\sin B=30\sin B</math><br />
Dividing by <math>\cos B</math>, we have <math>{8\sqrt{3}}= 22\tan B</math><br />
<br />
<math>\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}</math>. Our final answer is <math>4 + 3 + 11 = \boxed{018}</math>.<br />
<br />
=== Solution 4 ===<br />
(This solution avoids advanced trigonometry)<br />
<br />
Let <math>X</math> be the foot of the perpendicular from <math>D</math> to <math>\overline{BC}</math>, and let <math>Y</math> be the foot of the perpendicular from <math>E</math> to <math>\overline{BC}</math>.<br />
<br />
Now let <math>EY=x</math>. Clearly, triangles <math>EYB</math> and <math>DXB</math> are similar with <math>\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}</math>, so <math>DX=\frac{23}{15}x</math>.<br />
<br />
Since triangles <math>CDX</math> and <math>CEY</math> are 30-60-90 right triangles, we can easily find other lengths in terms of <math>x</math>. For example, we see that <math>CY=x\sqrt{3}</math> and <math>CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x</math>. Therefore <math>XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x</math>.<br />
<br />
Again using the fact that triangles <math>EYB</math> and <math>DXB</math> are similar, we see that <math>\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}</math>, so <math>BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}</math>.<br />
<br />
Thus <math>\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}</math>, and our answer is <math>4+3+11=\boxed{018}</math>.<br />
<br />
=== Solution 5 ===<br />
(Another solution without trigonometry)<br />
<br />
Extend <math>CD</math> to point <math>F</math> such that <math>\overline{AF} \parallel \overline{CB}</math>. It is then clear that <math>\triangle AFD</math> is similar to <math>\triangle BCD</math>.<br />
<br />
Let <math>AC=p</math>, <math>BC=q</math>. Then <math>\tan \angle B = p/q</math>.<br />
<br />
With the Angle Bisector Theorem, we get that <math>CD=\frac{8}{15}q</math>. From 30-60-90 <math>\triangle CAF</math>, we get that <math>AF=\frac{1}{\sqrt{3}}p</math> and <math>FD=FC-CD=\frac{2}{\sqrt{3}}p-\frac{8}{15}q</math>.<br />
<br />
From <math>\triangle AFD \sim \triangle BCD</math>, we have that <math>\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}</math>. Simplifying yields <math>\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1</math>, and <math>\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}</math>, so our answer is <math>4+3+11=\boxed{018}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2012|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Algebra star1234