https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Whirltwister&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T08:42:53ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_10&diff=451242012 AMC 10B Problems/Problem 102012-02-24T22:54:05Z<p>Whirltwister: /* Solution */</p>
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<div>== Problem 10 ==<br />
How many ordered pairs of positive integers (M,N) satisfy the equation <math>\frac {M}{6}</math> = <math>\frac{6}{N}</math><br />
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<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\10 </math><br />
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[[2012 AMC 10B Problems/Problem 10|Solution]]<br />
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== Solution ==<br />
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<math>\frac {M}{6}</math> = <math>\frac{6}{N}</math><br />
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is a ratio; therefore, you can cross-multiply.<br />
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<math>MN=36</math><br />
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Now you find all the factors of 36:<br />
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<math>1\times36=36</math><br />
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<math>2\times18=36</math><br />
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<math>3\times12=36</math><br />
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<math>4\times9=36</math><br />
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<math>6\times6=36</math>.<br />
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Now you can reverse the order of the factors for all of the ones listed above, because they are ordered pairs except for 6*6 since it is the same back if you reverse the order.<br />
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<math>4*2+1=\boxed{9}</math> <br />
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OR<br />
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<math> \textbf{(D)}</math></div>Whirltwisterhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_9&diff=451232012 AMC 10B Problems/Problem 92012-02-24T22:52:12Z<p>Whirltwister: /* Solutions */</p>
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<div>== Problem 9 ==<br />
Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of even integers among the 6 integers?<br />
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\5 </math><br />
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[[2012 AMC 10B Problems/Problem 9|Solution]]<br />
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== Solutions ==<br />
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Lets say that all 6 integers added are : <math>a,b,c,d,e,</math> and<math> f</math>.<br />
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If <math>a+b=26</math><br />
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and <math>a+b+c+d=41</math><br />
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Then,<br />
<math>c+d=15</math><br />
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Also, <br />
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<math>a+b+c+d+e+f=57</math><br />
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<math>a+b+c+d=41</math><br />
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Then,<br />
<math>e+f=16</math><br />
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So<br />
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<math>a+b=26</math><br />
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<math>c+d=15</math><br />
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<math>e+f=16</math><br />
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<math>a,b,e,f </math>can be all odd since odd + odd= even. And the sum of the two respective pairs are even.<br />
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However, either<math> c</math> or <math>d</math> has to be even to get a odd sum.<br />
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Therefore, there is <math>\boxed{1}</math> even integer<br />
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OR<br />
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<math> \textbf{(A)}</math></div>Whirltwisterhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_7&diff=451212012 AMC 10B Problems/Problem 72012-02-24T22:49:02Z<p>Whirltwister: /* Solutions */</p>
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<div>== Problem 7 ==<br />
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For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?<br />
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<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54 </math><br />
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== Solutions ==<br />
Let <math>x</math> be the number of acorns that both animals had.<br />
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So by the info in the problem:<br />
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<math>\frac{x}{3}=\left( \frac{x}{4} \right)+4</math><br />
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Subtracting <math> \frac{x}{4}</math> from both sides leaves<br />
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<math>\frac{x}{12}=4</math><br />
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<math>\boxed{x=48}</math><br />
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This is answer choice <math>\textbf{(D)}</math></div>Whirltwisterhttps://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10B_Problems/Problem_8&diff=451192012 AMC 10B Problems/Problem 82012-02-24T22:44:41Z<p>Whirltwister: /* Solutions */</p>
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<div>== Problem 8 ==<br />
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What is the sum of all integer solutions to <math>1<(x-2)^2<25</math>?<br />
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<math> \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\25 </math><br />
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[[2012 AMC 10B Problems/Problem 8|Solution]]<br />
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== Solutions ==<br />
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<math>(x-2)^2</math> = perfect square.<br />
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1< perfect square< 25<br />
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Perfect square can equal: 4, 9, or 16<br />
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Solve for x:<br />
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<math>(x-2)^2=4</math> <br />
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<math>x=4,0</math><br />
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and<br />
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<math>(x-2)^2=9</math><br />
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<math>x=5,-1</math><br />
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and<br />
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<math>(x-2)^2=16</math><br />
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<math>x=6,-2</math><br />
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''What is the sum of all integer solutions''<br />
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<math>4+5+6+0+(-1)+(-2)=\boxed{12}</math><br />
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OR<br />
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<math> \textbf{(B)}</math></div>Whirltwister