https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Poilio4236&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-13T03:28:15Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_I_Problems/Problem_14&diff=77404 2016 AIME I Problems/Problem 14 2016-03-05T20:17:00Z <p>Poilio4236: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> Centered at each lattice point in the coordinate plane are a circle radius &lt;math&gt;\frac{1}{10}&lt;/math&gt; and a square with sides of length &lt;math&gt;\frac{1}{5}&lt;/math&gt; whose sides are parallel to the coordinate axes. The line segment from &lt;math&gt;(0,0)&lt;/math&gt; to &lt;math&gt;(1001, 429)&lt;/math&gt; intersects &lt;math&gt;m&lt;/math&gt; of the squares and &lt;math&gt;n&lt;/math&gt; of the circles. Find &lt;math&gt;m + n&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> First note that &lt;math&gt;1001 = 143 \cdot 7&lt;/math&gt; and &lt;math&gt;429 = 143 \cdot 3&lt;/math&gt; so every point of the form &lt;math&gt;(7k, 3k)&lt;/math&gt; is on the line. Then consider the line &lt;math&gt;l&lt;/math&gt; from &lt;math&gt;(7k, 3k)&lt;/math&gt; to &lt;math&gt;(7(k + 1), 3(k + 1))&lt;/math&gt;. Translate the line &lt;math&gt;l&lt;/math&gt; so that &lt;math&gt;(7k, 3k)&lt;/math&gt; is now the origin. There is one square and one circle that intersect the line around &lt;math&gt;(0,0)&lt;/math&gt;. Then the points on &lt;math&gt;l&lt;/math&gt; with an integral &lt;math&gt;x&lt;/math&gt;-coordinate are, since &lt;math&gt;l&lt;/math&gt; has the equation &lt;math&gt;y = \frac{3x}{7}&lt;/math&gt;:<br /> <br /> &lt;cmath&gt; (0,0), (1, \frac{3}{7}), (2, \frac{6}{7}), (3, 1 + \frac{2}{7}), (4, 1 + \frac{5}{7}), (5, 2 + \frac{1}{7}), (6, 2 + \frac{4}{7}), (7,3). &lt;/cmath&gt;<br /> <br /> We claim that the lower right vertex of the square centered at &lt;math&gt;(2,1)&lt;/math&gt; lies on &lt;math&gt;l&lt;/math&gt;. Since the square has side length &lt;math&gt;\frac{1}{5}&lt;/math&gt;, the lower right vertex of this square has coordinates &lt;math&gt;(2 + \frac{1}{10}, 1 - \frac{1}{10}) = (\frac{21}{10}, \frac{9}{10})&lt;/math&gt;. Because &lt;math&gt;\frac{9}{10} = \frac{3}{7} \cdot \frac{21}{10}&lt;/math&gt;, &lt;math&gt;(\frac{21}{10}, \frac{9}{10})&lt;/math&gt; lies on &lt;math&gt;l&lt;/math&gt;. Since the circle centered at &lt;math&gt;(2,1)&lt;/math&gt; is contained inside the square, this circle does not intersect &lt;math&gt;l&lt;/math&gt;. Similarly the upper left vertex of the square centered at &lt;math&gt;(5,2)&lt;/math&gt; is on &lt;math&gt;l&lt;/math&gt;. Since every other point listed above is farther away from a lattice point (excluding (0,0) and (7,3)) and there are two squares with centers strictly between &lt;math&gt;(0,0)&lt;/math&gt; and &lt;math&gt;(7,3)&lt;/math&gt; that intersect &lt;math&gt;l&lt;/math&gt;. Since there are &lt;math&gt;\frac{1001}{7} = \frac{429}{3} = 143&lt;/math&gt; segments from &lt;math&gt;(7k, 3k)&lt;/math&gt; to &lt;math&gt;(7(k + 1), 3(k + 1))&lt;/math&gt;, the above count is yields &lt;math&gt;143 \cdot 2 = 286&lt;/math&gt; squares. Since every lattice point on &lt;math&gt;l&lt;/math&gt; is of the form &lt;math&gt;(3k, 7k)&lt;/math&gt; where &lt;math&gt;0 \le k \le 143&lt;/math&gt;, there are &lt;math&gt;144&lt;/math&gt; lattice points on &lt;math&gt;l&lt;/math&gt;. Centered at each lattice point, there is one square and one circle, hence this counts &lt;math&gt;288&lt;/math&gt; squares and circles. Thus &lt;math&gt;m + n = 286 + 288 = \boxed{574}&lt;/math&gt;.<br /> <br /> (Solution by gundraja)<br /> &lt;asy&gt;size(12cm);draw((0,0)--(7,3));draw(box((0,0),(7,3)),dotted);<br /> for(int i=0;i&lt;8;++i)for(int j=0;j&lt;4;++j){dot((i,j),linewidth(1));draw(box((i-.1,j-.1),(i+.1,j+.1)),linewidth(.5));draw(circle((i,j),.1),linewidth(.5));}<br /> &lt;/asy&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2016|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Poilio4236 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_16&diff=76928 2016 AMC 12B Problems/Problem 16 2016-02-22T03:45:37Z <p>Poilio4236: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> In how many ways can &lt;math&gt;345&lt;/math&gt; be written as the sum of an increasing sequence of two or more consecutive positive integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> ==Solution==<br /> We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.<br /> <br /> For the first case, we can cleverly choose the convenient form of our sequence to be<br /> &lt;cmath&gt;a-n,\cdots, a-1, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> <br /> because then our sum will just be &lt;math&gt;(2n+1)a&lt;/math&gt;. We now have <br /> &lt;cmath&gt;(2n+1)a = 345&lt;/cmath&gt;<br /> and &lt;math&gt;a&lt;/math&gt; will have a solution when &lt;math&gt;\frac{345}{2n+1}&lt;/math&gt; is an integer, namely when &lt;math&gt;2n+1&lt;/math&gt; is a divisor of 345. We check that <br /> &lt;cmath&gt;2n+1 = 3, 5, 15, 23&lt;/cmath&gt;<br /> work, and no more, because &lt;math&gt;2n+1=1&lt;/math&gt; does not satisfy the requirements of two or more consecutive integers, and when &lt;math&gt;2n+1&lt;/math&gt; equals the next biggest factor, &lt;math&gt;69&lt;/math&gt;, there must be negative integers in the sequence. Our solutions are &lt;math&gt;\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}&lt;/math&gt;.<br /> <br /> For the even cases, we choose our sequence to be of the form:<br /> &lt;cmath&gt;a-(n-1), \cdots, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> so the sum is &lt;math&gt;\frac{(2n)(2a+1)}{2} = n(2a+1)&lt;/math&gt;. In this case, we find our solutions to be &lt;math&gt;\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}&lt;/math&gt;.<br /> <br /> We have found all 7 solutions and our answer is &lt;math&gt;\boxed{\textbf{(E)} \, 7}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Poilio4236 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_16&diff=76927 2016 AMC 12B Problems/Problem 16 2016-02-22T03:44:22Z <p>Poilio4236: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> In how many ways can &lt;math&gt;345&lt;/math&gt; be written as the sum of an increasing sequence of two or more consecutive positive integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> ==Solution==<br /> We proceed with this problem by considering two cases, when: \par<br /> 1. There are an odd number of consecutive numbers \par<br /> 2. There are an even number of consecutive numbers<br /> <br /> For the first case, we can cleverly choose the convenient form of our sequence to be<br /> &lt;cmath&gt;a-n,\cdots, a-1, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> <br /> because then our sum will just be &lt;math&gt;(2n+1)a&lt;/math&gt;. We now have <br /> &lt;cmath&gt;(2n+1)a = 345&lt;/cmath&gt;<br /> and &lt;math&gt;a&lt;/math&gt; will have a solution when &lt;math&gt;\frac{345}{2n+1}&lt;/math&gt; is an integer, namely when &lt;math&gt;2n+1&lt;/math&gt; is a divisor of 345. We check that <br /> &lt;cmath&gt;2n+1 = 3, 5, 15, 23&lt;/cmath&gt;<br /> work, and no more, because &lt;math&gt;2n+1&lt;/math&gt; does not satisfy the requirements of two or more consecutive integers, and when &lt;math&gt;2n+1&lt;/math&gt; equals the next biggest factor, &lt;math&gt;69&lt;/math&gt;, there must be negative integers in the sequence. Our solutions are &lt;math&gt;\{114,115, 116\}, \{67, \cdots, 71\}, \{16, \cdots, 30\}, \{4, \cdots, 26\}&lt;/math&gt;.<br /> <br /> For the even cases, we choose our sequence to be of the form:<br /> &lt;cmath&gt;a-(n-1), \cdots, a, a+1, \cdots, a+n&lt;/cmath&gt;<br /> so the sum is &lt;math&gt;\frac{(2n)(2a+1)}{2} = n(2a+1)&lt;/math&gt;. In this case, we find our solutions to be &lt;math&gt;\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}&lt;/math&gt;.<br /> <br /> We have found all 7 solutions and our answer is &lt;math&gt;\boxed{\textbf{(E)} \, 7}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}</div> Poilio4236 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems/Problem_5&diff=76373 2012 AMC 12B Problems/Problem 5 2016-02-17T04:00:23Z <p>Poilio4236: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Two integers have a sum of 26. when two more integers are added to the first two, the sum is 41. Finally, when two more integers are added to the sum of the previous 4 integers, the sum is 57. What is the minimum number of even integers among the 6 integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> So, x+y=26, x could equal 15, and y could equal 11, so no even integers required here. 41-26=15. a+b=15, a could equal 9 and b could equal 6, so one even integer is required here. 57-41=16. m+n=16, m could equal 9 and n could equal 7, so no even integers required here, meaning only 1 even integer is required; A.<br /> <br /> ==Solution 2==<br /> <br /> Just worded and formatted a little differently than above.<br /> <br /> The first two integers sum up to &lt;math&gt;26&lt;/math&gt;. Since &lt;math&gt;26&lt;/math&gt; is even, in order to minimize the number of even integers, we make both of the first two odd.<br /> <br /> The second two integers sum up to &lt;math&gt;41-26=15&lt;/math&gt;. Since &lt;math&gt;15&lt;/math&gt; is odd, we must have at least one even integer in these next two.<br /> <br /> Finally, &lt;math&gt;57-41=16&lt;/math&gt;, and once again, &lt;math&gt;16&lt;/math&gt; is an even number so both of these integers can be odd.<br /> <br /> Therefore, we have a total of one even integer and our answer is &lt;math&gt;\boxed{(\text{A})}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2012|ab=B|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Poilio4236 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_3&diff=75812 2006 AMC 12B Problems/Problem 3 2016-02-13T17:39:01Z <p>Poilio4236: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?<br /> <br /> &lt;math&gt;<br /> \text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24<br /> &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> ===Solution 1===<br /> If the Cougars won by a margin of 14 points, then the Panthers' score would be half of (34-14). That's 10 &lt;math&gt;\Rightarrow \boxed{\text{(A)}}&lt;/math&gt;.<br /> <br /> ===Solution 2===<br /> Let the Panthers' score be &lt;math&gt;x&lt;/math&gt;. The Cougars then scored &lt;math&gt;x+14&lt;/math&gt;. Since the teams combined scored &lt;math&gt;34&lt;/math&gt;, we get &lt;math&gt;x+x+14=34 \\ \rightarrow 2x+14=34 \\ \rightarrow 2x=20 \\ \rightarrow x = 10&lt;/math&gt;, <br /> <br /> and the answer is &lt;math&gt;\boxed{\text{(A)}}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=B|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Poilio4236 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_20&diff=73251 2015 AMC 8 Problems/Problem 20 2015-11-27T05:59:27Z <p>Poilio4236: /* Solution 2*/</p> <hr /> <div>Ralph went to the store and bought 12 pairs of socks for a total of \$24. Some of the socks he bought cost \$1 a pair, some of the socks he bought cost \$3 a pair, and some of the socks he bought cost \$4 a pair. If he bought at least one pair of each type, how many pairs of \$1 socks did Ralph buy?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) } 4 \qquad<br /> \textbf{(B) } 5 \qquad<br /> \textbf{(C) } 6 \qquad<br /> \textbf{(D) } 7 \qquad<br /> \textbf{(E) } 8<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> So let there be &lt;math&gt;x&lt;/math&gt; pairs of &lt;math&gt;1&lt;/math&gt; dollar socks, &lt;math&gt;y&lt;/math&gt; pairs of &lt;math&gt;3&lt;/math&gt; dollar socks, &lt;math&gt;z&lt;/math&gt; pairs of &lt;math&gt;4&lt;/math&gt; dollar socks.<br /> <br /> We have &lt;math&gt;x+y+z=12&lt;/math&gt;, &lt;math&gt;x+3y+4z=24&lt;/math&gt;, and &lt;math&gt;x,y,z \ge 1&lt;/math&gt;.<br /> <br /> Now we subtract to find &lt;math&gt;2y+3z=12&lt;/math&gt;, and &lt;math&gt;y,z \ge 1&lt;/math&gt;.<br /> It follows that &lt;math&gt;y&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;2y&lt;/math&gt; is a multiple of &lt;math&gt;6&lt;/math&gt;, so since &lt;math&gt;0&lt;2y&lt;12&lt;/math&gt;, we must have &lt;math&gt;2y=6&lt;/math&gt;.<br /> <br /> Therefore, &lt;math&gt;y=3&lt;/math&gt;, and it follows that &lt;math&gt;z=2&lt;/math&gt;. Now &lt;math&gt;x=12-y-z=12-3-2=\boxed{\textbf{(D)}~7}&lt;/math&gt;, as desired.<br /> <br /> ==Solution 2==<br /> Since the total cost of the socks was &lt;math&gt;\$24&lt;/math&gt; and Ralph bought &lt;math&gt;12&lt;/math&gt; pairs, the average cost of each pair of socks is &lt;math&gt;\frac{\$24}{12} = \$2&lt;/math&gt;.<br /> <br /> There are two ways to make packages of socks that average to &lt;math&gt;\$2&lt;/math&gt;. You can have:<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; Two &lt;math&gt;\$1&lt;/math&gt; pairs and one &lt;math&gt;\$4&lt;/math&gt; pair (package adds up to &lt;math&gt;\$6&lt;/math&gt;)<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; One &lt;math&gt;\$1&lt;/math&gt; pair and one &lt;math&gt;\$3&lt;/math&gt; pair (package adds up to &lt;math&gt;\$4&lt;/math&gt;)<br /> <br /> So now we need to solve<br /> &lt;cmath&gt;6a+4b=24,&lt;/cmath&gt;<br /> where &lt;math&gt;a&lt;/math&gt; is the number of &lt;math&gt;\$6&lt;/math&gt; packages and &lt;math&gt;b&lt;/math&gt; is the number of &lt;math&gt;\\$4&lt;/math&gt; packages. We see our only solution (that has at least one of each pair of sock) is &lt;math&gt;a=2, b=3&lt;/math&gt;, which yields the answer of &lt;math&gt;2\times2+3\times1 = \boxed{\textbf{(D)}~7}&lt;/math&gt;.<br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=19|num-a=21}}<br /> {{MAA Notice}}</div> Poilio4236