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| − | == Problem==
| + | #REDIRECT [[2010 AMC 12B Problems/Problem 3]] |
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| − | A ticket to a school play cost <math>x</math> dollars, where <math>x</math> is a whole number. A group of 9th graders buys tickets costing a total of <math>\textdollar 48</math>, and a group of 10th graders buys tickets costing a total of <math>\textdollar 64</math>. How many values for <math>x</math> are possible?
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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>
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| − | ==Solution 1==
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| − | We see how many common integer factors <math>48</math> and <math>64</math> share.
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| − | Of the factors of <math>48</math> - <math>1, 2, 3, 4, 6, 8, 12, 16, 24, 48</math>; only <math>1, 2, 4, 8,</math> and <math>16</math> are factors of <math>64</math>.
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| − | So there are <math>\boxed{\textbf{(E)}\ 5}</math> possibilities for the ticket price.
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| − | ==Solution 2==
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| − | The difference between <math>48</math> and <math>64</math> is <math>16</math>. The number 16 has five factors: 1, 2, 4, 8, and 16. These are all possible ticket prices because the 10th graders bought a whole number <math>x</math> more tickets than the 9th graders (so <math>x</math> must evenly divide into 16). Therefore, there are <math>\boxed{\textbf{(E)}\ 5}</math> possibilities for the ticket price.
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| − | ==See Also==
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| − | {{AMC10 box|year=2010|ab=B|num-b=7|num-a=9}}
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| − | {{MAA Notice}}
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