Difference between revisions of "2010 AMC 10B Problems/Problem 8"
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| − | We see how many common integer factors 48 and 64 share. | + | == Problem== |
| − | Of the factors of 48 - 1, 2, 3, 4, 6, 8, 12, 16, 24, | + | |
| − | So there are <math>\boxed{\ | + | 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? |
| + | |||
| + | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5</math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | We see how many common integer factors <math>48</math> and <math>64</math> share. | ||
| + | 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>. | ||
| + | So there are <math>\boxed{\textbf{(E)}\ 5}</math> possibilities for the ticket price. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2010|ab=B|num-b=7|num-a=9}} | ||
Revision as of 00:43, 26 November 2011
Problem
A ticket to a school play cost 
dollars, where is a whole number. A group of 9th graders buys tickets costing a total of 
, and a group of 10th graders buys tickets costing a total of 
. How many values for are possible?
Solution
We see how many common integer factors 
and 
share.
Of the factors of - 
; only 
and 
are factors of .
So there are 
possibilities for the ticket price.
See Also
| 2010 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
