Difference between revisions of "2010 AMC 10B Problems/Problem 8"
Line 1: | Line 1: | ||
− | 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 01:43, 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 |