Difference between revisions of "2010 AMC 10B Problems/Problem 8"
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==Solution 2== | ==Solution 2== | ||
− | The difference between 48 and 64 is 16. 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. Therefore, there are <math>\boxed{\textbf{(E)}\ 5}</math> possibilities for the ticket price. | + | The difference between 48 and 64 is 16. 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. Therefore, there are <math>\boxed{\textbf{(E)}\ 5}</math> possibilities for the ticket price. |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2010|ab=B|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:59, 9 October 2017
Contents
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 1
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.
Solution 2
The difference between 48 and 64 is 16. 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 more tickets than the 9th graders. Therefore, 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 |
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