Difference between revisions of "2004 AMC 12A Problems/Problem 3"
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Chickendude (talk | contribs) (Clarified solution) |
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==Solution== | ==Solution== | ||
+ | Every integer value of <math>y</math> leads to an integer solution for <math>x</math> | ||
+ | Since <math>y</math> must be positive, <math>y\geq 1</math> | ||
− | + | Also, <math>y = \frac{100-x}{2}</math> | |
+ | Since <math>x</math> must be positive, <math>y < 50</math> | ||
+ | <math>1 \leq y < 50</math> | ||
+ | This leaves <math>49</math> values for y, which mean there are <math>49</math> solutions to the equation <math>\Rightarrow \mathrm{(B)}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}} | {{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}} |
Revision as of 15:44, 26 February 2008
Problem
For how many ordered pairs of positive integers is
?
Solution
Every integer value of leads to an integer solution for
Since
must be positive,
Also,
Since
must be positive,
This leaves
values for y, which mean there are
solutions to the equation
See Also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 12 Problems and Solutions |