Difference between revisions of "2004 AMC 12A Problems/Problem 3"
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<math>\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100</math> | <math>\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Every integer value of <math>y</math> leads to an integer solution for <math>x</math> | 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> | Since <math>y</math> must be positive, <math>y\geq 1</math> | ||
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<math>1 \leq 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> | This leaves <math>49</math> values for y, which mean there are <math>49</math> solutions to the equation <math>\Rightarrow \mathrm{(B)}</math> | ||
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| + | ==Solution 2== | ||
| + | If <math>x</math> and <math>2y</math> must each be positive integers, then we can say that <math>x</math> is at least 1 and <math>2y</math> is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer addends of 100). 98 identical objects can be distributed to two distinct bins in 99 ways (think stars and bars), yet this 99 is an overcount. Because <math>y</math> must be an integer, <math>2y</math> must be even; thus only <math>\left\lfloor \frac{99}{2} \right\rfloor = \boxed{ 49 \implies B}</math> ways exist to distribute these ones. | ||
==See Also== | ==See Also== | ||
Latest revision as of 00:35, 29 September 2018
Contents
Problem
For how many ordered pairs of positive integers 
is 
?
Solution 1
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 
Solution 2
If and 
must each be positive integers, then we can say that is at least 1 and is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer addends of 100). 98 identical objects can be distributed to two distinct bins in 99 ways (think stars and bars), yet this 99 is an overcount. Because must be an integer, must be even; thus only 
ways exist to distribute these ones.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
