2004 AMC 12A Problems/Problem 3

Problem

For how many ordered pairs of positive integers $(x,y)$ is $x + 2y = 100$ ?

$\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100$

Solution 1

Every integer value of $y$ leads to an integer solution for $x$ Since $y$ must be positive, $y\geq 1$

Also, $y = \frac{100-x}{2}$ Since $x$ must be positive, $y < 50$

$1 \leq y < 50$ This leaves $49$ values for y, which mean there are $49$ solutions to the equation $\Rightarrow \mathrm{(B)}$

Solution 2

If $x$ and $2y$ must each be positive integers, then we can say that $x$ is at least 1 and $2y$ 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 $y$ must be an integer, $2y$ must be even; thus only $\left\lfloor \frac{99}{2} \right\rfloor = \boxed{ 49 \implies B}$ ways exist to distribute these ones.

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 12 Problems and Solutions

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2004 AMC 12A Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

See Also