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2004 AMC 12A Problems/Problem 3

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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

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)}

See Also

2004 AMC 12A (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
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