2004 AMC 12A Problems/Problem 3

Revision as of 14:44, 26 February 2008 by Chickendude (talk | contribs) (Clarified solution)

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 (ProblemsAnswer KeyResources)
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