Difference between revisions of "2004 AMC 12A Problems/Problem 3"

Revision as of 14:44, 26 February 2008

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

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

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 $\Rightarrow \mathrm{(B)}$

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

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 ==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>
-The answer is (B) since x can't be 0, which leaves only 49 different values for y.
+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==
 {{AMC12 box|year=2004|ab=A|num-b=2|num-a=4}}