1998 AIME Problems/Problem 7

Problem

Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$

Solution

Define $x_i = 2y_i - 1$ . Then $2\left(\sum_{i = 1}^4 y_i\right) - 4 = 98$ , so $\sum_{i = 1}^4 y_i = 51$ .

So we want to find four integers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is $n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600$ , and $\frac n{100} = \boxed{196}$ .

1998 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1998 AIME Problems/Problem 7

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