1998 AIME Problems/Problem 7

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Problem

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

Solution

Define $\displaystyle x_i = 2y_i - 1$ . 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 $\frac n{100} = 196$ .

1998 AIME (Problems • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1998 AIME Problems/Problem 7

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Problem

Solution

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