# 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 1

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 natural numbers 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 $\frac n{100} = \boxed{196}$.

## Solution 2

Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into $4$ boxes so that each box has an odd number of stones. We then proceed by placing one stone in each box to begin with, ensuring that we have a positive number in every box. Now we have $94$ stones left. Because we want an odd number in each box, we pair the stones, creating $47$ sets of $2$. Every time we add a pair to one of the boxes, the number of stones in the box remains odd, because (an odd #) + (an even #) = (an odd #).

Our problem can now be restated: how many ways are there to partition a line of $47$ stones? We can easily solve this by using $3$ sticks to separate the stones into $4$ groups, and this is the same as arranging a line of $3$ sticks and $47$ stones. $$\frac{50!}{47! \cdot 3!} = 19600$$ $$\frac{50 * 49 * 48}{3 * 2} = 19600$$ Our answer is therefore $\frac{19600}{100} = \boxed{196}$

## Solution 3

Let $x = a + b$ and $y = c + d$. Then $x + y = 98$, where $x, y$ are positive even integers ranging from $2-98$.

-When $(x, y) = (2, 96)$, $(a, b) = (1, 1)$ and $(c, d) = (1, 95), (3, 93),...,(95, 1)$. This accounts for $48$ solutions.

-When $(x, y) = (4, 94)$, $(a, b) = (1, 3), (3, 1)$ and $(c, d) = (1, 93), (3, 91),...,(93, 1)$. This accounts for $94$ solutions.

We quickly see that the total number of acceptable ordered pairs $(a, b, c, d) = 1 \cdot 48 + 2 \cdot 47 + 3 \cdot 46 + ... + 48 \cdot 1 = (24.5 - 23.5)(24.5) + (24.5 - 22.5)(24.5 + 22.5) + ... + (24.5 + 23.5)(24.5 - 23.5) = 48(24.5)^2 - 2(0.5^2 + 1.5^2 + ... + 23.5^2) = 28812 - \frac{1^2 + 3^2 + ... + 47^2}{2} = 28812 - \frac{1^2 + 2^2 + ... + 47^2 - 4(1^2 + 2^2 + ... + 23^2)}{2} = 28812 - \frac{47(47 + 1)(2(47) + 1)/6 - 4(23)(23 + 1)(2(23) + 1)/6}{2} = 19600$. Therefore, $\frac{n}{100} = \frac{19600}{100} = \boxed{196}$.

(This solution uses the sum of squares identity to calculate $1^2 + 2^2 + ... + 47^2$ and $1^2 + 2^2 + ... + 23^2$.)

<baker77>

## Solution 4

We write the generating functions for each of the terms, and obtain $(x+x^3+x^5\cdots)^4$ as the generating function for the sum of the $4$ numbers. We seek the $x^{98}$ coefficient, or the $x^{94}$ coefficient in $(1+x^2+x^4...)^4.$ Now we simplify this as $\left(\frac{1}{1-x^2}\right)^4=\binom{3}{3} +\binom{4}{3}x^2+\binom{5}{3}x^4 \cdots$ and in general we want that the coefficient of $x^{2k}$ is $\binom{k+3}{3}.$ We see the $x^{94}$ coefficient so we let $k=47$ and so the coefficient is $\binom{50}{3}=19600$ in which $\frac{n}{100}=\boxed{196}.$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 