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 (Clever Stars and Bars Manipulation)

We want $x_1 +x_2+x_3+x_4 =98$. This seems like it can be solved with stars and bars, however note that the quadruples all need to be odd. This motivates us to set $x_i= 2y_i +1$, as for all integers $y_i$, $2y_i + 1$ will be odd. Substituting we get \[2y_1+2y_2+2y_3+2y_4 +4 = 98 \implies y_1+y_2+y_3+y_4 =47\] Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain stars as bars, which gives us $n= {50\choose3}$. Computing this and dividing by 100 gives us an answer of $\boxed{196}$. ~smartguy888

Solution 2

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 3

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 4

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 5

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

See also

1998 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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