Difference between revisions of "1998 AIME Problems/Problem 7"

Revision as of 18:38, 4 July 2013

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

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math>n</math> be the number of ordered quadruples <math>\displaystyle(x_1,x_2,x_3,x_4)</math> of positive odd [[integer]]s that satisfy <math>\sum_{i = 1}^4 x_i = 98.</math>  Find <math>\frac n{100}.</math>
+Let <math>n</math> be the number of ordered quadruples <math>(x_1,x_2,x_3,x_4)</math> of positive odd [[integer]]s that satisfy <math>\sum_{i = 1}^4 x_i = 98.</math>  Find <math>\frac n{100}.</math>
 == Solution ==
@@ Line 11: / Line 11: @@
 [[Category:Intermediate Combinatorics Problems]]
+{{MAA Notice}}

1998 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1998 AIME Problems/Problem 7"

Revision as of 18:38, 4 July 2013

Problem

Solution

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