Difference between revisions of "1998 AIME Problems/Problem 7"
(prob/solultion) |
Mathisfun7 (talk | contribs) (→Solution) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | Define <math> | + | Define <math>x_i = 2y_i - 1</math>. Then <math>2\left(\sum_{i = 1}^4 y_i\right) - 4 = 98</math>, so <math>\sum_{i = 1}^4 y_i = 51</math>. |
| − | 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 <math>n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600</math>, and <math>\frac n{100} = 196</math>. | + | 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 <math>n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600</math>, and <math>\frac n{100} = \boxed{196}</math>. |
== See also == | == See also == | ||
Revision as of 16:19, 25 May 2013
Problem
Let 
be the number of ordered quadruples 
of positive odd integers that satisfy 
Find 
Solution
Define 
. Then 
, so 
.
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 
, and 
.
See also
| 1998 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
