Difference between revisions of "1998 AIME Problems/Problem 7"
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== Solution == | == Solution == | ||
| + | === Solution 1 === | ||
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>. | 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} = \boxed{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>. | ||
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| + | === Solution 2 === | ||
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| + | Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into <math>4</math> 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 <math>94</math> stones left. Because we want an odd number in each box, we pair the stones, creating <math>47</math> sets of <math>2</math>. 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 #). | ||
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| + | Our problem can now be restated: how many ways are there to partition a line of <math>47</math> stones? We can easily solve this by using <math>3</math> sticks to separate the stones into <math>4</math> groups, and this is the same as arranging a line of <math>3</math> sticks and <math>47</math> stones. <cmath>\frac{50!}{47! \cdot 3!} = 19600</cmath> <cmath>\frac{50 * 49 * 48}{3 * 2} = 19600</cmath> Out answer is therefore <math>\frac{19600}{100} = \boxed{196}</math> | ||
== See also == | == See also == | ||
Revision as of 17:09, 16 August 2014
Problem
Let 
be the number of ordered quadruples 
of positive odd integers that satisfy 
Find 
Solution
Solution 1
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 
.
Solution 2
Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into 
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 
stones left. Because we want an odd number in each box, we pair the stones, creating 
sets of 
. 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 stones? We can easily solve this by using 
sticks to separate the stones into groups, and this is the same as arranging a line of sticks and stones. ![\[\frac{50!}{47! \cdot 3!} = 19600\]](http://latex.artofproblemsolving.com/1/1/c/11cc0307e137414841143ed322df60aadf79991f.png)
![\[\frac{50 * 49 * 48}{3 * 2} = 19600\]](http://latex.artofproblemsolving.com/b/5/5/b55ee3c984200525f16fc91fd7bb932ccf7fdee1.png)
Out answer is therefore 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
