Difference between revisions of "1993 AIME Problems/Problem 4"
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Solve them in tems of <math>c</math> to get | Solve them in tems of <math>c</math> to get | ||
− | <math>(a,b,c,d) = (c - 93,c - 92,c,c + 1),</math> <math>(c - 31,c - 28,c,c + 3),</math> <math>(c - 1,c + 92,c,c + 93),</math> <math>(c - 3,c + 28,c,c + 31)</math>. The last two | + | <math>(a,b,c,d) = (c - 93,c - 92,c,c + 1),</math> <math>(c - 31,c - 28,c,c + 3),</math> <math>(c - 1,c + 92,c,c + 93),</math> <math>(c - 3,c + 28,c,c + 31)</math>. The last two solutions don't follow <math>a < b < c < d</math>, so we only need to consider the first two solutions. |
The first solution gives us <math>c - 93\geq 1</math> and <math>c + 1\leq 499</math> <math>\implies 94\leq c\leq 498</math>, and the second one gives us <math>32\leq c\leq 496</math>. | The first solution gives us <math>c - 93\geq 1</math> and <math>c + 1\leq 499</math> <math>\implies 94\leq c\leq 498</math>, and the second one gives us <math>32\leq c\leq 496</math>. |
Revision as of 21:43, 29 August 2011
Problem
How many ordered four-tuples of integers with satisfy and ?
Contents
[hide]Solution
Solution 1
Let so . It follows that . Hence .
Solve them in tems of to get . The last two solutions don't follow , so we only need to consider the first two solutions.
The first solution gives us and , and the second one gives us .
So the total number of such four-tuples is .
Solution 2
Let and . From , .
Substituting , , and into , Hence, or .
For , we know that , so there are four-tuples. For , , and there are four-tuples. In total, we have four-tuples.
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |