Difference between revisions of "1993 AIME Problems/Problem 4"
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Hence, <math>(m,n) = (1,92)</math> or <math>(3,28)</math>. | Hence, <math>(m,n) = (1,92)</math> or <math>(3,28)</math>. | ||
− | For <math>(m,n) = (1,92)</math>, we know that <math>0 < a < a + 1 < a + 93 < a + 94 < 500</math>, so there are <math>405</math> four-tuples. For <math>(m,n) = (3,28)</math>, <math>0 < a < a + 3 < a + 31 < a + 34 < 500</math>, and there are <math>465</math> four-tuples. In total, we have <math>405 + 465 = 870</math> four-tuples. | + | For <math>(m,n) = (1,92)</math>, we know that <math>0 < a < a + 1 < a + 93 < a + 94 < 500</math>, so there are <math>405</math> four-tuples. For <math>(m,n) = (3,28)</math>, <math>0 < a < a + 3 < a + 31 < a + 34 < 500</math>, and there are <math>465</math> four-tuples. In total, we have <math>405 + 465 = \boxed{870}</math> four-tuples. |
== See also == | == See also == |
Revision as of 23:22, 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 |