Difference between revisions of "1993 AIME Problems/Problem 4"
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How many ordered four-tuples of integers <math>(a,b,c,d)\,</math> with <math>0 < a < b < c < d < 500\,</math> satisfy <math>a + d = b + c\,</math> and <math>bc - ad = 93\,</math>? | How many ordered four-tuples of integers <math>(a,b,c,d)\,</math> with <math>0 < a < b < c < d < 500\,</math> satisfy <math>a + d = b + c\,</math> and <math>bc - ad = 93\,</math>? | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | { | + | === Solution 1 === |
+ | Let <math>k = a + d = b + c</math> so <math>d = k-a, b=k-c</math>. It follows that <math>(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93</math>. Hence <math>(c - a,d - a) = (1,93),(3,31),(31,3),(93,1)</math>. | ||
+ | |||
+ | 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 solution doesnt 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>. | ||
+ | |||
+ | So the total number of such four-tuples is <math>405 + 465 = \boxed{870}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | Let <math>b = a + m</math> and <math>c = a + m + n</math>. From <math>a + d = b + c</math>, <math>d = b + c - a = a + 2m + n</math>. | ||
+ | |||
+ | Substituting <math>b = a + m</math>, <math>c = a + m + n</math>, and <math>d = b + c - a = a + 2m + n</math> into <math>bc - ad = 93</math>, | ||
+ | <cmath> | ||
+ | bc - ad = (1 + m)(1 + m + n) - a(a + 2m + n) = m(m + n). = 93 = 3(31) | ||
+ | </cmath> | ||
+ | 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. | ||
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=3|num-a=5}} | {{AIME box|year=1993|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 16:36, 20 April 2008
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 solution doesnt 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 |