Difference between revisions of "1990 AIME Problems/Problem 11"
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(last part of solution is weak, but I think it works) |
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== Problem == | == Problem == | ||
− | Someone observed that <math>6! = 8 \cdot 9 \cdot 10</math>. Find the largest positive integer <math>n^{}_{}</math> for which <math>n^{}_{}!</math> can be expressed as the product of <math>n - 3_{}^{}</math> consecutive positive integers. | + | Someone observed that <math>6! = 8 \cdot 9 \cdot 10</math>. Find the largest [[positive]] [[integer]] <math>n^{}_{}</math> for which <math>n^{}_{}!</math> can be expressed as the [[product]] of <math>n - 3_{}^{}</math> [[consecutive]] positive integers. |
== Solution == | == Solution == | ||
− | {{ | + | The product of <math>n - 3</math> consecutive integers can be written as <math>\frac{(n - 3 + a)!}{a!}</math> for some integer <math>a</math>. Thus, <math>n! = \frac{(n - 3 + a)!}{a!}</math>, from which it becomes evident that <math>a \ge 3</math>. Since <math>\displaystyle (n - 3 + a)! > n!</math>, we can rewrite this as <math>\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!</math>. For <math>a = 4</math>, we get <math>n + 1 = 4!</math> so <math>n = 23</math>. For greater values of <math>a</math>, we need to find the product of <math>a-3</math> consecutive integers that equals <math>a!</math>. <math>n</math> can be approximated as <math>\displaystyle ^{a-3}\sqrt{a!}</math>, which decreases as <math>a</math> increases. Thus, <math>n = 23</math> is the greatest possible value to satisfy the given conditions. |
== See also == | == See also == | ||
{{AIME box|year=1990|num-b=10|num-a=12}} | {{AIME box|year=1990|num-b=10|num-a=12}} |
Revision as of 11:21, 3 March 2007
Problem
Someone observed that . Find the largest positive integer for which can be expressed as the product of consecutive positive integers.
Solution
The product of consecutive integers can be written as for some integer . Thus, , from which it becomes evident that . Since , we can rewrite this as . For , we get so . For greater values of , we need to find the product of consecutive integers that equals . can be approximated as , which decreases as increases. Thus, is the greatest possible value to satisfy the given conditions.
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |