Difference between revisions of "2004 AMC 12A Problems/Problem 25"
(New page: ==Problem== For each integer <math>n\geq 4</math>, let <math>a_n</math> denote the base-<math>n</math> number <math>0.\overline{133}_n</math>. The product <math>a_4a_5...a_{99}</math> can ...) |
(LaTeX issue fixed) |
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− | <math>a_4a_5...a_{99}=\frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4*5*6*\cdots*99*(4^3-1)(5^3-1)\cdots(99^3-1)</math> | + | <math>a_4a_5...a_{99}=\frac{(5^3-1)(6^3-1)\cdots (100^3-1)}{4*5*6*\cdots*99*(4^3-1)(5^3-1)\cdots(99^3-1)}</math> |
<math>a_4a_5...a_{99}=\frac{999999}{4*5*6*\cdots*99*63}=\frac{13*37*33*6}{99!}</math> | <math>a_4a_5...a_{99}=\frac{999999}{4*5*6*\cdots*99*63}=\frac{13*37*33*6}{99!}</math> |
Revision as of 10:27, 4 December 2007
Problem
For each integer , let denote the base- number . The product can be expressed as , where and are positive integers and is as small as possible. What is the value of ?
Solution
Since isn't one of the answer choices, we need to get rid of some stuff:
Since only the two goes into 98, n is at it's minimum.