Difference between revisions of "2006 AIME I Problems/Problem 3"

Revision as of 13:59, 25 September 2007

Problem

Let $P$ be the product of the first $100$ positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k .$

Solution

The number can be represented as $10^na+b$ , where $a$ is the leftmost digit, and $b$ is the rest of the number. We know that $b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na$ . Thus $a$ has to be 7 since $10^n$ can not have 7 as a factor, and the smallest $10^n$ can be and have a factor of $2^2$ is $10^2=100.$ We find that $b$ is 25, so the number is 725.

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 == Problem ==
-Find the least [[positive]] [[integer]] such that when its leftmost [[digit]] is deleted, the resulting integer is <math>\frac{1}{29}</math> of the original integer.
+Let <math>P </math> be the product of the first <math>100</math> [[positive integer | positive]] [[odd integer]]s. Find the largest integer <math>k </math> such that <math>P </math> is divisible by <math>3^k .</math>
 == Solution ==

2006 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2006 AIME I Problems/Problem 3"

Revision as of 13:59, 25 September 2007

Problem

Solution

See also