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

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== Problem ==
 
== 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.
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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 ==
 
== Solution ==

Revision as of 12:59, 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.

See also

2006 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions