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 1/29 of the original integer.
 
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.
 
 
 
  
 
== Solution ==
 
== Solution ==
 
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The number can be represented as <math>10^na+b</math>, where <math> a </math> is the leftmost digit, and <math> b </math> is the rest of the number. We know that <math>b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na</math>. Thus <math> a </math> has to be 7 since <math> 10^n </math> can not have 7 as a factor, and the smallest <math> 10^n </math> can be and have a factor of <math> 2^2 </math> is <math> 10^2=100. </math> We find that <math> b </math> is 25, so the number is 725.
The number can be represented as <math>10^na+b</math>, where a is the leftmost digit, and b is the rest of the number. It satisfies <math>b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na</math>. 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 ==
 
== See also ==
 
* [[2006 AIME I Problems]]
 
* [[2006 AIME I Problems]]
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* [[Number Theory]]
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[[Category:Intermediate Number Theory Problems]]

Revision as of 17:14, 18 July 2006

Problem

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.

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