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

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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=25</math>, so the number is <math>725</math>.
 
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=25</math>, so the number is <math>725</math>.
  
*It is quite obvious that <math>n=2</math>, since the desired number can't be single or double digit, and cannot exceed 999. From <math>100a+b=29b</math>, proceed as above.
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*It is quite obvious that <math>n=2</math>, since the desired number can't be single or double digit, and cannot exceed <math>999</math>. From <math>100a+b=29b</math>, proceed as above.
  
 
== See also ==
 
== See also ==

Revision as of 05:44, 15 August 2011

Problem

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ 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 $b=25$, so the number is $725$.

  • It is quite obvious that $n=2$, since the desired number can't be single or double digit, and cannot exceed $999$. From $100a+b=29b$, proceed as above.

See also

2006 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AIME Problems and Solutions
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