Difference between revisions of "2006 AIME I Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | + | 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. | |
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== See also == | == See also == | ||
* [[2006 AIME I Problems]] | * [[2006 AIME I Problems]] | ||
Revision as of 23:49, 30 June 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 
, where a is the leftmost digit, and b is the rest of the number. It satisfies 
. 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.
