Difference between revisions of "2006 AIME I Problems/Problem 3"
m (2006 AIME I Problem 3 moved to 2006 AIME I Problems/Problem 3) |
|||
| Line 1: | Line 1: | ||
== 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 == | ||
| − | + | 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. | ||
== See also == | == See also == | ||
* [[2006 AIME I Problems]] | * [[2006 AIME I Problems]] | ||
| + | * [[Number Theory]] | ||
| + | |||
| + | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 16:14, 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 
, where 
is the leftmost digit, and 
is the rest of the number. We know that 
. Thus has to be 7 since 
can not have 7 as a factor, and the smallest can be and have a factor of 
is 
We find that is 25, so the number is 725.
