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/ | + | 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. |
== Solution == | == Solution == | ||
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== See also == | == See also == | ||
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* [[Number Theory]] | * [[Number Theory]] | ||
| + | {{AIME box|year=2006|n=I|num-b=2|num-a=4}} | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 20:18, 11 March 2007
Problem
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 
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.
See also
| 2006 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
