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 21:18, 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 |