2006 AIME A Problems/Problem 3
Revision as of 18:49, 25 September 2007 by Azjps (talk | contribs) (2006 AIME II Problems/Problem 3 moved to 2006 AIME A Problems/Problem 3)
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.
See also
2006 AIME II (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 |