2000 AIME I Problems/Problem 1
Problem
Find the least positive integer such that no matter how is expressed as the product of any two positive integers, at least one of these two integers contains the digit .
Solution
If a factor of has a and a in its prime factorization, then that factor will end in a . Therefore, we have left to consider the case when the two factors have the s and the s separated, so we need to find the first power of 2 or 5 that contains a 0.
For n = 1: n = 2: n = 3:
and so on, until,
n = 8: |
We see that contains the first zero, so n = .
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
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