1988 USAMO Problems/Problem 1
Problem
The repeating decimal , where and are relatively prime integers, and there is at least one decimal before the repeating part. Show that is divisble by 2 or 5 (or both). (For example, , and 88 is divisible by 2.)
Solution
Solution 1
First, split up the nonrepeating parts and the repeating parts of the decimal, so that the nonrepeating parts equal to and the repeating parts of the decimal is equal to .
Suppose that the length of is digits. Then Since , after reducing the fraction, there MUST be either a factor of 2 or 5 remaining in the denominator. After adding the fractions , the simplified denominator will be and since has a factor of or , must also have a factor of 2 or 5.
Solution 2
It is well-known that , where there are a number of 9s equal to the count of digits in , and there are a number of 0s equal to the count of digits in . Obviously is different from (which is itself the repeating part), so the numerator cannot have consecutive terminating zeros, and hence the denominator still possesses a factor of 2 or 5 not canceled out. This completes the proof.
See Also
1988 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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