1988 USAMO Problems/Problem 1
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.)
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.
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.
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