1978 IMO Problems/Problem 1
and are positive integers with . The last three decimal digits of are the same as the last three decimal digits of . Find and such that has the least possible value.
We have , or for some positive integer (if it is not positive just do ). Hence . So dividing through by we get . Observe that , so . So since , clearly the minimum possible value of is (and then ). We will show later that if is minimal then is minimal. We have . Hence, . Checking by hand we find that only works (this also shows that minimality of depends on , as claimed above). So . Consequently, with .
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