1978 IMO Problems/Problem 1
Let and be positive integers such that . In their decimal representations, the last three digits of are equal, respectively, to the last three digits of . Find and such that has its least 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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