1978 IMO Problems/Problem 1
Contents
Problem
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.
Solution
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 .
The above solution was posted and copyrighted by cobbler and Andreas. The original thread for this problem can be found here: [1] and [2]
Video Solution
https://www.youtube.com/watch?v=SRl4Wnd60os
See Also
1978 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |