1978 IMO Problems/Problem 1

Problem

Let $m$ and $n$ be positive integers such that $1 \le m < n$ . In their decimal representations, the last three digits of $1978^m$ are equal, respectively, to the last three digits of $1978^n$ . Find $m$ and $n$ such that $m + n$ has its least value.

Solution

We have $1978^m\equiv 1978^n\pmod {1000}$ , or $978^m-978^n=1000k$ for some positive integer $k$ (if it is not positive just do $978^n-978^m=-1000k$ ). Hence $978^n\mid 1000k$ . So dividing through by $978^n$ we get $978^{m-n}-1=\frac{1000k}{978^n}$ . Observe that $2\nmid LHS$ , so $2\nmid RHS$ . So since $2|| 978^n$ , clearly the minimum possible value of $n$ is $3$ (and then $489^n\mid k$ ). We will show later that if $n$ is minimal then $m$ is minimal. We have $978^{m-3}-1\equiv 0\pmod {125}\Leftrightarrow 103^{m-3}\equiv 1\pmod {125}$ . Hence, $m-3\mid \varphi(125)\Rightarrow m-3\mid 100$ . Checking by hand we find that only $m-3=100$ works (this also shows that minimality of $m$ depends on $n$ , as claimed above). So $m=103$ . Consequently, $m+n=106$ with $\boxed{(m,n)=(103,3)}$ .

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

1978 IMO (Problems) • Resources
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1978 IMO Problems/Problem 1

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