Problem 1

Let

m

and

n

be positive integers with

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.

Step 8 of 8: Assemble the minimal answer

In plain words

Both m and n-m were pinned to their smallest possible values, so this pair genuinely minimizes m+n.

m=3,\ n-m=100 \;\Rightarrow\; n=103,\quad m+n=106

Detailed analysis

Taking the smallest allowed $m=3$ and the exact order $n-m=100$ gives $n=103$ , so the least possible value of $m+n$ is $3+103=106$ .