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 3 of 8: Force m ≥ 3 from the factor of 8

In plain words

The odd factor cannot supply any power of 2, so all three factors of 2 must come from 1978^m.

1978 \equiv 2 \pmod 8 \;\Rightarrow\; 8 \mid 1978^m \iff m \ge 3

Detailed analysis

For the mod-8 part, $1978^{\,n-m}-1$ is odd, so $8$ must divide $1978^m$ outright. Since $1978\equiv 2\pmod 8$ , this happens exactly when $m\ge 3$ .