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 1 of 8: Translate the digit condition into a congruence

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In plain words

"Same last three digits" is just congruence mod 1000 in disguise.

1978^n \equiv 1978^m \pmod{1000} \iff 1978^m\left(1978^{\,n-m}-1\right)\equiv 0 \pmod{1000}

Detailed analysis

Two positive integers have the same last three decimal digits exactly when their difference is a multiple of $1000$ . Applying this to $1978^n$ and $1978^m$ and factoring out $1978^m$ gives the equivalent multiplicative condition.