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 4 of 8: Reduce the mod-125 part to an order condition

Previous step Next step

In plain words

1978^m is invertible mod 125, so it can be cancelled entirely from the congruence.

\gcd(1978,125)=1 \;\Rightarrow\; 1978^{\,n-m}\equiv 1 \pmod{125}

Detailed analysis

Since $1978$ is invertible modulo $125$ , the mod- $125$ condition $1978^m\left(1978^{\,n-m}-1\right)\equiv 0$ reduces directly to $1978^{\,n-m}\equiv 1\pmod{125}$ , independent of $m$ .