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 5 of 8: Euler's theorem bounds the multiplicative order

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In plain words

The order of an element always divides the group's exponent, here φ(125) = 100.

\varphi(125)=100 \;\Rightarrow\; 1978^{100}\equiv 1 \pmod{125},\quad \mathrm{ord}_{125}(1978)\mid 100

Detailed analysis

Euler's theorem gives $1978^{\varphi(125)}\equiv 1\pmod{125}$ with $\varphi(125)=100$ , so the least $n-m$ satisfying the congruence is the multiplicative order $\mathrm{ord}_{125}(1978)$ , which must divide $100$ .