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 7 of 8: Rule out the smaller divisors 4 and 20

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In plain words

Since the order must divide 100 and be a multiple of 4, checking just two values settles it completely.

1978^4\equiv 6,\ 1978^{20}\equiv 26 \pmod{125} \;\Rightarrow\; \mathrm{ord}_{125}(1978)=100

Detailed analysis

Direct computation modulo $125$ gives $1978^4\equiv 6$ and $1978^{20}\equiv 26$ , both different from $1$ , so neither $4$ nor $20$ is the order; the only remaining candidate is $100$ .