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 6 of 8: Lower-bound the order using mod 5

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In plain words

Any modulus-125 relation must also hold mod 5, so mod-5 behaviour trims the candidate orders.

1978\equiv 3 \pmod 5,\ \mathrm{ord}_5(3)=4 \;\Rightarrow\; 4 \mid \mathrm{ord}_{125}(1978)

Detailed analysis

If $1978^r\equiv 1\pmod{125}$ then also $1978^r\equiv 1\pmod 5$ . Since $1978\equiv 3\pmod 5$ and $3$ has order $4$ modulo $5$ (as $3^4=81\equiv 1$ ), the order modulo $125$ must be a multiple of $4$ , leaving only $4, 20, 100$ among the divisors of $100$ .