Every integer can be expressed in the following way: 5p+q, where p and q are certain integers and 0 £ q £ 4. Expanding (5p + q)¹⁰⁰ with the aid of the binomial theorem we get the general term:

ak = æ ç è 100 k ö ÷ ø 5100-kp100-kqk.

All the terms except the last one q¹⁰⁰ are divisible by 125. What we get is a number of this sort: 125×something+ q¹⁰⁰. But we know that 0 £ q £ 4 so the remainder when the 100th power is divided by 125 is the same as the remainder for q¹⁰⁰, with q = 0, 1, 2, 3, or 4.

If q=0 then q¹⁰⁰=0; if q=1 then q¹⁰⁰=1.

Let q=2; we want the remainder after 2¹⁰⁰ is divided by 125, so we work modulo 125. Now 2⁷=128 º 3 where the symbol ' º ' indicates that the numbers have the same remainder after division by 125. For q=3 it follows that 3⁵ = 243 º -7. Thus

2100 º 22×27×14

º 4×314

º 4×34 ×35×2

º 4×34 ×2432

º 4×81×(-7)2

= 4×81×49

= 15876 º 1

Similarly

3100 º 35×20

º 24320

º (-7)20

º (-32)6×(-7)2

º 230 ×49

º 27 ×4×4 ×49

º 34 ×4 ×49

= 15876 º 1

If q=4 then

q100 =

æ è

2100

ö ø

2

º 1.

So we can either get 0 or 1 as a remainder and we get 0 if the original number is a multiple of 5 and 1 otherwise.