Suppose the three-digit number with this property is n.

We are saying that the last three digits of n² – n will be 000.

In other words, n² – n is a multiple of 1000.

Therefore n (n – 1) = 1000 k for some integer k.

It is not possible for n and n – 1 to have any factors in common because they are consecutive numbers. (We say n and n – 1 are co-prime or relatively prime).

As n is a three-digit number, neither n not n – 1 can be divisible by 1000.

Therefore, either n is divisible by 2³ and n – 1 is divisible by 5³, or n – 1 is divisible by 2³ and n is divisible by 5³.

Now we can just list the cases: it will be more efficient to use the larger possible divisor.

If n – 1 is divisible by 5³,

Possible n	Does n have a factor of 2³= 8 ?
126	No
251	No
376	Yes : 376 = 8 x 47
501	No
626	No
751	No
876	No

So 376 is one possibility! Check: 376² = 141,376

If n is divisible by 5³,

Possible n	Does n – 1 have a factor of 2³= 8 ?
125	No
250	No
375	No
500	No
625	Yes: 624 = 8 x 78
750	No
875	No

So 625 is another possibility! Check: 625² = 390,625

We have considered all possible cases so there are no more 3-digit numbers like this.

By carrying out a similar analysis for 4-digit numbers you will find that there is just one possibility: 9376

These numbers are examples of automorphic numbers.

Challenge 243: Matching Tails