Prove that a natural number n is a multiple of 8 if and only if the number formed by the last 3 digits of n is a multiple of 8.

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Let n be a natural number that is a multiple of 8

If n<1000, then the last 3 digits of n is n, and n is a multiple of 8.

If n>=1000, there exists integer M such that n=8M.

Let q, r be the quotient and remainder when n is divided by 1000.

By division algorithm, n=1000q+r.

Note that r<1000, so r is the last 3 digits of n.

Then 8M=1000q+r

r=1000q-8M=8 (125q-M)

Since (125q-M) is an integer, r is a multiple of 3.

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Let n be a natrual number such that its last 3 digits is a multiple of 8.

Then there exists integers q and M such that n=1000q+8M.

Notice that n=8 (125q+M), where (125q+M) is an integer. Hence n is a multiple of 8.