If n is a positive integer, let d+1 be the number of digits in n (in base 10) and s be the sum of the digits. Let n(k) be the number formed by deleting the last k digits of n. Prove that n = s + 9 n(1) + 9 n(2) + ... + 9 n(d).
Solution
Let the digits of n be ad, ad-1, ... , a0, so that n = ad 10d + ... + a0. Then n(k) = ad 10d-k + ad-1 10d-k-1 + ... + ak. Obviously s = ad + ... + a0. Hence s + 9 n(1) + ... + 9 n(d) = ad(9.10d-1 + 9.10d-2 + ... + 9 + 1) + ad-1(9.10d-2 + ... + 9 + 1) + ... + ad-k(9.10d-k-1 + ... + 9 + 1) + a1(9 + 1) + a0 = ad 10d + ... + a0 = n.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002