IMO 1977 LL ROM35

Origin: ROM

Problem

Find all numbers N = a1a2 . . . an for which 9 \times a1a2 . . . an = an . . . a2a1 such that at most one of the digits a1, a2, . . . , an is zero.

Solution

The solutions are 0 and Nk = 10 99 . . .9    k 89, where k = 0, 1, 2, . . .. Remark. If we omit the condition that at most one of the digits is zero, the solutions are numbers of the form Nk1Nk2 . . . Nkr, where k1 = kr, k2 = kr−1 etc. The more general problem k \cdot a1a2 . . . an = an . . . a2a1 has solutions only for k = 9 and for k = 4 (namely 0, 2199 . . .978 and combinations as above).