IMO 1970 SL 2

Origin: ROM

Problem

Let a and b be the bases of two number systems and let An = x1x2 . . . xn (a), An+1 = x0x1x2 . . . xn (a), Bn = x1x2 . . . xn(b), Bn+1 = x0x1x2 . . . xn(b), be numbers in the number systems with respective bases a and b, so that x0, x1, x2, . . . , xn denote digits in the number system with base a as well as in the number system with base b. Suppose that neither x0 nor x1 is zero. Prove that a > b if and only if An An+1 < Bn Bn+1 .

Solution

Suppose that a > b. Consider the polynomial P(X) = x1Xn−1+x2Xn−2+ \cdot \cdot \cdot + xn−1X + xn. We have An = P(a), Bn = P(b), An+1 = x0an + P(a), and Bn+1 = x0bn + P(b). Now An/An+1 < Bn/Bn+1 becomes P(a)/(x0an + P(a)) < P(b)/(x0bn + P(b)), i.e., bnP(a) < anP(b). Since a > b, we have that ai > bi and hence xianbn−i \geqxibnan−i (also, for i \geq1 the inequality is strict). Summing up all these inequalities for i = 1, . . . , n we get anP(b) > bnP(a), which completes the proof for a > b.

On the other hand, for a < b we analogously obtain the opposite inequality An/An+1 > Bn/Bn+1, while for a = b we have equality. Thus An/An+1 < Bn/Bn+1 ⇔a > b.