MathDB

6

Part of 2003 Poland - Second Round

Problems(1)

Integer function, number of solutions of equation

Source: 54 Polish MO 2003 Second Round - Second Day Problem 6

4/21/2018
Each pair (x,y)(x, y) of nonnegative integers is assigned number f(x,y)f(x, y) according the conditions: f(0,0)=0f(0, 0) = 0; f(2x,2y)=f(2x+1,2y+1)=f(x,y)f(2x, 2y) = f(2x + 1, 2y + 1) = f(x, y), f(2x+1,2y)=f(2x,2y+1)=f(x,y)+1f(2x + 1, 2y) = f(2x, 2y + 1) = f(x ,y) + 1 for x,y≥0x, y \ge 0. Let nn be a fixed nonnegative integer and let aa, bb be nonnegative integers such that f(a,b)=nf(a, b) = n. Decide how many numbers satisfy the equation f(a,x)+f(b,x)=nf(a, x) + f(b, x) = n.
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