MathDB

6

Part of 1994 Swedish Mathematical Competition

Problems(1)

f(a+b) = f(f(a)+b), f(a+b) = f(a)+f(b) for a+b < 10, f(10) = 1

Source: 1994 Swedish Mathematical Competition p6

4/2/2021
Let NN be the set of non-negative integers. The function f:Nā†’Nf:N\to N satisfies f(a+b)=f(f(a)+b)f(a+b) = f(f(a)+b) for all a,ba, b and f(a+b)=f(a)+f(b)f(a+b) = f(a)+f(b) for a+b<10a+b < 10. Also f(10)=1f(10) = 1. How many three digit numbers nn satisfy f(n)=f(N)f(n) = f(N), where NN is the "tower" 2,3,4,52, 3, 4, 5, in other words, it is 2a2^a, where a=3ba = 3^b, where b=45b = 4^5?
number theoryfunctionalfunctional equation