+876 A function f(n) is defined as the product of its positive digits. For example, f(20)=2,f(12)=2,f(67)=42
We let N=∑99i=1f(i). Find f(N)
+2407 Hi :))
I'm going to group the numbers to make them easy to calulate.
i = 10, 20, 30, ..., 90. 1 + 2 + 3 + ... + 9
1 <= i <= 9. 1 + 2 + 3 + ... + 9
11 <= i <= 19. 1 + 2 + 3 + ... + 9
21 <= i <= 29. 2(1 + 2 + 3 + ... + 9)
31 <= i <= 39. 3(1 + 2 + 3 + ... + 9)
...
91 <= i <= 99. 9(1 + 2 + 3 + ... + 9)
Overall. there are 1 + 1 + 1 + 2 + 3 + ... + 9 = 2 + 10*9/2 = 47 groups of (1 + 2 + 3 + ... + 9)
47*(1 + 2 + 3 + ... + 9) = 47*45 = 2115.
I hope this helped. :))
=^._.^=
+876 This is correct. Nice job!
Perhaps, you notice in your answer that 2115=462−1? This is no coincidence.
Instead of letting f(0)=0, we let f(0)=1
And it follows that ∑99i=0f(i)=(1+1+2+3+⋯+9)2=462
Subtracting f(0), we have
462−f(0)=462−1=2116−1=2115
