+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 \(\mathbb{N} = \sum_{i=1}^{99}f(i)\). Find \(f(\mathbb{N})\)
+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 \(2115 = 46^2 - 1\)? This is no coincidence.
Instead of letting \(f(0)=0\), we let \(f(0)=1\)
And it follows that \(\sum_{i=0}^{99}f(i)=(1+1+2+3+\cdots+9)^2=46^2\)
Subtracting \(f(0)\), we have
\(\begin{align*} 46^2 - f(0) &= 46^2 - 1 \\ &= 2116 - 1 \\ &= \boxed{2115} \end{align*}\)
