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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 \(f(\mathbb{N})\)

 Dec 23, 2021
 #1
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N = 2158.

 Dec 23, 2021
 #3
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Oh no, it seems like we got different answers. 

Do you mind explaining what you did?

 

Thanks

=^._.^=

catmg  Dec 23, 2021
 #2
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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. :))

=^._.^=

 Dec 23, 2021
 #4
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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 

\(\begin{align*} 46^2 - f(0) &= 46^2 - 1 \\ &= 2116 - 1 \\ &= \boxed{2115} \end{align*}\)

MathProblemSolver101  Dec 23, 2021
edited by MathProblemSolver101  Dec 23, 2021

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