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# Simple Algebra + Counting

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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

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#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
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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