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Number Theory Level 4

The sum of the base- $10$ logarithms of the factors of $10^n$ is 792 ( $\log n_1+\log n_2+...\log n_k$ ) where $n_k$ are the factors of $10^n$ . What is $n$ ?

EDIT: The problem has been updated to $10^n$ instead of $10n$ .

The answer is 11.

1 solution

Bogdan Simeonov
May 19, 2014

The logarithm of a sum is the logarithm of the product.So

$log(\pi(10^n))=792$

But $\pi(x)=x^{0,5.\tau(x)}$

So $(10^n)^{0,5.(n+1)^2}=10^{792}$

This is equivalent to

$n.(n+1)^2=2.792$

From this we can obtain $\boxed{n=11}$

This question is exactly the same as this .

Shaan Vaidya - 7 years ago

Sums|smuS

The answer is 11.

1 solution

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