Add. Base. Clog. Divisors.

I don't know if this is the hard way to solve this, but I did it this way... First, let's prove a lemma...

Lemma : If $n$ is a positive integer with $k$ positive divisors, the product of all of it's positive divisors is $\sqrt{n^k}$ , which is... $\prod_{d\mid n}d = \sqrt{n^k}$

Proof : If $n$ is not a perfect square, then it has an even number of divisors. For every $a\mid n$ , there's a $b=\dfrac{n}{a} \neq a$ such that $b\mid n$ , and so $ab=n$ . So we can pair them up and obtain $n$ for each couple. There's $\dfrac{k}{2}$ such couples. Hence... $\prod_{d\mid n}d = n^{\frac{k}{2}}=\sqrt{n^k}$

If $n$ is a perfect square, then if has an odd number of divisors. For every $a\mid n$ except $\sqrt{n}$ , there's a $b=\dfrac{n}{a}\neq a$ such that $b\mid n$ and so $ab=n$ . So we can pair them up and obtain $n$ for each couple. There's $\dfrac{k-1}{2}$ such couples. We have one more divisor $\sqrt{n}$ . Hence we get... $\prod_{d\mid n}d = n^{\frac{k-1}{2}}\cdot \sqrt{n}=n^{\frac{k-1}{2}+\frac{1}{2}}=n^{\frac{k}{2}}=\sqrt{n^k}$

Therefore, for all $n\in \mathbb{N}$ , the product of their divisors... $\prod_{d\mid n}d = \sqrt{n^k}~~~~~~~~\text{[QED]}$

Moving on, according to the question, we get... $\sum_{d\mid n} \log_{10} d=\log_{10} \left(\prod_{d\mid n} d \right)=\log_{10}\sqrt{n^k}=792$ $\Longrightarrow ~~~ \dfrac{k}{2}\log_{10}n=792 ~~~ \Longrightarrow ~~~ k\cdot \log_{10}n=792\cdot 2$ $\Longrightarrow ~~~ k \cdot \log_{10} 10^N =kN= 1584$

Note that $10^N = 2^N 5^N$ , from the divisor theorem, we get $k=(N+1)(N+1)=(N+1)^2$ . Hence... $N(N+1)^2=1584 = 4^2\cdot 3^2\cdot 11 = 11\cdot 12^2 = 11\cdot (11+1)^2$

Therefore, $N=\fbox{11}$ .

Notice that: $log(a)+ log(b)= log(ab)$ so that means every pair of divisors of $10^{n}$ must simplify to $log(10^{n}) = n$ .

Now, we have to find the number of divisors of n.
$10^{n} = 2^{n} \times 5^{n}$ by exponent rules.
The number of divisors of: $2^{n} \times 5^{n}$ is $(n+1)(n+1)$ Note: This is an important fact to memorize.
This means the number of pairs of divisors is: $\frac{(n+1)(n+1)}{2}$ .
As stated in the first line, each pair of divisors will add $n$ value to the sum of the base-10 logarithms.

So, the final equation is: $\frac{(n+1)(n+1)}{2} \times n = 792$ .
With some simplification, this becomes: $n^{3} + 2n^{2} + n -1584 =0$ .
Using Descartes rule of signs there can only be one positive solution (which is consistent in the context of this problem.)
How to factor this? uhm... any way you want. I used the fancy guessing strategy (rational root theorem).
After lots of guessing and checking, you find that $n= \boxed{11}$ which is the final answer.

prime factors of 10= 2 X 5

so 10^N = 2^N x 5^N

so to divide 10^N , the factor/divisor should be of the form (2^m x 5^n)
where m,n are non negative integers ≤ N , ie, 0 ≤ m,n ≤ N & m,n ε Z

S=sum of log_10 (factors of 10^N)

S= (m,n = 0 to N) Σ Σ log(2^m . 5^n)

==>S = (m, =0 to N)ΣΣ (mlog2 + nlog 5) = (m=0 to N)Σ { (N+1)mlog2 + N(N+1)(log 5 )/2 }

==>S= (N+1)x{N(N+1)(log2)/2} + {N(N+1)(log 5)/2} x(N+1)

===>S=N(N+1)^2 {log2 + log 5}/2 ----------->log2 + log 5= log (2*5) =log 10 =1 (base is 10)

===>S= N(N+1)^2 /2 ---------> 2S =N(N+1)^2

here it is given sum ,S= 792

2S=N(N+1)^2 =====>2 x 792 =N(N+1)^2

===>N(N+1)^2 = 1584 as 11³>1584>12³

put N=11 ---> 11x12² =1584

Hence N=11 is the Answer......

First, let us recall that: $10^{n} = (5 \times 2 )^{n} = 5^{n} \times 2^{n}$

Therefore, thanks the tau formula for number of divisors, we know that $10^{n}$ has $( n + 1 ) \times (n+1) = ( n + 1 )^{2}$ total divisors.

Let these divisors be $D_{1} , D_{2}, . . . , D_{ ( n + 1 )^{2}}$ .

By the conditions of the problem we know that: $log(D_{1}) + log(D_{2}) + . . . + log(D_{ ( n + 1 )^{2}}) = 792$

Therefore, thanks to the property of sums of logarithms with equal bases: $log(D_{1}) + log(D_{2}) + . . . + log(D_{ ( n + 1 )^{2}}) = log(D_{1} \times D_{2} \times . . . \times D_{ ( n + 1 )^{2}}) = 729$

Therefore: $D_{1} \times D_{2} \times . . . \times D_{ ( n + 1 )^{2}} = 10^{729}$

So, now we need to find the product of the divisors of $10^{n}$ .

Now, we see that when we divide $10^{n}$ by one of it´s divisors, the result is another one of the divisors of $10^{n}$ . Let´s call this result "important result".

Now, let´s see the case when n is uneven: As I said at the beginning of the problem, $10^{n}$ has $( n + 1 )^{2}$ divisors. Therefore, if n is uneven, $10^{n}$ will have an even amount of divisors. Then, by this and by the important result (see above), the divisors can be paired up two by two, and the product of each of these pairs will be $10^{n}$ . It´s important to note, however, that since n is uneven, it does not have a square root, so all the divisors can be paired up perfectly, since when $10^{n}$ is divided by one of it´s divisors the result will be a divisor different from them (if the result of the division is the same divisor, that would mean that this divisor is the square root of $10^{n}$ ).

Then, since there are $( n + 1 )^{2}$ divisors, there will be $\frac{( n + 1 )^{2}}{2} pairs$ . Then, since the product of each pair is $10^{n}$ , the product of the divisors of $10^{n}$ , when n is uneven, will be $(10^{n})^{\frac{( n + 1 )^{2}}{2}} = 10^{\frac{n \times ( n + 1 )^{2}}{2}}$ .

Now, let´s see the case when n is even: As I said at the beginning of the problem, $10^{n}$ has $( n + 1 )^{2}$ divisors. Therefore, if n is even, $10^{n}$ will have an uneven amount of divisors. However, this means that we can not pair them all up, since there will be one remaining. Now, we see that since n is even, it does have a square root. Therefore, the only divisor that when $10^{n}$ is divided by it returns itself as an answer is $\sqrt{10^{n}} = 10^{\frac{n}{2}}$ . This means that we can take away this divisor, and pair the other ones two by two in such a way that the product of each pair is $10^{n}$ . Since there are $( n + 1 )^{2}$ , there will be $\frac{( n + 1 )^{2} - 1}{2}$ pairs.

This means that the product of the pairs will be $(10^{n})^{\frac{( n + 1 )^{2} - 1}{2}} = 10^{\frac{n \times [ ( n + 1 )^{2} - 1 ]}{2}} = 10^{\frac{n \times ( n + 1 )^{2} - n }{2}}$ .

However, we cant leave behind the divisor we excluded in order to form these pairs: $10^{\frac{n}{2}}$

Therefore, the product of the divisors of $10^{n}$ , when n is even, is: $10^{\frac{n \times ( n + 1 )^{2} - n }{2}} \times 10^{\frac{n}{2}} = 10^{\frac{n \times ( n + 1 )^{2} - n + n }{2}} = 10^{\frac{n \times ( n + 1 )^{2}}{2}}$

However, we see this was the same answer we got when n was uneven!

Therefore, the product of the divisors of $10^{n}$ , independently of the parity of n, will be $10^{\frac{n \times ( n + 1 )^{2}}{2}}$ .

Going back to the beginning of the problem, this newly found result implies that: $10^{\frac{n \times ( n + 1 )^{2}}{2}} = 10^{729}$ $\Rightarrow \frac{n \times ( n + 1 )^{2}}{2} = 729$ $\Rightarrow n \times ( n + 1 )^2 = 1458 = 2^{4} \times 3^{2} \times 11$

Then, playing around with the prime factors of 1458, we see that the only possible solution is $\boxed{n = 11}$

10^N=2^N 5^N,so,no. of divisors of 10^N=(N+1) (N+1)-1 If we pair the divisors in such a way that the base 10 logs of them add up to 10^N each pair,we get such (N+1)^2/2 pairs all of which has an equal value N.This pairing includes a single no. 10^N also. we write, (N+1)^2/2*N=792,solving we get N=11

The sum of the base-10 logarithms of 10^n is log(10^[(n+1)² n/2]) = (n+1)² n/2 (n+1)²*n/2 = 792 -> n=11