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The integer


has 6436343 divisors. Using only a scientific calculator, find a way to show it has exactly 5 prime divisors.

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up vote 31 down vote accepted

Calling $5685858885855807765856785858569666876865656567858576786786785$ as $n$, we have that $n^{22}$ has $6436343$ divisors. Let $n = p_1^{a_1}p_2^{a_2} \ldots p_k^{a_k}$, where $p_j$ are primes and $a_j \in \mathbb{Z}^+$. Then $$n^{22} = p_1^{22a_1}p_2^{22a_2} \ldots p_k^{22a_k}$$ Hence, the number of divisors is $$(1+22a_1)(1+22a_2)\cdots(1+22a_k) = 6436343 = 23^5$$ Hence, we get that $\require{enclose} \enclose{horizontalstrike}{k=5}$ and $\require{enclose} \enclose{horizontalstrike}{a_1 = a_2 = a_3 = a_4 = a_5 = 1}$. Hence, we get that $$\require{enclose} \enclose{horizontalstrike}{5685858885855807765856785858569666876865656567858576786786785 = p_1 p_2 p_3 p_4 p_5}$$ By adding the digits, we find that $3$ doesn't divide $n$. Hence, $p_1 = 5$.

The actual prime factorization (obtained using Wolframalpha) is $$5 \times 13 \times 647 \times 25414873859387 \times 5319740909859534399143659720463597175586901$$


A good point was raised by Mike in his comments. If we have $$(1+22a_1)(1+22a_2)\cdots(1+22a_k) = 23^5$$ This only implies $k \leq 5$. We also know that $k \geq 2$, since $5 \vert n$. So the answer is incomplete.


Some more updates after some very insightful comments from Mike, Henry and Pambos. Let us assume that we can use divisibility rules to check divisibility up-to $13$ and divide $n$ by numbers up-to $13$. Note that $5 \vert n$ and $13 \vert n$. Hence, $n = 65 \times m$. Let $m = \prod_{j=1}^k p_j^{a_j}$. Hence, we get that $23 \times 23 \times \prod_{j=1}^k (1+22a_j) = 23^5 \implies \prod_{j=1}^k (1+22a_j) = 23^3$. Hence, the number of distinct primes in $m$ can be at most $3$. We want to show that the number of distinct primes dividing $m$ is exactly $3$.

Case $1$: $m$ is a perfect prime power i.e. $m = p^a$ where $p\geq 17$. This means we have $(1+22a) = 23^3 \implies a = \dfrac{23^3-1}{22} = 553$. But the number $m$ is only $59$ digits long. But then $p^{553} \geq 17^{553}$ is more than $500$ digits long. Hence, not possible. This means that the number of distinct primes dividing $m$ is either $2$ or $3$.

Case $2$: $m$ is of the form $m=p_1^{a_1} p_2^{a_2}$. In this case, we have $$(1+22a_1)(1+22a_2) = 23^3$$ Without loss of generality, let us take $a_1 = 1$ and $a_2 = 24$ i.e. $m = p_1 p_2^{24}$.

$$p_2^{8} \equiv 1 \pmod{2^5}$$ $$p_2^{6} \equiv 1 \pmod{3^2}$$ $$p_2^{4} \equiv 1 \pmod{5^1}$$ $$p_2^{6} \equiv 1 \pmod{7^1}$$ $$p_2^{12} \equiv 1 \pmod{13^1}$$ Hence, we get that $$p_2^{24} \equiv 1 \pmod{131040}$$ But we have $m \equiv 74369 \pmod{131040}$. (Note that I have assumed that we have divisibility rules up-to $13$. Divisibility by $2^5$ is trivial since all we need to check is if the last $5$ digits of $m$ is divisible by $2^5$.)

This means $p_1 \equiv 74369 \pmod{131040}$. This gives us $p_2 < p_1$ and hence $p_2^{24} < 10^{54} \implies p_2 < 10^{0.25} 10^{2} < 178$.

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That's extraordinary. I was just typing up a far inferior response. +1 – mixedmath Dec 26 '12 at 5:14
@Ethan I'm not sure "lol" makes much sense in this context. – Alex Becker Dec 26 '12 at 5:18
Ethan uses lol like Jasper uses QED. I think it's a personal catchphrase or something. – anon Dec 26 '12 at 5:19
@Mike: The numbers $5\cdot17^{24}\cdot19^{24}$ and $5\cdot7\cdot11\cdot251^{24}$ have $61$ digits. – P.. Dec 26 '12 at 9:04
@Marvis: Also you can eliminate the case $k=3$ if you look $\pmod {10}$. Easily $5 \mid n$ and $25 \nmid n$. So in case $k=3$ necessarily $n=5\cdot p^{24}\cdot q^{24}$ for some primes $p,q$. The last digit of $\frac{n}{5}$ is $7$ but $1^{24}, \ 3^{24}, \ 7^{24}, \ 9^{24} \equiv 1 \pmod {10}$. – P.. Dec 26 '12 at 18:23

Im not really answering my own question, but its to large to put in a comment, I was thinking of using the fact that $$\lim_{s\to\infty}\frac{\ln(d(k^s))}{\ln(1+s)}=\text{number of prime divisors $k$ has}$$ $d(k)=$ number of divisors of $k$

And then using the fact $s$ is very large, I could say $$\frac{\ln(d(k^{22}))}{\ln(23)}\sim \text{number of prime divisors $k$ has}$$ It gives me $5$ also, but this isn't really a proof, it works with $s=2$, also btw.

Also I was just experimenting it looks like $\dfrac{\ln(d(k))}{\ln(2)}=$ (number of prime diviors of $k$) very often.

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