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The product of the proper positive integer factors of $n$ can be written as $n^{(ax+b)/c}$, where $x$ is the number of positive divisors $n$ has, $c$ is a positive integer, and the greatest common factor of the three integers $a$, $b$, and $c$ is 1. What is $a+b+c$?

I know that this works with the divisors of 24 24 has 8 divisors So x=8, a=1, b=10, c=6 $24^3=13824=24^{(8*1+10)/6}$

$\gcd(1,6,10)$

$a+b+c=1+10+6=17$

But it also works with 60 which with 60, $a+b+c=19$ I'm not sure what would be my final answer could someone help me? (also it seems like this could have infinite solutions)

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    $\begingroup$ Where the heck do you get this problem! I can tell you that:If $n = \prod p_i^{k_i}$ is the prime factorization of $n$ then $x = \prod (k_i + 1)-2$ is the number of proper factors. And $\frac {\prod p_i^{\sum_{j=1}^{k_i} j}}n$ is the product of the proper divisors. BUT to convert that and solve for $\frac {\prod p_i^{\sum_{j=1}^{k_i} j}}n=n^{\frac{ax + b}c}$ for some integers $a,b,c$ just seems like a pointless and directionless monster. $\endgroup$
    – fleablood
    Feb 14, 2019 at 17:43
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    $\begingroup$ So if you can figure $n = \prod p_i^{k_i}$ and $x = \prod(k_i + 1) -2$ and $n^{(ax+b)/c} =\prod p_i^{k_i(ax+b)/c} \prod p_i^{\frac{k_i(k_i -1)}2}$ you are done. Good luck $\endgroup$
    – fleablood
    Feb 14, 2019 at 17:53
  • $\begingroup$ @RossMillikan: Intuitively I would expect neither $1$ nor the number itself to count as a "proper" factor. This matches the definition in Mathworld, which contrasts it to "proper divisor". $\endgroup$ Feb 14, 2019 at 19:01
  • $\begingroup$ @HenningMakholm: I had never seen that distinction. I will update my answer. It won't change much. $\endgroup$ Feb 14, 2019 at 19:12
  • $\begingroup$ Is this from Brilliant.org? It's the type of problem you would find there. If so, consider including a link. $\endgroup$ Feb 14, 2019 at 20:03

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You just showed one case where $n=24$ where it will naturally have infinite solutions. Look for a solution in all cases. Spoiler below.

$a=1,b=-2,c=2$ solves all the cases. So $a+b+c=1$ is the solution.

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    $\begingroup$ That's indeed the correct answer, once the question is interpreted as: find $a,b,c$ such that this holds for all $n$. However, you should explain this a little bit, especially the less obvious case where $n$ is a square: for nonsquare $n$, each divisor $d$ can be paired with $n/d$, whereas for square $n$, the divisor $\sqrt{n}$ is alone and accounts for the additional factor $\sqrt{n}$ in the product of divisors, hence the missing $1/2$ in the exponent. $\endgroup$ Jan 11, 2021 at 15:58
  • $\begingroup$ @Jean-ClaudeArbaut Not quite sure why you say "$\sqrt{n}$ accounts for the missing 1/2 in the exponent". $\quad$ Instead, I argue that it's "$n$ that accounts for it, regardless of whether $n$ is a square": The square of the product of all positive factors by your pairing is clearly $ n^x$, so the product of all positive factors is $n^{x/2}$ and so the product of all proper positive factors is $ n^{x/2 -1}$. $\endgroup$
    – Calvin Lin
    Jan 8, 2022 at 11:55
  • $\begingroup$ @CalvinLin Here is what I meant. The pairing does not work for square $n$: $x$ is even, except when $n$ is square (since $x=\prod (k_i+1)$ where the $k_i$ are the exponents of the prime divisors of $n$, $x$ is odd iff all $k_i$ are even). Therefore, when $n$ is not square, factors are paired together and their product (including $1\times n$) is $n^{x/2}$. When $n$ is square $x-1$ factors can be paired, and there remains $\sqrt n$ alone, so you get $n^{(x-1)/2}\times\sqrt{n}$, again including $1\times n$. All in all, the answer is always $n^{x/2-1}$. $\endgroup$ Jan 8, 2022 at 12:50
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As long as $n$ is neither prime nor a perfect square, the product of its proper factors is of the form $n^y$ for some positive integer $y$. (In fact $y=x/2-1$). Then we need to solve $$ \frac{ax+b}{c} = y $$ for coprime $a,b,c$. But that is always possible: You can choose any prime $p > x$ and then set $$ a=1, \quad b = yp - x, \quad c = p $$ Then $a,b,c$ will even be pairwise coprime.

When $x=2y+2$, we get $$ a+b+c % = 1 + yp - 2y - 2 + p = (1 - 2y - 2) + (y+1)p $$ which means that there is an infinity of solutions for each (non-square, non-prime) $n$.

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As you say $24$ has $8$ positive divisors, so you need $\frac {8a+b}c=3$, which you can satisfy with $a=1,b=10,c=6$ or with $a=1,b=4,c=4$ and many other combinations

$60$ has $12$ positive divisors, so you need $\frac {12a+b}c=5$, which you can satisfy with $a=1,b=3,c=3$ or with $a=2,b=1,c=5$ and many other combinations.

I would just display these four solutions and punch the setter in the nose complain to the setter for making a poor problem.

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  • $\begingroup$ But the "proper factors" were the ones that were multiplied together on the left-hand side of the equation -- so on further thought it doesn't even matter whether we consider $1$ to be among them. The $x$ on the right-hand side counts "positive divisors" which includes both $1$ and $n$. (Yeah, this is a horribly inconsistent, confusing, patently unmotivated problem). $\endgroup$ Feb 14, 2019 at 19:21
  • $\begingroup$ @HenningMakholm: I see you are right. I'll fix again. $\endgroup$ Feb 14, 2019 at 19:27
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    $\begingroup$ The recent answer has the correct interpretation, I think, hence the unique solution: with $p^k$ for prime $p$ it's easy to get $b=-2a$ and $c=-b$, then by factoring out the common factor $a$, it's done. And it's then easy to prove it's valid for all $n$. $\endgroup$ Jan 11, 2021 at 16:15

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