Given the number of all positive divisors of an unknown positive integer and the product of those divisors, how can one find the unknown integer?

Sunday, April 22, 2012

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This is an original puzzle from cotpi.

## quasi solved this puzzle:

Theorem.Let \(n\) be a positive integer. Let \(k\) be the number of all positive divisors of \(n\). Let \(m\) be the product of all positive divisors of \(n\). Then \(n = m^{\frac{2}{k}}\).Proof.If \(n\) is not a perfect square, the \(k\) positive divisors of \(n\) can be grouped in \(\frac{k}{2}\) pairs \((a, b)\) with \(a \lt b\) and \(ab = n\). It follows that the product of the factors of n is \(n^{\frac{k}{2}}\).If \(n\) is a perfect square, the \(k - 1\) positive integer factors of \(n\) other than \(\sqrt{n}\) can be grouped in \(\frac{k - 1}{2}\) pairs \((a, b)\) with \(a \lt b\) and \(ab = n\). The product of those factors is \(n^{\frac{k - 1}{2}}\), hence the product of all the factors is \(n^{\frac{k - 1}{2}} \sqrt{n}\), which simplifies to \(n^{\frac{k}{2}}\).

In both cases, \(m = n^\frac{k}{2}\), hence \(n = m^\frac{2}{k}\).

## Gabriel Elpers solved this puzzle:

Suppose there are \(D\) divisors of the natural number \(N\) and that \(P\) is the product of all the divisors. For every divisor \(d\) there is a unique quotient \(q\) such that \(dq = N\). The product of all divisors and their quotients is \(P^2\) since each divisor appears in this expression exactly twice. Rearranging the factors so that each divisor is paired with its quotient, we see that \(P^2 = N^D\). Therefore, \(N = P^{\frac{2}{D}}\).

## Siddhesh Chaubal solved this puzzle:

Consider the case in which the number of divisors of the positive integer \(n\) is even. Let it be \(2k\) where \(k\) is a positive integer. Then the product of all divisors is \(n^k\), since each divisor of \(n\) has corresponding to it another divisor of \(n\) such that their product is \(n\). Since there are \(k\) such pairs, their product is \(n^k\). Thus the \(k\)th root of the product is \(n\).

Now consider the case in which the number of divisors is odd. Let it be \(2k + 1\) where \(k\) is a non-negative integer. Then, the number is a perfect square. The product of all divisors in this case is \(n^k \sqrt{n}\), i.e. \(\sqrt{n}^{2k + 1}\). To find \(n\), take the \((2k + 1)\)th root of the product and square it.

## Gnanasenthil G solved this puzzle:

Theorem.Let \(n\) be a positive integer, let the number of all positive divisors of \(n\) be \(k\), and let the product of all positive divisors of \(n\) be \(m\). Then \(m = n^{\frac{k}{2}}\).Proof.Let \(A\) be the set of all positive divisors of \(n\). When a positive integer is divided by its divisor it gives a quotient which is another divisor of the positive integer.If \(n\) is a perfect square, there is an element \(x \in A\) such that \(x^2 = n\), so this element doesn't pair with a different element in \(A\) such that their product is \(n\). In this case, \(|A|\) is an odd number since for every other element in \(A\), there exists another element in \(A\) such that their product is \(n\). We have \(\sqrt{n}\) and \(\frac{k - 1}{2}\) pairs of numbers in \(A\) such that the product of the numbers in each pair is \(n\). Thus, the product of all elements in \(A\) is

\begin{equation*} m = \sqrt{n} \cdot n^{\frac{k - 1}{2}} = n^{\frac{k}{2}}. \end{equation*}

If \(n\) is not a perfect square, \(|A|\) is even since each element in \(|A|\) has a pair in \(A\) such that their product is \(n\). In this case, we have \(\frac{k}{2}\) pairs of numbers in \(A\) such that the product of the numbers in each pair is \(n\). Thus, the product of all elements in \(A\) is

\begin{equation*} m = n^{\frac{k}{2}}. \end{equation*}

## Nilabja Ray solved this puzzle:

Let \(n\) be the positive integer, \(k\) be the number of all positive divisors of \(n\) and \(m\) be the product of those divisors.

Unless \(n\) is a perfect square, \(k\) is even, i.e. \(k = 2l\) for a positive integer \(l\). There are \(l\) pairs of divisors and the product of the two numbers in each pair is \(n\). Therefore, the product of all \(l\) pairs is

\begin{equation*} m = n^l = n^{\frac{k}{2}}. \end{equation*}

If \(n\) is a perfect square, \(k\) is odd, i.e. \(k = 2l + 1\) for a non-negative integer \(l\). One of the divisors is \(\sqrt{n}\) and there are \(l\) pairs of divisors and the product of the two numbers in each pair is \(n\). Therefore

\begin{equation*} m = n^{l + \frac{1}{2}} = n^{\frac{k}{2}}. \end{equation*}

## Tim Ophelders solved this puzzle:

Let \(D\) be the unknown set of divisors of an unknown positive integer \(A\). The goal is to find \(A\), given \(N = |D|\) and \(P = \prod D\).

Let \(D_\lt\), \(D_\gt\) and \(D_=\) be the sets of numbers in \(D\) smaller, greater and equal to \(\sqrt A\), respectively. Their union is again \(D\).

\[ D_\lt \cup D_\gt \cup D_= = D. \]

Note that there is a bijection between \(D_\lt\) and \(D_\gt\).

\[ d \in D_\lt \Longleftrightarrow \frac Ad \in D_\gt. \]

Let \(m = |D_\lt| = |D_\gt|\). Then \(|D| = |D_=| + |D_\lt| + |D_\gt| = |D_=| + 2m = N\). Note that \(|D_=|\) has the same parity as \(N\), and \(m = \frac{N - |D_=|}{2}\), so

\[ D_= = \begin{cases} \emptyset & \text{ if } N \text{ is even} \\ \{\sqrt{A}\} & \text{ if } N \text{ is odd}. \end{cases} \] \[ \prod D_= = \begin{cases} 1 & \text{ if } N \text{ is even} \\ \sqrt{A} & \text{ if } N \text{ is odd}. \end{cases} \] \[ (\prod D_\gt) \cdot (\prod D_\lt) = \prod_{d \in D_\gt} d \cdot \frac{A}{d} = \prod_{d \in D_\gt} A = A^m \] \[ P = (\prod D_\gt) \cdot (\prod D_\lt) \cdot (\prod D_=) = \begin{cases} A^{\frac{N}{2}} & \text{ if } N \text{ is even} \\ A^{\frac{N - 1}{2}} \sqrt{A} = A^{\frac{N}{2}} & \text{ if } N \text{ is odd}. \end{cases} \] Therefore, \(P = A^{\frac{N}{2}}\), so \(A = \sqrt[N]{P^2}\).