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Find the smallest positive multiple $n$ of 1994 such that the produc of the divisors of $n$ is $n^{1994}$

Attempt

Let $n\in \mathbb{N}$ and denote by $d(n)$ the number of divisors of $n$ and $p(n)$ the product of all the divisors of $n$ including $n$.

Consider the factorization of $n$ given by $n=p_1^{\alpha_1} p_2^{\alpha_2}\cdots p_n^{\alpha_n}$ for $\alpha_i\in \mathbb{N}$. Is known the fact that $p(n)=n^{d(n)/2}=n^{1994}$ it is $d(n)=997 \cdot 2^2$ in addition $n=997 \cdot 2 \cdot k\, $ for some $k\in \mathbb{N}$.

since $n$ is the smallest multiple of $1994$ with $997 \cdot 2^2$ divisors then $n$ should be $2^{998} \cdot 997$

Any comment or corretion was helpful

  • No, but Thank you I appreciate your help –  Jan 12 '21 at 23:25
  • This question appeared in the literature about twenty seven years ago. A new, much more interesting and exciting version appeared the next year, featuring the constant 1995. – mjw Jan 12 '21 at 23:28

2 Answers2

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As you correctly identify, since the product of the divisors of $n$ is $n^{1994}$, $n$ must have $1994$ factor pairs, i.e. $3988$ divisors. We also know that $n$ must be divisible by prime $997$, as must the number of divisors. So to get this we need a factor of $p^{\large{\color{red}{996}}}$ which you narrowly miss, and obviously this should use $p=2$ to get as small as possible.

However you have also missed the second factor $2$ in the number of divisors, which we can arrange with a multiple of $3$ to get the smallest $n$. So $n = 2^{996}\cdot 3 \cdot 997$, which has $(996+1)(1+1)(1+1) = 3988$ divisors as required, adding one to each prime exponent.

Joffan
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Your idea is really good, but at the end you want the smallest $k$ such that $d(997\cdot 2\cdot k)=997\cdot 2^{2}$. You said that $k=2^{997}$ does the job, but this doen't, because $d(997\cdot 2^{998})=(1+1)(998+1)\neq 997\cdot 2^{2}$.

To fix this, note that if you write $k=2^{m}\cdot p$, where $p$ doen't have factors of $2$, then $$d(\color{blue}{997}\cdot \color{red}{2\cdot 2^{m}}\cdot p)=\color{blue}{(1+1)}\cdot\color{red}{(m+2)}\cdot d(p)$$ so you want $(m+2)\cdot d(p)=2\cdot 997$. Here are two possibilities: $m+2=2$ of $m+2=997$.

  • If $m+2=2$ then $m=0$, therefore $d(p)=997$, and the lowest $p$ possible in this case is $p=3^{996}$.
  • If $m+2=997$ then $m=995$, therefore $d(p)=2$, and the lowest $p$ possible in this case is $p=3$.

So the candidates are $997\cdot 2\cdot 3^{996}$ and $997\cdot 2^{996}\cdot 3$, but clearly the last is smaller, so $n=997\cdot 2^{996}\cdot 3$.

Nah
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  • $n$ should be a multiple of $1994$ – J. W. Tanner Jan 12 '21 at 23:28
  • whoops, I meant $n=997\cdot 2\cdot 2^{498}\cdot 3$. – Nah Jan 12 '21 at 23:33
  • it seems to me that the factors of your $n$ divisible by $3$ are $3, 3\cdot2, 3\cdot2^2, 3\cdot2^3,...,3\cdot2^{499}$ and $3\cdot997, 3\cdot997\cdot2, 3\cdot997\cdot2^2, 3\cdot997\cdot2^3,...,3\cdot997\cdot2^{499}$, so the product of the factors has $1000$ factors of $3$, where as your $n^{1994}$ has $1994$ factors of $3$ – J. W. Tanner Jan 12 '21 at 23:50
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    I had misread the question, I just edited the answer, thanks. – Nah Jan 12 '21 at 23:58