The Least Natural number $n$ which has $18$ divisors

Question

My Try:: $18 = 2\times 3 \times 3 = 3^2 \times 2$

Now How can I solve it

Thanks

If $p_i$ are distinct primes, then the number of divisors of $\prod_i p_i^{a_i}$ is equal to $\prod_i(1+a_i)$. So match the $(1+a_i)$:s with some factorization of $18$, and use the $a_i$:s as exponents of smallest possible primes. Caveat: It might happen that $6\times 3$ works better than $3\times3\times2$. You check it out! — Jyrki Lahtonen, Feb 11 '13 at 15:09

score 4 · Answer 1 · edited Feb 11 '13 at 18:48

We know the number of divisors of $\prod_{1\le r\le m}p_i^{m_i}$ is $\prod_{1\le r\le m}(m_i+1)$ where $p_i$s are distinct primes, and power $m$s positive integers

As $18=2\cdot3^2$, $m$ can not be $\ge4$

If $m=1$, $m_1+1=18\implies m_1=17\implies n_\text{min}=2^{17}$

If $m=2$, $(m_1+1)(m_2+1)=18=6\cdot3=2\cdot9$

So, either $m_1=2$, $m_2=5\implies n_\text{min}=2^5 3^2=32\cdot9=288<2^{17}$

or $m_1=1$,$m_2=8 \implies n_\text{min}=2^8 3=768$

If $m=3$, $m_1,m_2.m_3$ can only be $1,2,2$ so in this case, $n_\text{min}=2^2 3^2 5=180<288$

So, the minimum natural number with $18$ divisors is $180.$

score 4 · Accepted Answer · answered Feb 11 '13 at 15:23

The smallest prime power $n=p^k$ with 18 divisors would need $k=17$ and hence be $2^{17}$.

If $n=p^kq^m$ with two primes $p,q$ and $k\ge m\ge1$, then it has $(k+1)(m+1)$ divisors, which allows $k=8,m=1$ or $k=5,m=2$ and suggests $n=2^83$ or $n=2^53^2$. (Of course, it is best to take the smallest possible primes and use the biggest exponent with the smallest prime).

If $n=p^kq^mr^l$ correspondingly with $k\ge m\ge l\ge 1$, then we have $(k+1)(m+1)(l+1)$ divisors, which allows $k=m=2$, $l=1$ only and suggests $n=2^23^25$.

Four or more distinct prime divisors are not possible as that would produce either $2^4=16$ or at least $3\cdot 2^3=24$ divisors, but never $18$.

Choose the smallest number among $2^{17}$, $2^83$, $2^53^2$ and $2^23^25$.

Abhra Abir Kundu · Answer 3 · 2013-02-11T15:33:52.137

$18=2\times 3\times 3$

So the power of the prime factors can be

case $1: 1,8$

case $2: 1,2,2$

case $3: 5,2$

case $4:17$

In the first case the smallest no. is $2^8\times3^1=256\times 3=768$

In the second case the smallest one is $5^12^23^2=36\times 5=180$

In the third case it is $3^22^5=9*32=288$

In the 4th case it is $2^{17}$

So clearly the smallest one is $180$.

Here I have used the fact that if $n=\prod_{i=1}^k p_i^{\alpha_i} $ then the no. of divisors of n are $\prod_{i=1}^k(\alpha_i+1)$

This can easily be verified as follows:

Any divisor of n can be of the form $\prod_{i=1}^k p_i^{\beta_i}$ where each $\beta _i$ can take values in $\{0,1,\dots,\alpha_i\}$ so there are $(\alpha_i+1)$ choices for each $\beta_i$.So the total no of such divisors =$\prod_{i=1}^k(\alpha_i+1)$

You missed 17 as a case, giving $2^{17}=133072$, even worse. — Ross Millikan, Feb 11 '13 at 15:19

score 0 · Answer 4 · answered Feb 11 '13 at 15:08

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This isn't clear. There are three possible questions in my opinion. One is "What is the smallest natural number for which 18 is a divisor?"; the second is "What is the smallest natural number that has 18 divisors?"; and third, "What is the smallest natural number that has 18 distinct divisors?". Can you clarify, please?

answered Feb 11 '13 at 15:08

Trevor

382

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Making a distinction between the 2nd and 3rd question is a bad idea. The distinctness is implied in the 2nd, for otherwise, any natural number could be said to have almost any number of divisors, and there'd be no point to the question at all. – Cameron Buie Feb 11 '13 at 15:31

The Least Natural number $n$ which has $18$ divisors

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