A number has 24 factors.The multiplication of its prime factors is 30(taking 1 at a time).What is the ratio of the square roots of the highest and the smallest such number possible?

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- Jun 23rd 2009, 07:24 AMjashansinghala difficult problem
A number has 24 factors.The multiplication of its prime factors is 30(taking 1 at a time).What is the ratio of the square roots of the highest and the smallest such number possible?

- Jun 23rd 2009, 08:23 AMSwlabr

The hard part is the first sentence. You know the primes that divide such a number (as these are the primes that multiply to get 30), and the powers of the primes sum to 24 and are all non-zero (as otherwise the product would not be 24). So, what is the largest number possible? What is the smallest?

Big hint: let $\displaystyle x=p_1^{a_1}p_2^{a_2} \ldots p_n^{a_n}$ be such a number, with $\displaystyle p_1<p_2 < \ldots < p_n$. Then the largest is the one with $\displaystyle a_1=a_2= \ldots = a_{n-1} = 1$ and $\displaystyle a_n = 24-(n-1)$.

(Unless I'm missing something here?) - Jun 23rd 2009, 08:56 AMTheAbstractionist
I took

**jashansinghal**’s first line to mean 24 factors, not 24 prime factors. But then I’m unable to find any integer with exactly 24 factors. Maybe I’m missing something as well. (Worried) - Jun 23rd 2009, 09:01 AMSwlabr
- Jun 23rd 2009, 11:26 AMMoo
Hello,

For example, 6 has 4 factors : 1,2,3,6

If I'm not mistaking (hasn't seen this for a while) : the number of factors of a number $\displaystyle n=p_1^{\alpha_1}\cdot p_2^{\alpha_2} \cdots p_m^{\alpha_m}$ is $\displaystyle \prod_{k=1}^m \{\alpha_k+1\}$

So here, we have $\displaystyle \prod_{k=1}^m \{\alpha_k+1\}=24$

And the product of its prime factors is 30 : $\displaystyle \prod_{k=1}^m p_k=30$

Then, we have to be able to find the smallest and greatest possible numbers responding to these conditions... So I'm pretty sure we have to study the prime decomposition of 30 :D

But now... I have to take a shower (Rofl) - Jun 23rd 2009, 11:59 AMPlato
Each of these has 24 factors:$\displaystyle 2^3\cdot 3^2\cdot 5=360~\&~2\cdot 3^2\cdot 5^3=2250$.

If I understand the bit about the product being 30 then they are smallest and largest. - Jun 23rd 2009, 12:05 PMTheAbstractionist
- Jun 23rd 2009, 12:43 PMMoo
$\displaystyle 2\cdot 3\cdot 5^5>2\cdot 3^2 \cdot 5^3$ and satisfies all the conditions...

- Jun 23rd 2009, 01:09 PMPlato
- Jun 23rd 2009, 01:27 PMMoo
- Jun 24th 2009, 01:25 AMjashansinghal
- Jun 24th 2009, 10:44 AMTheAbstractionist
All right guys, I just remembered this: the tau function $\displaystyle \tau(n)$ which gives the number of factors of each integer $\displaystyle n > 1.$ $\displaystyle \tau$ has the following property: if $\displaystyle n=p_1^{k_1}\cdots p_r^{k_r}$ where the $\displaystyle p_i$ are distinct primes, then

$\displaystyle \tau(n)\ =\ \prod_{i\,=\,1}^r(1+k_i)$

So for this particular problem, the distinct primes are 2, 3, and 5 and the number is of the form $\displaystyle 2^a3^b5^c$ where $\displaystyle (1+a)(1+b)(1+c)=24.$ The possibilities are thus

(i) $\displaystyle a=1,\,b=1,\,c=5$

(ii) $\displaystyle a=1,\,b=5,\,c=1$

(iii) $\displaystyle a=5,\,b=1,\,c=1$

(iv) $\displaystyle a=1,\,b=2,\,c=3$

(v) $\displaystyle a=1,\,b=3,\,c=2$

(vi) $\displaystyle a=2,\,b=1,\,c=3$

(vii) $\displaystyle a=2,\,b=3,\,c=1$

(viii) $\displaystyle a=3,\,b=1,\,c=2$

(ix) $\displaystyle a=3,\,b=2,\,c=1$

The largest number is therefore $\displaystyle 2^13^15^5$ and the smallest is $\displaystyle 2^33^25^1.$ Hence the answer to the question is $\displaystyle \sqrt{\frac{2^13^15^5}{2^33^25^1}}\,=\,\frac{25}{2 \sqrt3}.$ - Jun 24th 2009, 11:41 AMMoo
- Jun 24th 2009, 02:25 PMTheAbstractionist