# a difficult problem

• Jun 23rd 2009, 07:24 AM
jashansinghal
a 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 AM
Swlabr
Quote:

Originally Posted by jashansinghal
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?

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 AM
TheAbstractionist
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 AM
Swlabr
Quote:

Originally Posted by TheAbstractionist
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)

Well, I wasn't sure what "24 factors" meant otherwise. 24 ways of factorising it, perhaps? 5 has one way, 6 has two ways (6 and 2.3) etc, but I thought that a bit far fetched...
• Jun 23rd 2009, 11:26 AM
Moo
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 AM
Plato
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 PM
TheAbstractionist
Quote:

Originally Posted by Plato
Each of these has 24 factors:$\displaystyle 2^3\cdot 3^2\cdot 5=360~\&~2\cdot 3^2\cdot 5^3=2250$.

So they do! Now how in the world did I miss those? (Surprised)
• Jun 23rd 2009, 12:43 PM
Moo
$\displaystyle 2\cdot 3\cdot 5^5>2\cdot 3^2 \cdot 5^3$ and satisfies all the conditions...
• Jun 23rd 2009, 01:09 PM
Plato
Quote:

Originally Posted by Moo
$\displaystyle 2\cdot 3\cdot 5^5>2\cdot 3^2 \cdot 5^3$ and satisfies all the conditions...

I am willing to bet that who ever wrote that question also missed that solution.
Because it almost makes this question pointless: “What is the ratio of the square roots of the highest and the smallest such number possible?”
• Jun 23rd 2009, 01:27 PM
Moo
Quote:

Originally Posted by Plato
I am willing to bet that who ever wrote that question also missed that solution.
Because it almost makes this question pointless: “What is the ratio of the square roots of the highest and the smallest such number possible?”

Completely agree.
Yours is aesthetically more beautiful lol (Rofl)
• Jun 24th 2009, 01:25 AM
jashansinghal
Quote:

Originally Posted by Plato
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.

how can you get the solution without doing hit and trial method
• Jun 24th 2009, 10:44 AM
TheAbstractionist
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 AM
Moo
Quote:

Originally Posted by TheAbstractionist
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 $\displaystyle \sqrt{\frac{2^13^15^5}{2^33^25^1}}\,=\,\frac{25}{2 \sqrt3}.$

Well, basically, you're just rewording and explaining the steps of what has been said above (Rofl)
• Jun 24th 2009, 02:25 PM
TheAbstractionist
Quote:

Originally Posted by Moo
Well, basically, you're just rewording and explaining the steps of what has been said above (Rofl)

Yes, and I should have done it much sooner. (Evilgrin)