If the product of all the factors of a positive integer, N

The question is not easy if you do not understand the concept well.
Check this post to fully understand the concept: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/08 ... questions/
Once you go through the 3 posts properly (including the two for which the link is given at the link given above), the solution should make complete sense. Feel free to get back if there are still doubts remaining.

Hello

I understand the step till the total factors (12) calculated for derived N=72.
But can someone explain how again we are narrowing down to 'B' as an answer.
Also is there an example of any other 'N' for which we have more than one solution to clarify the approach discussed on the thread.

Sorry my doubt might be silly here.

Hi...
there is only solution for N, and that is 72, and option B says 1 possible value..
So..thts why B is the answer for Q1.
I guess I could answer your question..

please read my explanation in this thread... it is quite explanatory...

https://www.veritasprep.com/blog/2015/08/finding-the-product-of-factors-on-gmat-questions/
Q) The product of all factors is given as 2^(16) * 3^(14). Will there be any value of N in such a case?
N^(f/2) = 2^16*3^14 = (2^a*3^b)[(a+1)(b+1)/2]
Therefore, a(a+1)(b+1)/2 = 16 & b(a+1)(b+1)/2 = 14
Dividing a/b = 8/7, thus a = 8b/7
b(b+1)(8b/7 + 1) = 14*2
b(b+1)(8b+7) = 2^2*7^2
for b=1, 1*2*15 = 2^1*3^1*5^1
for b=2, 2*3*23 = 2^1*3^1*23^1
for b = 3, 3*4*31 = 3^1*4^1*31^1
Thus as we aren’t able to find any specific value of b for which a/b=8/7, the product of 216*314 will not have a unique value of N.

Hello Karishma/Team, is my solutions for the question you posted on QWQW post correct?

Thanks in advance.

[url]GMAT club : if-the-product-of-all-the-factors-of-a-positive-integer-n-103612.html?fl=similar[/url]
Given: 2^16*3^14 any specific value of N?
sqrt(N)^[(a+1)*(b+1)/2] = 2^16*3^14 = 6^14*2^2 = 6^14*2^(7*2/7).
Thus as 2^16*3^14 can't be expressed as sqrt(N)^[(a+1)*(b+1)/2], it doesn't have a unique value of N.

Given: 2^16*3^13 any specific value of N?
sqrt(N)^[(a+1)*(b+1)/2] =6^13*2^3 = 6^13*2^(13*3/13)
Thus as 2^16*3^13 can't be expressed as sqrt(N)^[(a+1)*(b+1)/2], it doesn't have a unique value of N.

Given: 2^12*3^12 any specific value of N?
sqrt(N)^[(a+1)*(b+1)/2] =6^12 = 6^(24/2) = sqrt(6)^24
Thus the unique value for 2^12*3^12 is 36. However, we don't get unique values of a and b such that we can have 4*3 ~ (3+1)*(2+1) because both powers of 2 and 3 are 12 (i.e. equal!!!). Here is where @Karishma's trial and error method comes useful:
2^12 * 3^12
common factor:2, f/2=2 => f=4; N=2^6*3^6, therefore number of factors =>(6+1)*(6+1) = 49 |= 4 number of factors f
common factor:3, f/2=3 => f=6; N=2^4*3^4, therefore number of factors =>(4+1)*(4+1) = 25 |= 6 number of factors f
common factor:4, f/2=4 => f=8; N=2^3*3^3, therefore number of factors =>(3+1)*(3+1) = 16 |= 8 number of factors f
common factor:6, f/2=6 => f=12; N=2^2*3^2, therefore number of factors =>(2+1)*(2+1) = 9 |= 12 number of factors f
common factor:12, f/2=12 => f=24; N=2^1*3^1, therefore number of factors =>(1+1)*(1+1) = 4 |= 24 number of factors f
Thus as there is no value of "f" for which we can get unique values of power a and b of prime factors 2 and 3, we can't have unique value of factor for 2^12*3^12.

Hence for question like this we need to address 2 conditions:
1) for given 2^a*3^b we should be able to represent the value in terms of sqrt(N)^f where f = (a+1)*(b+1)*...
AND
2) once we are able to do (1) we should be able to HAVE values of a and b for which value of f matches in both cases of sqrt(N)^f and f = (a+1)*(b+1)*...
Unless both these conditions are met we CANNOT have a unique product for a given combination of 2^a*3^b*...

Hello Karishma/Team, continuing with previous questions, are these solutions to the question posted on given url correct?

Thanks in advance.

Use the hit and trial method. It is far easier.

The product of all factors is \((\sqrt{N})^f = 2^{16} * 3^{14}\)

Assume values of f/2 from the common factors of 16 and 14. The only common factor they have is 2.
So can f/2 be 2 i.e. can f be 4? In that case N will be 2^8 * 3^7. The number of factors will be f = (8+1)*(7+1) = 72, not 4.

There is no other possible value for f/2 so no solution.

Dear Karishma,

Can you please provide algebraic solution of this question? I'm stuck at the cubic equation to find values of 'a' and 'b'.

Hi Vyshak
Do you think we need to understand this concept of product of all the factors of positive integer n?
Just a tad too Out of bound for my liking.


Thanks
Stone Cold
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Note that this is not a concept that forms the basics of other concepts and hence it is not critical. That said, if you do get a question based on this, you will find it extremely hard if you do not understand this concept well.

Apologies for bumping an old question but could someone give me an example of a situation where 'N' has more than one value?

Also, another question, why cannot I equate like this below:

\(2^9\) . \(3^9\) = \(n^{f/2}\)

\(6^9\)= \(n^{f/2}\)

9=\(\frac{f}{2}\)

Therefore, f = 18

this is a very interesting point.....can an expert confirm it?

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