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

Answer is B

lets say the number is 2^m * 3^n

Lets say

2^0 *3^0 * 2^0 * 3^1....2^0 * 3^n ...A0 >> 2^{0*(N+1)} * 3^{N(N+1)/2}
2^1 *3^0 * 2^1 * 3^1....2^1 * 3^n ...A2 >> 2^{1*(N+1)} * 3^{N(N+1)/2}
.
.
.
2^m *3^0 * 2^m * 3^1....2^m * 3^n ...AM >> 2^{M*(N+1)} * 3^[N(N+1)/2}

So product of all the factors are: A0 * A1 *...*AM = 2^{M(M+1)*(N+1)/2} * 3^{N(N+1)*(M+1)/2}

From the equation:

M(M+1)*(N+1)/2 = 18 --- X
N(N+1)*(M+1)/2 = 12 --- Y

dividing X, Y, we get M/N = 3/2
Lets say M = 3X, N = 2X

substituting the value in X OR Y we got X = 1 is the only integer satisfying

Hence the number is 2 ^ 3 * 3 ^ 2

SAME WAY for Q2:Answer is B

From the equation:

M(M+1)*(N+1)/2 = 9 --- X
N(N+1)*(M+1)/2 = 9 --- Y

dividing X, Y, we get M/N = 1/1
Lets say M = X, N = X

substituting the value in X OR Y we got X = 2 is the only integer satisfying

Hence the number is 2 ^ 2 * 3 ^ 2

: Consider giving me a kudo....

B for both, because prime factorization is unique, isn't it?

In other words, if you're given a prime factorization, it will correspond to one and only one number?

Looks like this question turned out to be way too easy for people of your intellect! I will keep it in mind next time I make a question for GMATClub!

You are absolutely correct regarding your answers and explanations and you have solved using algebra so I am not going to repeat that method. (If someone wants to get the complete algebra based solution from me, drop in a post or pm me)

But I would like to suggest that if the numbers are small and easy to work with, using hit and trial could be a real time saver.
If product is \(2^{18}.3^{12}\), this is equal to \((2^a.3^b)^{f/2}\) (where f is the total number of factors)
If N is \(2^9.3^6\), then f/2 = 2 and f = 4 but total number of factors of N will be much more than 4
If N is \(2^6.3^4\), then f/2 = 3 and f = 6 but total number of factors of N will be much more than 6 (edited)
If N is \(2^3.3^2\), then f/2 = 6 and f = 12. Total number of factors of N is 4.3 = 12. A match.

Similarly for \(2^9.3^9\). For perfect squares, you will have to take f as odd.
If f/2 = 9/2, f = 9 which means \(N = 2^2.3^2\) (A Match)

The reason hit and trial isn't a bad idea is that there will be only one such number (Yes, TehJay, you are right). Look at your equations to convince yourself that at most 1 solution is possible. If I can quickly find it, I am done.
Why should I then bother to find it at all. Shouldn't I just answer with option 'B' in both cases? Think of a case in which the product of all factors is given as \(2^{16}.3^{13}\). Will there be any value of N in such a case?

Responding to a pm:

First check out these 2 posts on factors:

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/12 ... ly-number/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/12 ... t-squares/

Now you know that the factors equidistant from the center multiply to give the number:
Factors of 6:
1, 2, 3, 6

1*6 = 6
2*3 = 6

6 has 4 factors and when we multiply them, we will get \(6*6 = 6^2\)
So to get the product of all the factors of N, all we need is \(N^{f/2}\) where f is the number of factors.

If we know that the product is \(2^{18}*3^{12} = N^{f/2} = (2^a*3^b)^{f/2}\)
N will have two prime factors 2 and 3 and they will have some power which we assume to be a and b.

Now we are using hit and trial to find the values of a and b.
Total number of factors = f

If f/2 = 2 i.e. f = 4, then N is \(2^9.3^6\) but total number of factors of N will be much more than 4. They will be (9+1)*(6+1) so f cannot be 4.
If f/2 = 3 i.e. f = 6, then N is \(2^6.3^4\), but total number of factors of N will be much more than 6. They will be (6+1)*(4+1) so f cannot be 6
If f/2 = 6 i.e. f = 12, then N is \(2^3.3^2\). Total number of factors of N is (3+!)(2+1) = 12. This is the value of f that we assumed.
Hence the number of factors must be 12 and N must be \(2^3.3^2\)

Q) What is the product of the factors of 432?
a) 2^90
b) 3^78
c) 2^40 * 3^30
d) 2^25 * 3^75
e) 2^5 * 3*10 * (2^25 * 3^30 - 1)

what about this ?
need help and more elaborative solution?
@ bunuel ???

This is quite straight forward. Look at the formula I discussed above.

\(432 = 2^4 * 3^3\)
Number of factors = 5*4 = 20

Product of all the factors \(= (2^4 * 3^3)^{20/2} = 2^{40}*3^{30}\)

I believe the same question can be asked for sum of the factors as well .Just like the case of multiplication, the case of addition will have the answer as B.

Consider the number 8 whose factors are (1,2,4,8)

Product of factors = 1* 2*4*8 =64

Sum of the factors = 1+2 +4+8 = 15

Both the above results can we have only 1 unique way in which they can be written. (Think about it!!!)

The trick in such kind of questions is to do trial with smaller numbers like 4,8, 9 etc so that we are able to generalize.

Using the above explanation the question can also be thought as a DS question.
Reason: We needed only to generalize and not really deal with the actual number presented in the question.

Hope this helps....

This is not correct. A given sum may not be unique to a single factor.
The factors of 10 are 1, 2, 5, 10 adding up to 18.
The factors of 17 are 1, 17 adding up to 18.

There could be many other such cases. To generalize something, taking a few examples is not enough. After observing the pattern, you have to figure out the logic to generalize the concept. No logic - no generalization.

There is a method to find the factors of a number as well its products.
Let us take an example for better understanding.
e.g. let us find the factors of 18

Step I:18= 2*3*3 (prime factorization)
18= (2^1) * (3^2)

Step II: Increment the powers of each prime factor by 1. i.e. power of 2 is 1, increment it by 1. Also, power of 3 is 2, increase it by 1.

Step III: Number of factors of 18 = (1+1)*(2+1) = 2 * 3 = 6

now let us try to find out the product of these factors

Step IV: product of factors of 18 = 18^ (number of factors/2)
= 18 ^ (6/2)
= 18 ^ 3
= 5832

REASON:
The factors of 18 are 1, 2, 3, 6, 9, 18.
Product of factors will be = 1*2*3*6*9*18
(we can rewrite this as..)
Product of factors = (1*18)*(2*9)*(3*6) = 18*18*18 = 18^3

So....
if we want to write a generic formula, then, it can be written as
" If a positive integer 'n' has 'x' factors,
then the product of all the factors = n ^ (x/2)"

SOLUTION FOR THE PROBLEM No.1
2^18* 3^12 = Product of factors of N
(2^6)*(2^12)*(3^12) = prod of factors of N
(2^12/2)*(6^12) = (6* 2^1/2)^12 = ([72]^1/2)^12 = 72 ^ (12/2)

Hence number of factors = 12
= (a+1)*(b+1)
= (2+1) * (3+1)

Hence there can be only one possible N, i.e. 72
Ans: B

SOLUTION FOR THE PROBLEM No.2
(2^9)* ( 3^9) = Product of factors of N
= 6^9
= {(36)^1/2} ^9 (this sqr root sign does not work... )
= (36)^ 9/2
hence N= 36

9 = (a+1)*(b+1)
= (2+1) * (2+1)
a= 2, b=2
Hence only possible answer, i.e.1
Ans: B

Looking at Q1, why does not a=5 b=1 satisfy it?

Also can someone show me an example of 2 digits that can take on more then 1 value of N? I find this topic very confusing, especially for a basic level question.

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.

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.

I have a simple doubt. Does there exist two different positive numbers which has the same product of all factors?
I was not able to find any.

Posted from my mobile device

­No. Two different positive integers will have different product of their positive factors. 

But then, by default the answer of this question should be 1 right?­