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

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Re: If the product of all the factors of a positive integer, N  [#permalink]

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26 Oct 2015, 11:26
nravi549 wrote:
If a number, N, can be expressed as: $$2^a$$ * $$3^b$$ * $$5^c$$...
Then, the product of all factors of N is:
$$\sqrt{N}^(^1^+^a^)^*^(^1^+^b^)^*^(^1^+^c^) ^.^.^.^.$$

For example: 36 = $$2^2$$ * $$3^2$$
Then the product of all factors of 36 is equal to: $$\sqrt{36}^(^1^+^2^)^*^(^1^+^2^)$$
==> $$6^3*^3$$
==> $$6^9$$
==> $$2^9$$*$$3^9$$

Now, lets come back to (Q.i):
Given, Product = $$2^1^8$$*$$3^1^2$$
==> $$2^6$$*$$2^1^2$$*$$3^1^2$$
==> $$2^6$$*$$6^1^2$$
==> ($$\sqrt{2}^1^2$$) * ($$6^1^2$$)
==> ($$(6\sqrt{2})^1^2$$)
==> ($$\sqrt{72}^1^2$$) = $$\sqrt{N}^(^1^+^a^)^*^(^1^+^b^)^*^(^1^+^c^) ^.^.^.^.$$
==> N = 72, (1+a)*(1+b) = 12
==> The only valid solution for N=72 is: a = 3, b = 2
==> Ans: B

(Q.ii):
Given, Product = $$2^9$$*$$3^9$$
==> Product = $$6^9$$
==> Product = $$(\sqrt{36})^9$$ = $$\sqrt{N}^(^1^+^a^)^*^(^1^+^b^)^*^(^1^+^c^) ^.^.^.^.$$
==> N = 36, a = 2, b = 2
==> 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.
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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26 Oct 2015, 20:51
1
GMATDemiGod wrote:
nravi549 wrote:
If a number, N, can be expressed as: $$2^a$$ * $$3^b$$ * $$5^c$$...
Then, the product of all factors of N is:
$$\sqrt{N}^(^1^+^a^)^*^(^1^+^b^)^*^(^1^+^c^) ^.^.^.^.$$

For example: 36 = $$2^2$$ * $$3^2$$
Then the product of all factors of 36 is equal to: $$\sqrt{36}^(^1^+^2^)^*^(^1^+^2^)$$
==> $$6^3*^3$$
==> $$6^9$$
==> $$2^9$$*$$3^9$$

Now, lets come back to (Q.i):
Given, Product = $$2^1^8$$*$$3^1^2$$
==> $$2^6$$*$$2^1^2$$*$$3^1^2$$
==> $$2^6$$*$$6^1^2$$
==> ($$\sqrt{2}^1^2$$) * ($$6^1^2$$)
==> ($$(6\sqrt{2})^1^2$$)
==> ($$\sqrt{72}^1^2$$) = $$\sqrt{N}^(^1^+^a^)^*^(^1^+^b^)^*^(^1^+^c^) ^.^.^.^.$$
==> N = 72, (1+a)*(1+b) = 12
==> The only valid solution for N=72 is: a = 3, b = 2
==> Ans: B

(Q.ii):
Given, Product = $$2^9$$*$$3^9$$
==> Product = $$6^9$$
==> Product = $$(\sqrt{36})^9$$ = $$\sqrt{N}^(^1^+^a^)^*^(^1^+^b^)^*^(^1^+^c^) ^.^.^.^.$$
==> N = 36, a = 2, b = 2
==> 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: http://www.veritasprep.com/blog/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.
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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08 Dec 2015, 03:06
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.
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Posts: 28
Re: If the product of all the factors of a positive integer, N  [#permalink]

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09 Dec 2015, 02:22
seemachandran wrote:
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...
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If the product of all the factors of a positive integer, N  [#permalink]

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Updated on: 05 Sep 2016, 07:09
http://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.

Originally posted by manishtank1988 on 29 Aug 2016, 19:37.
Last edited by manishtank1988 on 05 Sep 2016, 07:09, edited 1 time in total.
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If the product of all the factors of a positive integer, N  [#permalink]

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Updated on: 05 Sep 2016, 07:24
[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.

Originally posted by manishtank1988 on 29 Aug 2016, 19:38.
Last edited by manishtank1988 on 05 Sep 2016, 07:24, edited 3 times in total.
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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29 Aug 2016, 22:05
1
manishtank1988 wrote:
http://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) = 216*314 = (2a*3b)[(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.

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.
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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30 Aug 2016, 01:34
VeritasPrepKarishma wrote:
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?

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'.
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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03 Dec 2016, 20:39
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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Re: If the product of all the factors of a positive integer, N  [#permalink]

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04 Dec 2016, 10:12
stonecold wrote:
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

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.
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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01 Jun 2017, 11:37
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
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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19 Aug 2018, 02:14
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Re: If the product of all the factors of a positive integer, N &nbs [#permalink] 19 Aug 2018, 02:14

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

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