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

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
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If the product of all the factors of a positive integer, N  [#permalink]

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Updated on: 11 Aug 2012, 00:48
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This question (which is in two parts) tests your fundamentals of a very interesting topic - Factors

Q.i. If the product of all the factors of a positive integer, N, is $$2^{18}*3^{12}$$, how many values can N take?
(A) None
(B) 1
(C) 2
(D) 3
(E) 4

Q.ii. If the product of all the factors of a positive integer, N, is $$2^{9}*3^{9}$$, how many values can N take?
(A) None
(B) 1
(C) 2
(D) 3
(E) 4
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Karishma
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Originally posted by VeritasKarishma on 25 Oct 2010, 08:24.
Last edited by Bunuel on 11 Aug 2012, 00:48, edited 1 time in total.
Edited the question.
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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25 Oct 2010, 12:12
24
11
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
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##### General Discussion
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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25 Oct 2010, 09:27
2

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....
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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25 Oct 2010, 10:16
2
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?
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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25 Oct 2010, 10:33
krushna wrote:

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....

Not sure but there has to be a shorter and easier way !

Isn't the product of the all the factors of the positive integer N, the positive integer N itself ? And N can only be N
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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25 Oct 2010, 11:04
No. The product of all the factors of N is NOT EQUAL to N.
lets say N = 4.
The factors are 1,2,4. Hence the product will be 8.

It is easy to find out the answer. But to get the number, I did all those calculations. Hope someone can come up with a shorter calculation
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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25 Oct 2010, 11:51
krushna wrote:
No. The product of all the factors of N is NOT EQUAL to N.
lets say N = 4.
The factors are 1,2,4. Hence the product will be 8.

It is easy to find out the answer. But to get the number, I did all those calculations. Hope someone can come up with a shorter calculation

My bad, I though it was prime factors.

However, I still think that it can only be ONE value of N. Based on your example, N=4 ....... prime factors 1, 2, 2 ........ number of factors = 3 ......... which are 1,2 , 4 ......... the product is 8. Which integer other than 4 can generate factors that when multiplied will give out 8 ?
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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25 Oct 2010, 13:24
3
VeritasPrepKarishma wrote:
This question (which is in two parts) tests your fundamentals of a very interesting topic - Factors

Q.i. If the product of all the factors of a positive integer, N, is $$2^{18}.3^{12}$$, how many values can N take?
(A) None
(B) 1
(C) 2
(D) 3
(E) 4

The product of factors of a number n is $$n^{\frac{f}{2}}$$ where f is the number of factors. This is proved here. Let the number be $$n=2^a3^b$$, then $$f=(a+1)(b+1)$$
$$2^{18}3^{12}=(2^a3^b)^{(a+1)(b+1)/2}$$
Hence, $$a(a+1)(b+1)=36$$ & $$b(b+1)(a+1)=24$$
Dividing, $$a/b=3/2$$ or $$2a=3b$$
$$b(b+1)(3b+2)=48$$
b is a positive integer, so its easy to check for all positive solutions.
b=1, doesnt work
b=2, works
b>=3, doesnt work
b=2 means a=3. Hence only possible n=2^3 * 3^2 = 72

VeritasPrepKarishma wrote:
Q.ii. If the product of all the factors of a positive integer, N, is $$2^{9}.3^{9}$$, how many values can N take?
(A) None
(B) 1
(C) 2
(D) 3
(E) 4

Same method yields :
$$a(a+1)(b+1)=18$$ & $$b(b+1)(a+1)=18$$
This time dividing tells us a=b
$$b(b+1)^2=18$$
b=1, doesnt work
b=2, works
b>=3, doesnt work
So the only solution is 2^2*3^2 or 36
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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Updated on: 13 Aug 2012, 22:46
1
2
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?
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Originally posted by VeritasKarishma on 25 Oct 2010, 18:09.
Last edited by VeritasKarishma on 13 Aug 2012, 22:46, edited 1 time in total.
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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11 Aug 2012, 00:45
If N is 2^6.3^4, then f/2 = 4 and f = 8 but total number of factors of N will be much more than 8

Hi karishma,
Is there a typo here cos f/2=4 will not yield 2^6.3^4

Pls let me know
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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13 Aug 2012, 22:46
shankar245 wrote:
If N is 2^6.3^4, then f/2 = 4 and f = 8 but total number of factors of N will be much more than 8

Hi karishma,
Is there a typo here cos f/2=4 will not yield 2^6.3^4

Pls let me know

Yes, it should be: if $$N = 2^6*3^4$$, f/2 = 3 and f = 6
You cannot take 4 common out of 18 and 12 so f/2 will not be 4.
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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18 Oct 2012, 21:05
2
Responding to a pm:

First check out these 2 posts on factors:

http://www.veritasprep.com/blog/2010/12 ... ly-number/
http://www.veritasprep.com/blog/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 $$2^3.3^2$$
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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16 May 2013, 20:14
Product of factors of N is given by N^(no. of factors of N/2)
Case A:
2^18 * 3^12
take HCF of powers will get 6
we can write above equation as (2^3*3^2)^6. ie, no. of factors = 6*2=12
so N= 2^3 * 3^2 (only one way to form a number N, since it is a multiple of prime factors)

Case B:
2^9*3^9
take HCF will get 9
so we can write above equation as (2*3)^9. Since equation is perfect square, no. of factors be odd number. therefore N=2^2*3^2 (only one way to form a number N)
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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05 Oct 2013, 00:03
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

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)

need help and more elaborative solution?
@ bunuel ???
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Re: Another Question - 700 Level, Number Properties  [#permalink]

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05 Oct 2013, 05:20
1
sunny3011 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

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)

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 $$= (2^4 * 3^3)^{20/2} = 2^{40}*3^{30}$$
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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08 Oct 2013, 08:51
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....
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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08 Oct 2013, 21:04
gayatri200 wrote:
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.
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If the product of all the factors of a positive integer, N  [#permalink]

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14 Sep 2014, 23:46
3
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
Ans: B
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Re: If the product of all the factors of a positive integer, N  [#permalink]

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03 Feb 2015, 09:01
1
"Product of factors of N is given by N^(no. of factors of N/2)
Case A:
2^18 * 3^12
take HCF of powers will get 6
we can write above equation as (2^3*3^2)^6. ie, no. of factors = 6*2=12
so N= 2^3 * 3^2 (only one way to form a number N, since it is a multiple of prime factors)

Case B:
2^9*3^9
take HCF will get 9
so we can write above equation as (2*3)^9. Since equation is perfect square, no. of factors be odd number. therefore N=2^2*3^2 (only one way to form a number N)"

VeritasPrepKarishma is this correct?The HCF part?This takes a lot less time to do but I want to take expert opinion before coming to any conclusion.

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

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03 Feb 2015, 22:01
Ralphcuisak wrote:
"Product of factors of N is given by N^(no. of factors of N/2)
Case A:
2^18 * 3^12
take HCF of powers will get 6
we can write above equation as (2^3*3^2)^6. ie, no. of factors = 6*2=12
so N= 2^3 * 3^2 (only one way to form a number N, since it is a multiple of prime factors)

Case B:
2^9*3^9
take HCF will get 9
so we can write above equation as (2*3)^9. Since equation is perfect square, no. of factors be odd number. therefore N=2^2*3^2 (only one way to form a number N)"

VeritasPrepKarishma is this correct?The HCF part?This takes a lot less time to do but I want to take expert opinion before coming to any conclusion.

I am not sure how HCF has relevance here except that it gives you the highest value of f/2.

Try $$N = 2^{12} * 3^{12}$$ using your method.
_________________
Karishma
Veritas Prep GMAT Instructor

Re: If the product of all the factors of a positive integer, N   [#permalink] 03 Feb 2015, 22:01

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