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All factors of a positive integer A are multiplied and the product

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All factors of a positive integer A are multiplied and the product  [#permalink]

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New post Updated on: 03 Oct 2014, 11:43
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All factors of a positive integer A are multiplied and the product obtained is A^3. If A is greater than 1, how many factors does A have?

A. 2
B. 3
C. 5
D. 6
E. 8

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Originally posted by aadikamagic on 03 Oct 2014, 11:28.
Last edited by Bunuel on 03 Oct 2014, 11:43, edited 1 time in total.
Edited the question.
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Re: All factors of a positive integer A are multiplied and the product  [#permalink]

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New post 02 Jan 2015, 00:49
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aadikamagic wrote:
All factors of a positive integer A are multiplied and the product obtained is A^3. If A is greater than 1, how many factors does A have?

A. 2
B. 3
C. 5
D. 6
E. 8


The question can be done in seconds if you understand the properties of factors of a number. Note that every distinct factor of a number which is not a perfect square has a complementary factor.

For example,
Factors of 6: 1, 2, 3, 6
1 has a complementary factor 6 such that 1*6 = 6 (the original number)
2 has a complementary factor 3 such that 2*3 = 6 (the original number)

So if you have a number, n, which is not a perfect square, and start multiplying its factors, you will get n to a certain power. the power will half of the number of factors. For example, 6 has 4 factors and when you multiply all 4 factors, you get 6^2 (2 is half of 4).
So if you know that the product of all factors is A^3, it means there must have been twice the number of factors i.e. 3*2 = 6 factors.

This is explained in detail in this post:
http://www.veritasprep.com/blog/2010/12 ... ly-number/

and this post deals with the special case of perfect squares:
http://www.veritasprep.com/blog/2010/12 ... t-squares/
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Re: All factors of a positive integer A are multiplied and the product  [#permalink]

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New post 01 Jan 2015, 05:40
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The product of factors of a number n is n^(f/2) where f is the number of factors.

A^(f/2) = A^3

f/2 = 3 ==> f = 6
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Re: All factors of a positive integer A are multiplied and the product  [#permalink]

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New post 03 Oct 2014, 22:07
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Answer = D = 6

Number ................ Factors ........................ Product

2 ............................. 1, 2 ................................. 2

3 ............................... 1, 3 ............................... 3

4 ................................ 1, 2, 4 ............................... 8

5 ............................... 1, 5 .................................. 5

6 ............................... 1, 2, 3, 6 ......................... 36

7 ................................ 1, 7 ................................. 7

8 ................................. 1, 2, 4, 8 ........................... 64

9 ............................... 1, 3, 9 ............................ 27

10 ................................ 1, 2, 5 ............................ 10

11.................................... 1, 11 ........................... 11

12 ................................... 1, 2, 3, 4, 6, 12 ................. 12 * 12 * 12

Number which we are looking for = 12

Answer = D
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Re: All factors of a positive integer A are multiplied and the product  [#permalink]

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New post 04 Oct 2014, 23:25
Thanks Paresh for the solution. But this approach is basically a hit and trial. Can anyone suggest a generic approach?
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Re: All factors of a positive integer A are multiplied and the product  [#permalink]

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New post 01 Jan 2015, 18:12
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Hi All,

This question is more about Number Properties and how math "works" than anything else.

Since the question tells us to multiply the factors of a number A and end up with A^3, you have to first thing about what A^3 "means."

A^3 = (A)(A)(A)

So each of these individual "A's" has to be the same number

When you're dealing with factors, this means that each "A" should be a PAIR of factors that are multiplied. Since there are 3 PAIRS of factors, then the total number of factors MUST be 3(2) = 6

The proof can be seen if we use A = 12

The factors of 12 are: (1 and 12), (2 and 6), (3 and 4).

A= 12
A^2 = 12^3
(1x12)(2x6)(3x4) = A^3

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Re: All factors of a positive integer A are multiplied and the product  [#permalink]

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New post 27 Jan 2019, 00:54
aadikamagic wrote:
All factors of a positive integer A are multiplied and the product obtained is A^3. If A is greater than 1, how many factors does A have?

A. 2
B. 3
C. 5
D. 6
E. 8


One way to solve this problem is as follows :

The three pairs of factors can be

A * 1

A/2 * 2

A/3 * 3

Note : Product of each Pair is A and the product of all the factors is \(A^3\)

Each factor should be distinct. The minimum value for A ( which should be a multiple of both 2 and 3 ) will be 12.

There are six distinct factors (12,1), (6,2), and (4,3)

Choice D
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Re: All factors of a positive integer A are multiplied and the product   [#permalink] 27 Jan 2019, 00:54
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