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

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

Final Answer:

GMAT assassins aren't born, they're made,
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Re: All factors of a positive integer A are multiplied and the product [#permalink]
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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

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]
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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

We only have five answer choices. Let's use that to our advantage.

A) What's a number with 2 factors? How about 2? 1*2 = 2. Is that 2^3? Nope. Eliminate.

B) What's a number with 3 factors? How about 4? 1*2*4 = 8. Is that 4^3? Nope. Eliminate.

C) What's a number with 5 factors? Maybe you see that 16 does. Or maybe you don't and you skip C.

D) What's a number with 6 factors? How about 12? 1*2*3*4*6*12 = (1*12)*(2*6)*(3*4). Heeyyyyy, that's 12^3!

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