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# number theory and factors

Author Message
Manager
Joined: 01 Jun 2015
Posts: 207
Location: India
GMAT 1: 620 Q48 V26

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26 Mar 2017, 05:40
1
5
00:00

Difficulty:

55% (hard)

Question Stats:

49% (01:18) correct 51% (01:15) wrong based on 81 sessions

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A number N^2 has 15 factors. How many factors can N have?

A. 5 or 7 factors
B. 6 or 8 factors
C. 4 or 6 factors
D. 9 or 8 factors
E. 3 or 5 factors

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GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4015
Re: number theory and factors  [#permalink]

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26 Mar 2017, 06:42
2
Top Contributor
3
techiesam wrote:
A number N² has 15 factors. How many factors can N have?

A. 5 or 7 factors
B. 6 or 8 factors
C. 4 or 6 factors
D. 9 or 8 factors
E. 3 or 5 factors

IMPORTANT RULE: If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = (5)(4)(2) = 40

----NOW ONTO THE QUESTION-------------------
Scanning the answer choices, I can see that there must be TWO ways in which N² can have 15 positive factors.

Since 15 = (3)(5), we can see that, if N² = (some prime^2)(some other prime^4), then ...
...the total number of divisors of N² = (2+1)(4+1) = (3)(5) = 15
If N² = (some prime^2)(some other prime^4), then N = (some prime^1)(some other prime^2)
If N = (some prime^1)(some other prime^2), then the number of positive divisors of N = (1+1)(2+1) = (2)(3) = 6
ELIMINATE A, D and E

Also recognize that if N² = (some prime^14), then ...
...the total number of divisors of N² = (14+1) = 15
If N² = (some prime^14), then N = (some prime^7)
If N = (some prime^7), then the number of positive divisors of N = (7+1) = 8

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Re: number theory and factors  [#permalink]

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21 Oct 2018, 18:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: number theory and factors   [#permalink] 21 Oct 2018, 18:30
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