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prime factorization

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prime factorization

by rahul.s » Tue Feb 23, 2010 1:35 am
If K is a positive integer, how many different prime numbers are factors of the expression K^2?

(1) Three different prime numbers are factors of 4K^4
(2) Three different prime numbers are factors of 4K

OA: E
Source: Knewton

i was stunned after reading the statements. i didn't know how to approach them. how do i simplify the statements? what's the approach?
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by ajith » Tue Feb 23, 2010 1:41 am
rahul.s wrote:If K is a positive integer, how many different prime numbers are factors of the expression K^2?

(1) Three different prime numbers are factors of 4K^4
(2) Three different prime numbers are factors of 4K

OA: E
Source: Knewton

i was stunned after reading the statements. i didn't know how to approach them. how do i simplify the statements? what's the approach?
K, K^2, K^4, K^n where n is an integer, all have equal no of prime factors.

1) Three prime factors are there for 4K^4
Insufficient to determine how many prime factors are there for K^4 since 2 may or may not be a prime factor of K^4
thus K^4 and thus K will have either 2 or 3 prime numbers as factors

2) Same is the case, we do not know whether 2 is a factor of K, Insufficient

Combined also insufficient, E
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by Pedros » Tue Feb 23, 2010 3:49 am
ajith wrote:
rahul.s wrote:If K is a positive integer, how many different prime numbers are factors of the expression K^2?

(1) Three different prime numbers are factors of 4K^4
(2) Three different prime numbers are factors of 4K

OA: E
Source: Knewton

i was stunned after reading the statements. i didn't know how to approach them. how do i simplify the statements? what's the approach?
K, K^2, K^4, K^n where n is an integer, all have equal no of prime factors.
2 could be a factor of K as it is for 4 , Good catch.....

1-You mean equal number of distinct prime factors, right ?
2- how hard you guys rate this question?

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by chaya009 » Tue Feb 23, 2010 1:55 pm
Good explanation ajith, thanks.
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by ajith » Wed Feb 24, 2010 4:53 am
Pedros wrote: 2 could be a factor of K as it is for 4 , Good catch.....

1-You mean equal number of distinct prime factors, right ?
2- how hard you guys rate this question?

Thanks
1. Yes I do mean distinct prime factors
2. It is a question which is not very difficult when you know the concept. Yet, it is a bit tricky
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