If k is a positive integer, is k the square of an integer?

This is what I was confused by - so even though you have multiple prime numbers as in (2^2*3*5*7)^2 this is still considered exactly four different prime numbers? Therefore stat 2 is insuff?

Or put differently if I were to ask someone which scenario has exactly four different prime numbers
1 - 2^2*3*5*7
2 - (2^2*3*5*7)^2
the answer would be that they both have four different prime numbers?

Thank you

Yes, both 2^2*3*5*7 and (2^2*3*5*7)^2 have 4 different primes, namely; 2, 3, 5, and 7. How else?

Consider this: 2, 4, 8, and 16 all have only one distinct prime, which is 2.

Hope it's clear.

Exactly 4 different prime numbers means divisible by only 4 different prime numbers and no other factors. But again this is not possible as if you have 4 different prime factors then you can also have non prime factors. Also 1 is a factor of all the numbers.

That's not correct.

For example: 6 is divisible by exactly two distinct prime factors 2 and 3, but this doesn't mean that 6 divisible by ONLY two factors each of which is a prime.

Thanks Bunuel. Another kudos to you, fine sir.

I guess I was thinking that squaring changes TOTAL factors but not different prime factors.

If \(a\) and \(b\) are positive integers then \(a^b\) will have as many different prime factors as \(a\) itself, exponentiation doesn't "produce" primes.

For example: 6, 6^2, 6^3, ..., 6^100 will all have only two distinct primes: 2 and 3. Though total # of factors will naturally be different: # of factors of 6=2*3 is 4, # of factors of 6^2=2^2*3^2 is (2+1)(2+1)=9, ...

For more on this check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.

Hi Bunnel,

I have a doubt. following are the details

1.As in your previous post you have suggested a perfect square will always have odd number of multiples. Now we have 4 different prime numbers as factor

ex. 2^1 *3^1* 5^1*7^1 so number of factors=( 2*2*2*2) = 16 can not be perfect square we can take diff. powers such as 2,3,4 and verify

we can consider this or not?

Please clarify

Thanks.

Firs of all a prefect square has odd number of factors, not multiples, the number of multiples is not limited for any integer.

Next, I don't quite understand how are you trying to use that in your solution. Anyway, in my post there are two examples given: one is not a prefect square (\(k=2^2*3*5*7\)) and another is (\(k=(2^2*3*5*7)^2\)).

Rephrase Question: Is k a perfect square?

(1)
If k = 4, the answer to the rephrase question is YES.
But if k = 8, the answer is NO.
So, INSUFFICIENT.


(2)
This means that k has four different prime factors, but we don't know how many times those factors appear in the prime factorization of k.
so, for example, if k is 2 x 3 x 5 x 7 (which is divisible by the four primes 2, 3, 5, 7), it's not the square of an integer(perfect square);
if k is 2 x 2 x 3 x 3 x 5 x 5 x 7 x 7 (divisible by the same four primes), it's the square of the integer 2 x 3 x 5 x 7.
so, INSUFFICIENT.

(together)
you need the 4 prime factors (because of statement 2), and you also need to have at least two '2's in the prime factorization (because of statement 1).
the aforementioned perfect square (2 x 2 x 3 x 3 x 5 x 5 x 7 x 7) still works.
to create a number that satisfies the criteria yet isn't a perfect square, just add another copy of one of these prime factors (e.g., 2 x 2 x 3 x 3 x 5 x 5 x 7 x 7 x 7).
INSUFFICIENT Together.

answer = e

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I think option two of this question is ambiguous.

When it says 'exactly' four different prime numbers, it may mean only four prime numbers will be dividers. If you raise the power of any prime, it will become more than 'exactly 4'.

IMO, because of this ambiguity the question got discontinued later on and in newer sets of OG questions we do not see such ambiguity.

I have observed such ambiguity/meaning issues in some other older questions as well. I believe as a batural process of evolution, GMAC too has evolved towards clearer meanings of its questions.

What do you think?

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