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tarek99
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If k is a positive integer, is k the square of an integer? Updated on: 25 Dec 2021, 09:11
Originally posted by tarek99 on 21 Nov 2007, 16:23.
Last edited by Bunuel on 25 Dec 2021, 09:11, edited 2 times in total.
Edited the question and added the OA
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If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4.

(2) k is divisible by exactly four different prime numbers.
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If k is a positive integer, is k the square of an integer? 01 Mar 2012, 01:11
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If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4. If \(k=4\) answer is YES, but if \(k=8\) answers is NO. Not sufficient.

(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.

(1)+(2) Again if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.
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If k is a positive integer, is k the square of an integer? 07 Dec 2009, 13:57
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if K is a positve integer, is K the square of an integer?
1-K is divisible by 4
2-K is divisible by exactly four different prime numbers
Show more

Question Stem : K is a positive integer. Is K the square of an integer?
We are looking for a Yes/No answer.

St.(1) : K is divisible by 4

K can be 4 which answers the question stem as Yes.

K can be 8 which answers the question stem as No.

Contradicting solutions means this statement is insufficient.

St.(2) : K is divisible by exactly 4 different prime numbers.

K can be (2 x 3 x 5 x 7) = 210. This answers the question stem as No.

K can be (2 x 3 x 5 x 7)^2 = (210)^2. This answers the question stem as Yes.

Note : St.(2) just puts a restriction on the number of prime factors that are there. Not on how many times those prime factors can occur.

Since this statement too provides contradicting solutions, it is insufficient.

St.(1) and (2) together : K must have four different prime factors and it must be divisible by 4.

Thus we need four prime factors and at least two 2's.

K can be (2 x 2 x 3 x 5 x 7) = 420. This answers the question stem as No.

K can be (2 x 2 x 3 x 3 x 5 x 5 x 7 x 7) = (210)^2. This answers the question stem as Yes.

Hence even together the statements are insufficient.

Answer : E
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If k is a positive integer, is k the square of an integer? 10 Apr 2012, 12:21
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(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.

Show more

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
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If k is a positive integer, is k the square of an integer? 10 Apr 2012, 12:28
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Bunuel


(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.


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
Show more

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.
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If k is a positive integer, is k the square of an integer? 10 Apr 2012, 12:30
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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.
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If k is a positive integer, is k the square of an integer? 10 Apr 2012, 12:36
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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.
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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.
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If k is a positive integer, is k the square of an integer? 10 Apr 2012, 12:45
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Bunuel


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.
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Thanks Bunuel. Another kudos to you, fine sir.

I guess I was thinking that squaring changes TOTAL factors but not different prime factors.
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If k is a positive integer, is k the square of an integer? 10 Apr 2012, 12:56
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Bunuel


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.

Thanks Bunuel. Another kudos to you, fine sir.

I guess I was thinking that squaring changes TOTAL factors but not different prime factors.
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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.
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If k is a positive integer, is k the square of an integer? 26 Apr 2014, 05:16
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If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4.

(2) k is divisble by exactly four different prime numbers.

Responding to a pm.

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

(1) k is divisible by 4. If \(k=4\) answer is YES, but if \(k=8\) answers is NO. Not sufficient.

(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.

(1)+(2) Again if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.
Show more

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.
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If k is a positive integer, is k the square of an integer? 26 Apr 2014, 09:48
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Bunuel
tarek99
If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4.

(2) k is divisble by exactly four different prime numbers.

Responding to a pm.

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

(1) k is divisible by 4. If \(k=4\) answer is YES, but if \(k=8\) answers is NO. Not sufficient.

(2) k is divisible by exactly four different prime numbers.

We don't know the powers of these primes, so if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES (\(k\) equals to square of some integer). Not sufficient.

(1)+(2) Again if \(k=2^2*3*5*7\) the answer is NO, but if \(k=(2^2*3*5*7)^2\) the answer is YES. Not sufficient.

Answer: E.

Hope it's clear.

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.
Show more

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 (\(k=(2^2*3*5*7)^2\)).
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If k is a positive integer, is k the square of an integer? 16 Dec 2019, 12:55
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If k is a positive integer, is k the square of an integer?

(1) k is divisible by 4.

(2) k is divisble by exactly four different prime numbers.
Show more


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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If k is a positive integer, is k the square of an integer? 11 Aug 2022, 22:34
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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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If k is a positive integer, is k the square of an integer? 20 Nov 2024, 12:20
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