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Intern  B
Joined: 13 Jul 2018
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GMAT 1: 650 Q47 V33 Is the positive integer n the square of an integer?  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 59% (01:59) correct 41% (01:50) wrong based on 196 sessions

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Is the positive integer n the square of an integer?

1) 4n is the square of an integer

2) n^3 is the square of an integer
examPAL Representative P
Joined: 07 Dec 2017
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Re: Is the positive integer n the square of an integer?  [#permalink]

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2
This is a bit abstract, it's best to use the Alternative approach and plug in numbers:
1) the condition holds and answer is "yes" for 1 and 4, "yes" for 4 and 16, yes for 16 and 64... and so on! sufficient!
2)this condition holds, and the answer is "yes" for 1 & 1, 4 and 64, for whom the answer is "yes" - sufficient!
D.
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Re: Is the positive integer n the square of an integer?  [#permalink]

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If a number is the square of an integer, then in the prime factorization of that number, all of the exponents will be even. For example, (3^8)(5^6) is the square of an integer (it is the square of (3^4)(5^3) ) whereas (3^7)(5^6) is not.

Statement 1 tells us that in the prime factorization of (2^2)(n), all of the exponents are even. The prime factorization of n is identical to that of (2^2)(n) except that in n, the exponent on the '2' is smaller by two, and subtracting two doesn't change an even number to an odd number. So if the exponents in (2^2)(n) are even, so are the exponents in n, and Statement 1 is sufficient.

Statement 2 tells us that in the prime factorization of n^3, all of the exponents are even. The exponents in the prime factorization n^3 are all 3 times as big as the exponents in the prime factorization of n. Multiplying by 3 doesn't change evenness or oddness, so if all the exponents in n^3 are even, so are the exponents in n, and Statement 2 is also sufficient.

The answer is D.
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Re: Is the positive integer n the square of an integer?  [#permalink]

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Pdirienzo wrote:
Is the positive integer n the square of an integer?

1) 4n is the square of an integer

2) n^3 is the square of an integer

prime factorization is key. too hard problemn

4n =a^2
prime factorization a^2= x^2k=x^2 or x^4 or x^6...
4n=x^2k
x^2k= even
x=even
x=(2y)^2k
n must be a square

go to 2 nd condition
n^3=a^2
a^2 = x^2a*y^2b, with x,y are prime
n^3=x^2a*y^2b , this mean a and b must be multiple of 3
this mean 2a and 2b are multiple of 6
n^3=x^6k*y^6m
this mean
n=x^2k*y^2m
this mean n is a squre of integer.

too hard.
Senior Manager  P
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Re: Is the positive integer n the square of an integer?  [#permalink]

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Hi,

Here are my two cents for this question

lets say if we write n in term of its prime factors then n = $$A^{P}*B^{Q}*C^{R}$$, where A, B , C are prime numbers and P, Q R are non negative integers.

Now if the powers of prime factors of n are even we can say that n is perfect square

Now Statement 1:
4n = $$I^{2}$$
which means
2^{2}[/m]n = $$I^{2}$$
taking square root on both the sides we have
$$\sqrt{2^{2}n}$$= $$\sqrt{I^{2}}$$
2 $$\sqrt{n}$$ = I
For LHS to be equal to RHS $$\sqrt{n}$$ has to be integer, This can be possible only when the powers of prime factors of n are even) , which implies that n is perfect square

Now Statement 2
$$n^{3}$$= $$A^{3P}*B^{3Q}*C^{3R}$$,
we are told that $$n^{3}$$=$$I^{2}$$
$$A^{3P}*B^{3Q}*C^{3R}$$=$$I^{2}$$
taking square root on both sides we have

$$\sqrt{(A^{3P}*B^{3Q}*C^{3R})}$$ = I
this means P Q R are even
this means that n is a perfect square

Hence Answer Choice D
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Re: Is the positive integer n the square of an integer?  [#permalink]

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Probus wrote:
Hi,

Here are my two cents for this question

lets say if we write n in term of its prime factors then n = $$A^{P}*B^{Q}*C^{R}$$, where A, B , C are prime numbers and P, Q R are non negative integers.

Now if the powers of prime factors of n are even we can say that n is perfect square

Now Statement 1:
4n = $$I^{2}$$
which means
2^{2}[/m]n = $$I^{2}$$
taking square root on both the sides we have
$$\sqrt{2^{2}n}$$= $$\sqrt{I^{2}}$$

2 $$\sqrt{n}$$ = I
For LHS to be equal to RHS $$\sqrt{n}$$ has to be integer, This can be possible only when the powers of prime factors of n are even) , which implies that n is perfect square

Now Statement 2
$$n^{3}$$= $$A^{3P}*B^{3Q}*C^{3R}$$,
we are told that $$n^{3}$$=$$I^{2}$$
$$A^{3P}*B^{3Q}*C^{3R}$$=$$I^{2}$$
taking square root on both sides we have

$$\sqrt{(A^{3P}*B^{3Q}*C^{3R})}$$ = I
this means P Q R are even
this means that n is a perfect square

Hence Answer Choice D

One Doubt ... In explanation of first statement Can square root of n = decimal.....and then 2 * decimal equal to integer
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Re: Is the positive integer n the square of an integer?  [#permalink]

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DavidTutorexamPAL wrote:
This is a bit abstract, it's best to use the Alternative approach and plug in numbers:
1) the condition holds and answer is "yes" for 1 and 4, "yes" for 4 and 16, yes for 16 and 64... and so on! sufficient!
2)this condition holds, and the answer is "yes" for 1 & 1, 4 and 64, for whom the answer is "yes" - sufficient!
D.

Hi DavidTutorexamPAL, Situation 1 is not true for integer 3. The square of 3 is 9. Hence, 4n equals 9 which gives n as 9/4. Thus it implies that n is not the square of an integer.
OA being D, can you please clarify what is wrong in my approach?
GMAT Tutor G
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Re: Is the positive integer n the square of an integer?  [#permalink]

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shobhitkh wrote:
Situation 1 is not true for integer 3. The square of 3 is 9. Hence, 4n equals 9 which gives n as 9/4. Thus it implies that n is not the square of an integer.
OA being D, can you please clarify what is wrong in my approach?

The question tells us that n is an integer, so n cannot be 9/4.
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Joined: 23 Feb 2015
Posts: 1257
Re: Is the positive integer n the square of an integer?  [#permalink]

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The question says: "n" is a positive integer. So, 9/4 is positive integer? No. Whatever you do in statement 1, n must be positive integer.
shobhitkh wrote:
DavidTutorexamPAL wrote:
This is a bit abstract, it's best to use the Alternative approach and plug in numbers:
1) the condition holds and answer is "yes" for 1 and 4, "yes" for 4 and 16, yes for 16 and 64... and so on! sufficient!
2)this condition holds, and the answer is "yes" for 1 & 1, 4 and 64, for whom the answer is "yes" - sufficient!
D.

Hi DavidTutorexamPAL, Situation 1 is not true for integer 3. The square of 3 is 9. Hence, 4n equals 9 which gives n as 9/4. Thus it implies that n is not the square of an integer.
OA being D, can you please clarify what is wrong in my approach?

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