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Pdirienzo
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.
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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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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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DavidTutorexamPAL
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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shobhitkh
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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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
DavidTutorexamPAL
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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avigutman gmatophobia my approach - please let me know if my reasoning is valid:

Question rephrased:
Is n a perfect square? OR
Does n contain factors in even exponents only?

S1. 4n = integer^2
n = integer^2 / 2^2

Because we know that n is a positive integer, 2^2 must divide evenly into integer^2. And because integer^2 is a perfect square, it must contain factors in even exponents only. Therefore, at the very least, integer^2 = 2^2 or it also contains other perfect squares, making n a perfect square.
SUFFICIENT

S2. n^3 = integer^2
n * n^2 = integer^2
n = integer^2 / n^2

Similar logic to S1. Because we know that n is a positive integer, n^2 must divide evenly into integer^2. And because integer^2 is a perfect square, it must contain factors in even exponents only. Therefore, at the very least, integer^2 = n^2 or it also contains other perfect squares, making n a perfect square.
SUFFICIENT

Therefore, D.
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achloes
avigutman gmatophobia my approach - please let me know if my reasoning is valid:

Question rephrased:
Is n a perfect square? OR
Does n contain factors in even exponents only?

S1. 4n = integer^2
n = integer^2 / 2^2

Because we know that n is a positive integer, 2^2 must divide evenly into integer^2. And because integer^2 is a perfect square, it must contain factors in even exponents only. Therefore, at the very least, integer^2 = 2^2 or it also contains other perfect squares, making n a perfect square.
SUFFICIENT

S2. n^3 = integer^2
n * n^2 = integer^2
n = integer^2 / n^2

Similar logic to S1. Because we know that n is a positive integer, n^2 must divide evenly into integer^2. And because integer^2 is a perfect square, it must contain factors in even exponents only. Therefore, at the very least, integer^2 = n^2 or it also contains other perfect squares, making n a perfect square.
SUFFICIENT

Therefore, D.
Well done achloes, looks good to me!
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Hi KarishmaB,

Can you please share your approach for Statement 2?

Thanks in advance. :)
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Pdirienzo
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

For n to be a square of an integer, all its distinct prime factors must have an even exponent.

If you are not sure why, check this video
https://youtu.be/DxIH8rjhpKY
or these posts
https://anaprep.com/number-properties-f ... -a-number/
https://anaprep.com/number-properties-f ... ct-square/

Assume \(n = 2^a * 3^b * 5^c\)
For n to be a perfect square, all a, b and c must be even integers.

(1) 4n is the square of an integer

So 4n will have even exponents of all distinct prime factors. 4n is simply \(2^2 * n = 2^2 * 2^a * 3^b * 5^c \)
Here, all (a+2), b and c are given to be even integers. This means that a, b and c are even integers.

Hence, n is a perfect square. Sufficient alone.

(2) n^3 is the square of an integer

\(n^3 = (2^a * 3^b * 5^c)^3 = 2^{3a} * 3^{3b} * 5^{3c}\)
We are given that 3a, 3b and 3c are all even integers. Then a, b and c must be even integers too (the 2 must come from a, b and c in 3a, 3b and 3c).

Hence, n is a perfect square. Sufficient alone.

Answer (D)
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It was really a tough one, thanks for the explanation.
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