Is the positive integer n the square of an integer?

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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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.

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.

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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One Doubt ... In explanation of first statement Can square root of n = decimal.....and then 2 * decimal equal to integer

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?

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.


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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.

Well done achloes, looks good to me!

Hi KarishmaB,

Can you please share your approach for Statement 2?

Thanks in advance.
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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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