N is a positive integer. Is n the square of an integer?

Thanks Chetan!!!

Although in that case I seem to have invented a new question

and also in that case my answer choice is Option D instead of A

But thank you

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

N is a positive integer. Is n the square of an integer?
1) 4n is the square of an integer
2) n^3is the square of an integer

In the original condition, there is 1 variable(n), which should match with the number of equation. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), in 4n=m^2 (m is some integer), 4n is an even number so it should be m^2=even. Then, m^2=(2k) ^2 where k is some integer and therefore 4n=(2k) ^2=4k^2 -> n=k^2, which is yes and sufficient.
For 2), n=3√(t^2 ) is derived from n^3=t^2 where t is some integer. Since n is a positive integer, cube root should be removed. Therefore, n=3√(t^2 )=3√{s^3}^2 )=3√(s^6 )=s^2 is also yes and sufficient. Therefore, the answer is D.


 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

IMO D

I think it is a simple question if one know the properties of Perfect squares:
a. They have even powers of prime factors
b. They have odd no of distinct factors
c. The sum of the factors is odd.
d. They have odd number of odd factors and even number of even factors.

Now moving to Question:

St1 : 1) 4n is the square of an integer
>> 4 is a perfect square (2^2). This when multiplied by n gives us another perfect square implying that n is also a perfect square since prime factor of 4 has even power, n also should have even powers for its prime factors to make 4n a perfect square.

So st 1 sufficient.

St2: 2) n^3 is the square of an integer

This is interesting. Now lets go to roots and powers concept and loan some formulas from there. (A^m)^n = a^(m*n)

If n^3 is a perfect square, all its prime factors have even power. The power 3 is odd and when will a product be even when multiplied with an odd number >>> when the second number is even.

Which means n has to have even powers for its prime factors to make n^3 a perfect square, meaning N is also a perfect square.

St 2 is sufficient.

target check whether n is square of an integer
#1
4n is the square of an integer
for the given condition ; √4n ; 2√n ; n has to be square of integer eg ; n = 1,4,9,16...
sufficient
#2
n^3 is the square of an integer
value of n^3 (1,4) = 1,64, where n = 1,8 ; which is perfect square ; sufficient
option D

N is a positive integer. Is n the square of an integer?

1) 4n is the square of an integer.
4n=x^2
n=(x^2)/4
n=(x^2)/(2^2)
given n is an integer, and n is equal to x^2 / n^2. (x/2)^2
Sufficient.

2) n^3 is the square of an integer
n^3=x^2
n=x^2/n^2
n=(x/n)^2
Sufficient.

First glance -- this is testing squares and what we know about the factors of squares.

1) If 4n is the square of an integer, then n's prime factors must be even or 1. Prime factors of squares must have an "even" exponent (greater than 1). Sufficient.

2) We know that squares have an even exponent (greater than 1) for prime factors, except for 1. So to result in a square -- n must be an even exponent greater than 1 (or 1) in order to result in a square (since an odd x even = even). This also means that the square root of n^3 is an integer, which means that n's prime factors must be to an even power. Sufficient.

Answer is D -- both statements are sufficient.

1) 4n is the square of an integer
for the given condition; 4*1 ; 4*4; 4*9; sufficient

2) n^3 is the square of an integer
value of n^3 (1,4) = 1,64, where n = 1,8 ; which is perfect square ; sufficient

Ans. is D

Is N = the Square of an Integer?

----the "Square of an Integer" means the Question is asking whether N is a PERFECT SQUARE ----

Does N = (Integer)^2?


Rule: All Perfect Squares will have a Prime Factorization in which every Prime Base will be raised to an EVEN Power



Stmt.1: 4N = (Integer)^2

because a Perfect Square only has EVEN Exponents of its Prime Bases:

4N = (p)^2 * (q)^2 * (r)^2 etc.....

----where---- p , q , and r = Prime Bases


Since we are told that N = (+)Pos. Integer, the Square of an Integer on the Right Hand Side of the Equation in Statement 1 MUST be Divisible by 4

in other words, AT THE VERY LEAST, the Perfect Square must have (2)^2 in its Prime Factorization


MIN Possible Value of 4N:

4N = (2)^2

N = (2)^2 / (4)

n = 1 -----> YES, 1 is a Perfect Square


Because we are Given:

(Fact 1) ---- N = (+)Positive Integer

and

(Statement 1) --- 4N = (Integer)^2



The Prime Factorization of 4N must always look like the following:

4N = (2)^2 * (Possibly more of Prime Base 2)^EVEN Power * (Other Prime Bases)^EVEN Power

when we Divide the R.H.S. by 4, we will still be left with ALL Prime Bases raised to an Even Power


thus

N will always be a Perfect Square



Stmt. 2:

(N)^3 = (Integer)^2


Rule: the Prime Factorization of a PERFECT Cube must Always have Prime Bases raised to Exponents that are MULTIPLES OF 3

thus, in order for (N)^3 = PERFECT Square ------>

(N)^3 = [Integer^2] ^3k

(N)^3 = (Integer)^6k

----where K = some (+)Positive Integer----

When we take the CUBE ROOT of Both Sides of the Equation:

N = (Integer)^2k

which means N will always be equal to a Perfect Square

examples:

(N)^3 = 64 = (2^2)^3 = (4)^3 --------> N = 4

(N)^3 = 729 = (3^2)^3 = (9)^3 -------> N = 9



-D- Both Statements are Sufficient

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