Prime & Square

[joyseychow]

Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.

I got this correct. But I'm still not sure about Stmt 1. Pls. Explain. Thks!

[shahdevine]

Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.

I got this correct. But I'm still not sure about Stmt 1. Pls. Explain. Thks!

statement 1)

restate:

for every prime number that is a factor of n then p^2 is also a factor of n.

let's make our prime number(p) = 5. N at least would have to equal 25 in order for p^2 to be a factor of N. So n=p^2. However, n could have additional primes that are not p. For instance n could equal 50, whose factors are 5^2 and 2. If this were the case n does not equal the square of an integer. It equals the square of an integer times another prime number. So insufficient.

Statement 2)

sqroot n = an integer
n = integer squared

sufficient

B is answer.

pls confirm.

you got this!

[navalpike]

My answer is D.

A.) I am translating this to mean that if there is “any” prime number in N, then the p^2 version of that prime is also in N. Meaning – Every prime is squared.

If this is the correct interpretation, then this is sufficient. A perfect square, when reduced to primes, will have even numbers as exponents of those primes. For Example.

2*3*5*7 = Not a perfect square

2^2*3^2*5^2*7^2 = perfect square.

Sufficient.

[zeenab]

I think the answer is D as well.

If a number is a perfect square, all of its prime factors have to occur an even number of times in its factorization. Which means, if the number is divisible by a prime number n, it should be divisble by n^2 as well.
So if the corollary holds good as well.
Example -

Take 100 which is 10^2

(5*2)^2 - so the number is divisible by 5 and 5^2 and 2 and 2^2.

OA please?

[shahdevine]

I think the answer is D as well.

If a number is a perfect square, all of its prime factors have to occur an even number of times in its factorization. Which means, if the number is divisible by a prime number n, it should be divisble by n^2 as well.
So if the corollary holds good as well.
Example -

Take 100 which is 10^2

(5*2)^2 - so the number is divisible by 5 and 5^2 and 2 and 2^2.

OA please?

you may be right. my interpretation is wrong. we need OAs...

[mehravikas]

Answer shouldn't be D...

Statement 1 -

n = 45, p = 3

p is a divisor of n
also p^2 is a divisor of n

but n in this case is not a square of an integer.

I think the answer is D as well.

If a number is a perfect square, all of its prime factors have to occur an even number of times in its factorization. Which means, if the number is divisible by a prime number n, it should be divisble by n^2 as well.
So if the corollary holds good as well.
Example -

Take 100 which is 10^2

(5*2)^2 - so the number is divisible by 5 and 5^2 and 2 and 2^2.

OA please?

you may be right. my interpretation is wrong. we need OAs...

[tohellandback]

Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.

I got this correct. But I'm still not sure about Stmt 1. Pls. Explain. Thks!

edited..
IMO B- look for scoobydooby's example
1) it says that all the prime factors of n have even powers- i was wrong
it says that the powers are greater than 2. Can be even. Can be odd
Not sufficient

2) Sufficient

@mehravikas,
1) says every prime number.
In your example 45= 3^2*5.
you haven't considered 5.

[scoobydooby]

would go with B

stmnt 1 is not sufficient by itself.

if N=9, divisible by 3 and 3^2 . yes N perfect square
if N=27, divisble by 3 and 3^2. no N not perfect square

stmnt 2 sufficient
all perfect squares have integers as their square roots.

[shahdevine]

would go with B

stmnt 1 is not sufficient by itself.

if N=9, divisible by 3 and 3^2 . yes N perfect square
if N=27, divisble by 3 and 3^2. no N not perfect square

stmnt 2 sufficient
all perfect squares have integers as their square roots.

cool i'm not crazy...for a minute i was racking my brains. I thought I was from another planet when everyone was getting Ds instead of B.

[arorag]

should be B
For A lets have n=8, p= 2, p2= 4

[joyseychow]

Yes OA is indeed B

[vittalgmat]

The question is a bit confusing.. esp stmt 1.

But this is a good discussion.
I got B with my interpretation.


thanks everyone.
-V

[vikram_k51]

Is the positive integer n equal to the square of an integer?
(1) For every prime number p, if p is a divisor of n, then so is p2.
(2) sqrt root (n) is an integer.

D

[dikku07]

in favour of B

[jjk]

B

You can test statement 1 by setting n equal to 9 or 18 and p equal to 3.

If n = 9 and p = 3, 9/3 and 9/9 yield integers.

If n = 18 and p = 3, 18/3 and 18/9 yield integers.

However, 18 is NOT a square of an integer. Statement 1 is thus insufficient.