If k is an integer, is k the square of an integer?

[Guest]

If k is a positive integer, is K the square of an integer?

1) k is divisible by 4

2) k is divisible by exactly four primes

[rfernandez]

We'd be more than happy to write out a solution to this problem, but is there something in particular you're struggling with?

[Guest]

We'd be more than happy to write out a solution to this problem, but is there something in particular you're struggling with?

Sir, This is a maths Problem, Dont you think you are going into Critical Reasoning, "Drawing an Conclusion, based on Asuumption"?

[mdh3000]

I'll give it a try...

Rephrase the question, is the square root of k an integer?

Statement 1:

We know k is a + integer so...
k = 4, 8, 12, 16, 20, 24 ....

INSUFFICIENT since the square root of some of those numbers is an integer (4, 16), but others aren't (8, 12, etc).

Statement 2:

k is divisible by 4 primes (label them p1-4), so we know 4 primes are factors of k...

1) so k could equal p1*p2*p3*p4. The square root of the product would not be an integer.
2) however, if k was equal to (p1*p2*p3*p4)^2, then the square root would be an integer.

INSUFFCIENT.

Statement 1 + 2:

For a number to be divisible by four, it must be divisible by 2 TWICE. So we know that 2 is one of the four prime factors. However, that doesn't provide us with any more info, so...

The answer is E.

mdh

[RonPurewal]

If k is a positive integer, is K the square of an integer?

1) k is divisible by 4

2) k is divisible by exactly four primes

here we go:

(1)
mdh3000's explanation is perfect.

(2)
this means that k has four different prime factors, but we don't know how many times those factors appear in the prime factorization of k.
so, for example, if k is 2 x 3 x 5 x 7 (which is divisible by the four primes 2, 3, 5, 7), it's not the square of an integer;
if k is 2 x 2 x 3 x 3 x 5 x 5 x 7 x 7 (divisible by the same four primes), it's the square of the integer 2 x 3 x 5 x 7.
so, insufficient.

(together)
you need the 4 prime factors (because of statement 2), and you also need to have at least two '2's in the prime factorization (because of statement 1).
the aforementioned perfect square (2 x 2 x 3 x 3 x 5 x 5 x 7 x 7) still works.
to create a number that satisfies the criteria yet isn't a perfect square, just add another copy of one of these prime factors (e.g., 2 x 2 x 3 x 3 x 5 x 5 x 7 x 7 x 7).
insufficient

answer = e