Does P have a factor X where 1

Question

Does P have a factor X where 1 < X < P, and X and P are positive integers?

(1) GCD (P^2, k) = k, where k is a prime number

(2) 36*20 + 2 < P < 36*20+6

Bunuel · Accepted Answer

Does P have a factor X where 1 < X < P, and X and P are positive integers?

The question basically asks whether p is a prime number. If it is, then it won't have a factor x such that 1 < x < p (definition of a prime number).

(1) GCD (P^2, k) = k, where k is a prime number. Can p be a prime? Yes, consider p = k = 2. Can p be a non-prime? Yes, consider p = 4 and k = 2. Not sufficient.

(2) 36*20 + 2 < P < 36*20+6 --> p could be 36*20 + 3 = 723, 36*20 + 4 = 724, or 36*20 + 5 = 725. None of these numbers is prime: 723 is a multiple of 3 (sum of its digits is a multiple of 3), 724 is even and 725 is a multiple of 5. So, we can give a definite answer that p is not a prime number, therefore it must have a factor which is greater than 1 and less than p itself. Sufficient.

Answer: B.

Bunuel · Answer

Similar questions to practice: if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html does-the-integer-k-have-a-factor-p-such-that-1-p-k-126735.html for-any-integer-n-greater-than-1-n-denotes-the-product-of-168575.html does-integer-n-have-2-factors-x-y-such-that-1-x-y-n-165983.html if-z-is-an-integer-is-z-prime-128732.html Hope this helps.

vadam · Answer

but what about X itself. for example, if X=721 and in this case it is not a factor of P?

Bunuel · Answer

Your question is not clear. First of all the question asks "Does P have a factor X where 1 <  X < P ..." plus, an integer is a factor of itself. For example, 721 is a factor of 721.

mvictor · Answer

(1) GCD (P^2, k) = k, where k is a prime number
from this, we know that k is a factor of p^2. but we don't know whether k is a factor of P or not. what if k=p? what if k=p=1? p^2=1. since we need integers, there might be no x at all here. 
suppose we have p=9
k=3. then yes. it might be a number x that is a factor of p.

(2) 36*20 + 2 < P < 36*20+6
722 < P < 726
p can be 723, 724, 725.
724 is even, so it is divisible 
725 ends in 5 so is divisible by 5
7+2+3=12, so divisible by 3 
we see that all possible options of P have other factors. we can give a definite answer.
B.

stonecold · Answer

Here all the values between 36*20+2 and 36*20+6 are non primes as it it possible to obtain the factors.
Hence B is sufficient
statement 2 can be discarded as p can be 3 and p can be 81 (for k=3)
thus B is sufficient

zanaik89 · Answer

Hi Bunuel..

How did u conclude that P is a prime number?

Pls help..thanks in advance

Bunuel · Answer

Where does it say that p is a prime?

bumpbot · Answer

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Does P have a factor X where 1<X<P , and X and P are positive integers

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