If p is completely divided by the number 17, and p=x^2*y

Question

If “p” is completely divided by the number 17, and  p = x^2 ∗ y , where x and y are distinct prime numbers, which of these numbers must be divisible by 289?

A.  x^2 
B.  y^2 
C.  xy 
D.  x^2y^2 
E.   x^3y

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gmatophobia · Answer

We know that p is completely divisible by 17. Hence, p consists of 17 as its factor.

Also its given that

p =  x^2 * y

Therefore, either x = 17 or y = 17 (of course both cannot be 17, as its given that the numbers are distinct).

289 =  17^2 . Its a must be true question, hence we need an answer which is definitely true given the conditions.

A.  x^2

Between x and y we don't know which number was 17. If x is17, then  x^2  is divisible by 17, however if x is any other prime number except 17, then  x^2  is not divisible by 289. Hence we can eliminate this option.

B.  y^2

Similar to option A , in this option as well we are not sure if  y^2  must be divisible by  17^2 .

If y is 17, then  y^2  is divisible by 17, however if y is any other prime number except 17, then  x^2  is not divisible by 289. Hence we can eliminate this option.

C. xy

Out of x and y we know only one number is 17, hence xy can never be divisible by  17^2

D.  x^2y^2

This is correct ! Whether x is 17 or y is 17, the product of the squares of two will definitely be divisible by  17^2 .

E.  x^3y

Similar to A and B, we cannot guarantee if  x^3*y   will be a divisible by  17^2 .

If x = 17, yes
If y = 17, no

Hence eliminate this.

Option D

Jainam24 · Answer

OA: D
p is a multiple of 17. so let p=17a 
p=x^2*y. 
either x=17 and y=a or vice versa 
So either x=17 or y=17 
289=17*17

Now the question states "must" so conservatively assume that y=17 for it to satisfy the divisbility of 289 y needs to be min y^2. In case of x to it needs to be x^2 
Thus D is the ans

av1901 · Answer

\frac{p}{17}  is an integer => p has 17 as one of its factors

Also,

p = x^2 ∗ y  : Prime factorization of p is a combination of two different prime numbers and one of them is 17

We need to determine which, amongst the options, must be divisible by  289 (= 17^2) 
Since this is a must be true question, we need to find the answer which is true in any given circumstance

Let's jump into the options

A.  x^2