If x and y are positive integers and xy, is (x-y)^3+(3y+1)^5 odd?

Question

If x and y are positive integers and x>y, is (x-y)^3+(3y+1)^5 odd?

(1) x is odd.
(2) y is even.

TestPrepUnlimited · Answer

We may reduce the powers down to 1 as that does not change each terms' parity. Thus we can ask if  (x - y) + (3y + 1) = x + 2y + 1  is odd.
And 2y is always even so we are just asking if  x + 1  is odd, or if  x  is even.

Statement 1:

Sufficient.

Statement 2:

Insufficient.

Ans: A

VIVA1060 · Answer

Hi,  TestPrepUnlimited  I didn't understand how can we remove the bases and multiply. It didn't strike me, just wanted to understand the logic here.

TestPrepUnlimited · Answer

Hi  VIVA1060  no problem!

Whatever you see here is not proper math of course, I am only reducing the exponents because it does not change if the terms are odd or even.

If x was even then  x^2, x^3, x^4, x^5, .... , x^n  are all even, where n is any positive integer.

If x was odd then  x^2, x^3, x^4, x^5, .... , x^n  are all odd, where n is any positive integer.

Thus we have this one-to-one relation we can conclude, on the other hand we know if  x^n  is even/odd, then  x  must be even/odd.

Now let's look at the problem again,  (x - y)^3  can be reduced to  x - y  for the sake of this problem. If  (x - y)^3  was even/odd, then  x - y  will still be even/odd. Similarly  (3y + 1)^5  I reduce it to  3y + 1  and work with that instead. Now we can ask if  x - y + 3y + 1 = x + 2y + 1  is odd instead of asking if  (x - y)^3 + (3y + 1)^5  is odd, because these two expressions must have the same parity (odd/even attribute).

AT1982 · Answer

Given x>y>0
S1.
x = 5(odd) & y =4(even)
(5-4)^3 + (3 x 4 +1)^5 = Even
x = 5 & y =3
(5-3)^3 + (3 x 3 + 1)^5 = Even

Confirmed NO

S2.
only Y is even but both options would change
if x is even/odd

Answer A

If x and y are positive integers and x>y, is (x-y)^3+(3y+1)^5 odd?

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