Nice DS problem

[netigen]

Whats the value of X^10+Y^10?

1. X and Y are integers
2. Product of X and Y is 89

[jasonc]

To answer the question, we need to have some idea of what x & y are.

1 is obviously insufficient so we can rule out a & d.

Looking that 2, we see that it also gives us no information about what x & y are, because we can put in a ton of different values for x, y and still get 89 (x=1 y=89, x=2 y=89/2... etc). This eliminates b.

So we're left with c & e.

Looking at both statements together, we see that if x & y are integers, the possible values for x & y while still having xy = 89 decreases to 1 set of distinct values IFF 89 is prime.

So the question becomes is 89 prime?
Yes it is, so we choose C as our answer.


Note, to find if a number is prime, do the following:
take the square root of 89, =~9.xxx (since 9 squared is 81 & 10 squared is 100). Now divide 89 by every prime number below 9.xx, which would be 2, 3, 5, 7. You'll see that 89 is not divisible by any of them and thus is prime. (I won't explain why this shows that 89 is prime, but can go into more details if you'd like)

EDIT:
forgot to mention that negative values for x/y won't matter since we're looking for x^10 +y^10

[rajivmatta]

I am not sure I fully comprehend

From second condition I get
89^10 + 1^10 or 9^10 + 9^10..which result in two different numbers..Therefore both conditions should be insufficient..

[AleksandrM]

raji,

9 x 9 = 81, not 89. I am not sure where you are going with the second part?????

89 is a prime number. It looks like the product of 89 could only be a result of 89 x 1 = 89. However 178 x 1/2 is also 89. So I would agree that you need both statements in order to solve this one.