HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte

Let M = 10x+y and N = 10y + x (where x and y are single digit whole numbers)
=> M^2 + N^2 = 101(x^2+y^2) + 40xy

Now, let's look at the answer choices. 40 and 101 stand out because they appear in our equation as well.
Let's consider 101... it divides the first term ( 101(x^2+y^2) )completely (without remainder). So, the only way the entire number will be divisible by 101 is if the 2nd term (40xy) is ALSO divisible by 101. 40 and 101 have no common factors because 101 is a prime number. Hence, the only way for 40xy to be divisible by 101 is if xy is divisible by 101. That will require x = 101 and y = 1 or vice-versa. Now, either of those options aren't possible because x and y are single digit numbers. So, 101 is definitely not a factor of M^2 + N^2.

If we do a similar test for 40... (this is just for understanding, and not a required step because we already found our answer)
The 2nd term is divisible.
For the first term to be divisible as well, (x^2 + y^2) has to be divisible by 40. A possibility is (2,6). Another is (6,2). Either of these will satisfy all the required conditions without violating any assumptions.
Hence, 40 can be a factor

For the other answer choices, neither 101 nor 40 is divisible directly. So, we may be able to find combination of x and y to satisfy the divisibility overall. But more importantly, we DON'T HAVE TO FIND ANY OF THOSE. Reason: This is a single option correct question! Once we found 101 is not a factor, we are done!!

Answer: E