The difference between the squares of two consecutive odd integers
is a square<1000 whose root equals the combined sum of the digits
of the two integers. What is the sum of the digits of the larger integer?  Select:
A) 2
B) 5
C) 8
D) 10
E) 11
Assume two odd integers to be (2x-1) and (2x+1).   Difference of their squares = (2 x+1)^2 - (2x-1)^2   = 4x^2 + 1 + 4x - (4x^2 + 1 - 4x)   = 4x^2 + 1 + 4x - 4x^2 - 1 + 4x   = 8x   Now root of 8x needs to be an integer such that it is equal to the sum of the digits of the two odd integers.   8 = 2^3, so x needs to be such that it "completes" a perfect square.   If we find x, we can find the value of 2x+1 (larger integer) and then sum of its digits.   Let's check the options, starting with c.   For sum to be 8, few possibilities are: 17, 35, 53, 71   If we take 17, the pair is 15 and 17, meaning x = 8.   8x = 64   root of 64 = 4 but 4 is not equal to 1+5+1+7.   Reject.   If we take 35, the pair is 33 and 35, meaning x = 17.   8x = 8*17   Reject since perfect square is not possible.   If we take 53, the pair is 51 and 53, meaning x = 26.   8x = 8*26   Reject since perfect square is not possible.   If we take 71, the pair is 69 and 71, meaning x = 35.   8x = 8*35   Reject since perfect square is not possible.   I tried each option and the possibilities and then got lost.   Then, it occurred to me that I was only checking 2-digit integers. What about 3-digits?   Starting with option a, only 2-digit odd integer whose sum is 2 is 101.   If we take 101, the pair is 99 and 101, meaning x = 50.   8x = 8*50= 
<gadget id="calculator">8*50</gadget>
<output>400</output>
400   root of 400= 20 which is indeed equal to 9+9+1+1.   2nswer (2).
<result>A</result>
The difference between the squares of two consecutive odd integers
is a square<1000 whose root equals the combined sum of the digits
of the two integers. What is the sum of the digits of the larger integer?