gracie wrote:

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?

A. 2

B. 5

C. 8

D. 10

E. 11

source:self

Really hard. But let's start with few concepts:

1. Square of an odd integer will be odd.

2. Difference of odd and odd will be even.

3. Any odd integer can be expressed as difference of two squares.

4. An even integer can be expressed as difference of two squares only if that even integer is a multiple of 4.

Assume two odd integers to be (2x-1) and (2x+1).

Difference of their squares = (2x+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 = 400

root of 400 = 20 which is indeed equal to 9+9+1+1.

Phew!

Answer (A).

Does anyone have a better approach?

Is this even a GMAT level question?