This question shows up often on standardized tests (SAT, ACT, GRE, etc) and can be solved in about two seconds 99.9% of the time if you memorize your square tables up to about 25 or so. Here's how to solve it:

1. Add 1 to the given product.

2. The answer is the square root of this number -/+ 1.

Why it works:

Any pair of consecutive odd or even numbers can be represented by X and X + 2, but also X - 1 and X + 1. We use this latter representation because it is in conjugate pair form, and the product will be X^2 - 1. So, all you are doing is reversing the process. Step 1 eliminates the -1 term. Taking the square root leaves you with X, a number which will be 1 greater and 1 less than the two numbers you are asked for in the original problem. Knowing this, you can memorize a list of common consecutive odd/even integer products, which will also equal the number in between^2 - 1:

11 x 13 = 12^2 - 1 = 143

12 x 14 = 13^2 - 1 = 168

13 x 15 = 14^2 - 1 = 195

14 x 16 = 15^2 - 1 = 224

15 x 17 = 16^2 - 1 = 255

16 x 18 = 17^2 - 1 = 288

17 x 19 = 18^2 - 1 = 323

18 x 20 = 19^2 - 1 = 359

19 x 21 = 20^2 - 1 = 399

20 x 22 = 21^2 - 1 = 440

21 x 23 = 22^2 - 1 = 483

22 x 24 = 23^2 - 1 = 528

23 x 25 = 24^2 - 1 = 575

24 x 26 = 25^2 - 1 = 624

You can also extend this same reasoning to products of numbers with even difference multiples: (X) * (X + 2n). In this case you would just add n^2 the product first, then take the square root. The answer would be this number -/+ n. It still works because you are looking for a conjugate pair in the form (X + n)(X – n) to represent the two unknown numbers. For example, if the product of two consecutive multiples of 6 = 391, n = 6/2 = 3 in this case.

Step 1. Add 3^2 to 391 to get 400.

Step 2. Answers are sqrt(400) -/+ n = 20 -/+ 3 = 17 and 23.