# Foiling the "Product of two consecutive odd or even numbers" problem.

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.

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Andrew M.

Engineer and Inventor Specializing in Math and Test Prep

100+ hours
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