Explanation of Multiplication Trick for 6, 7, 8, 9

This multiplication trick was posted on Facebook. Here's why it works (and what its limitations are). The images below are from the original link.

In case the link above breaks, here's the trick:

Here's Why It Works:

Suppose you want to multiply $\,x\,$ by $\,y\,,$ where these two numbers are $\,6\,,$ $\,7\,,$ $\,8\,,$ or $\,9\,.$ That is, you want to find the product $\,xy\,.$

How to Handle the Exceptions—an Easy ‘Carry’

The exceptions ($\,6\cdot 6\,$ and $\,6\cdot 7\,$) aren't much harder. Take the extra ‘$\,1\,$’ from the product, and add it to the sum, like this:

$6\cdot 6\,$:

$6\cdot 7\,$:


If the product $\,(10-x)(10-y)\,$ is a two-digit number, then when you subtract $\,10\,,$ the result is a one-digit number. So, add zero in the form ‘$\,10 - 10\,$’, as follows:

$$ \begin{align} 10(&\text{SumOnAndBelow}) + \text{ProductAbove}\cr &= \ \ \overbrace{[10(x+y-10) \color{blue}{+ 10}]}^{\text{tens-place digit increases by one}} + \overbrace{[(10-x)(10-y) \color{blue}{- 10}]}^{\text{now a one-digit #}} \end{align} $$ Yep—the tens-place digit increases by $\,1\,,$ since you're adding ONE more ten!