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).

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:

$6\cdot 6\,$:

$6\cdot 7\,$:

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