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:
 hold your hands, with palms facing you
 on each hand, number the fingers 6 through 9 (from the bottom up; pinky to index finger)
 touch the fingers you want to multiply:
for example, to multiply 6 by 9, touch the pinky on one hand to the index finger on the other hand
 COUNT the two fingers that touch, plus any fingers below;
this sum becomes the TENS digits in the product
 COUNT the fingers above the touching fingers on the left.
COUNT the fingers above the touching fingers on the right.
MULTIPLY these two (small) numbers together.
This becomes the ONES digit in the product.
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 $\,xy\,$.

The number of fingers strictly below $\,x\,$ is $\,x  6\,$.
For example, if $\,x = 6\,$ then there are $\,x  6 = 6  6 = 0\,$ fingers below.
If $\,x = 7\,$, there is $\,x  6 = 7  6 = 1\,$ finger below.

The number of fingers strictly above $\,x\,$ is $\,10  x\,$.
For example, if $\,x = 6\,$ then there are $\,10  x = 10  6 = 4\,$ fingers above.
If $\,x = 7\,$, there are $\,10  x = 10  7 = 3\,$ fingers above.

The SUM OF THE FINGERS strictly below, plus the
two fingers that are touching, is:
$$(x6) + (y  6) + 2\ \ =\ \ x + y  10 \quad \quad \quad \text{(SumOnAndBelow)}$$
Observe that $\,x + y  10\,$ is always between $\,2\,$ and $\,8\,$ when $\,x\,$ and
$\,y\,$ only take on values in the set $\,\{6,7,8,9\}\,$.
In particular, SumOnAndBelow is always a singledigit number.

The PRODUCT OF THE FINGERS strictly above is
$$
(10x)(10y) \quad \quad \quad (\text{ProductAbove})
$$
Unfortunately, this is not always a singledigit number!
If $\,x = 6\,$ and $\,y = 6\,$, then we get (for example) $\,(106)(106) = 16\,$.
Or, if $\,x = 6\,$ and $\,y = 7\,$ (or vice versa), then we get $\,(106)(107) = 4\cdot 3 = 12\,$.
All other combinations do indeed give a singledigit number.
 They want us to put the SumOnAndBelow digit in the tens place.
How much does this contribute to the final number?
Answer:
$10\cdot(\text{SumOnAndBelow}) = 10(x+y  10)$

So, here's why this works (when ProductAbove is a singledigit number):
$$
\begin{align}
10(\text{SumOnAndBelow}) + \text{ProductAbove} \ \ &= \ \ 10(x+y10) + (10x)(10y)\cr
&=\ \ 10x + 10y  100 + 100  10x  10y + xy\cr
&=\ \ xy
\end{align}
$$
Cool! I love it!
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:
$6\cdot 6\,$:
 $\text{SumOnAndBelow} = 2$
 $\text{ProductAbove} = 4\cdot 4 = \color{red}{1}6$
 Take the ‘$\,\color{red}{1}\,$’ from the tens place and add it to the $\,2\,$, giving $\,3\,$.
 Result: $\,36\,$
$6\cdot 7\,$:
 $\text{SumOnAndBelow} = 3$
 $\text{ProductAbove} = 4\cdot 3 = \color{red}{1}2$
 Take the ‘$\,\color{red}{1}\,$’ from the tens place and add it to the $\,3\,$, giving $\,4\,$.
 Result: $\,42\,$
Why?
If the product $\,(10x)(10y)\,$ is a twodigit number, then when you
subtract $\,10\,$, the result is a onedigit number.
So, add zero in the form ‘$\,10  10\,$’, as follows:
$$
10(\text{SumOnAndBelow}) + \text{ProductAbove} \ \ = \ \ \overbrace{[10(x+y10) \color{blue}{+ 10}]}^{\text{tensplace digit increases by one}} + \overbrace{[(10x)(10y) \color{blue}{ 10}]}^{\text{now a onedigit #}}
$$
Yep—the tensplace digit increases by $\,1\,$, since you're adding ONE more ten!