# “Digit”al Multiplication

So here’s another musing I had on a beach visit. I don’t recall where I learned this trick, but I’ve had it for decades.  I suspect most of you already know how you can multiply by 9 using your fingers, but I’ll briefly explain just in case.  An extension follows.

Start by laying out both hands with all fingers outstretched.  Number your fingers from 1 to 10 from left to right.

To compute $9*n$ for integer values of $n$ between 1 and 10, fold down the $n^{th}$ finger and count the number of still-extended fingers before and after the folded finger. Thinking of those two numbers as a two-digit number gives your answer. For example, to compute $9*9$, fold down the $9^{th}$ finger as shown below.

Because there are 8 fingers before the fold and 1 after, $9*9=81$. Simple.

On the beach, I recalled this little trick and first extended it to two other simple multiples of 9. Computing $9*1$ is simple, but folding down the $1^{st}$ finger confirms the answer by showing 0 before and 9 fingers after the fold, so $9*1=09=9$.

Computing $9*10$ works the same way.  Folding down the $10^{th}$ finger shows 9 before and 0 fingers after the fold, so $9*10=90$.

As I lay on the sand, I marveled at how nice this worked, but was saddened that such a cute approach had such limited applicability.  Thinking about what was going on in this problem, the obvious observation was that all were products of 9 and worked from 10 initial fingers.  So what would happen if you used a different number of fingers?

Starting with 7 fingers, numbered as before, I thought I might be able to multiply by 6.  As a first attempt, I tried $6*6$, but the next image shows that my approach gives $6*6=51$–obviously not the correct result.

Then inspiration struck.  Perhaps the answer was good, but my interpretation was off.  If multiplication by 9s (using 10 fingers) gave an answer in base-10, perhaps multiplication by 6s (using 7 fingers) needed to be interpreted in base-7.  That is, $6*6=51_7=(5*7^1+1*7^0)_{10}=(35+1)_{10}=36_{10}$.  Eureka!  But does it always work?

Testing once more, I tried using 5 fingers meaning I would be multiplying by one less (4) and getting an answer in the base of the number of fingers I was using.  The next image shows $4*2=13_5=(1*5^1+3*5^0)_{10}=(5+3)_{10}=8_{10}$.

Generalizing, imagine that you could hold out any number of fingers.  The image below suggests that you extended $n$ “digits”, so you could use this to multiply by $(n-1)$.

The next image supposes that you hold down the $k^{th}$ finger leaving $k-1$ fingers before and $n-k$ fingers after.

That suggests $(n-1)*k=(k-1)(n-k)_n$ where $(k-1)(n-k)$ is a 2-digit number whose left digit is $(k-l)$ and whose right digit is $(n-k)$.  Expanding, $(k-1)(n-k)_n=((k-1)*n^1+(n-k)*n^0)_{10}$ and $(n-1)*k=(k-1)(n-k)_n=(k*n-k)_{10}$.  QED

So, if you just had enough “digits” (pun intended) and didn’t mind working in different number bases, the result of every single multiplication could be known with one simple finger fold!

### 2 responses to ““Digit”al Multiplication”

1. chrisharrow

Reblogged this on CAS Musings.