# Multiplication verification by adding digits, how does this work?

When I was still in elementry school (4-12 years old in Holland) we were tought to do multiplication without the use of a calculator, for example 128 * 415. When I came home with some homework my father recalled a trick he learned when he was younger. I remembered this recently, but can't figure out how it works.

First you calculate the result:

    128
415 x
-------
640
1280
51200 +
--------
53120


Then the trick: Take the separate digits of the first number: 1,2 and 8. Add these together: 1 + 2 + 8 = 11 While the result is more than one digit, do it again: 1 + 1 = 2. The first "checksum" is 2.

Do the same for the second number: 4 + 1 + 5 = 10, then our second checksum is 1: 1 + 0 = 1.

Multiply these checksums: 1 * 2 = 2. The checksum of the input in this case is 2.

Now we do the same for the answer we calculated: 5 + 3 + 1 + 2 + 0 = 11, This means our answer checksum is 2: 1 + 1 = 2.

Finally compare both "checksums": we found 2 for the input and 2 for the answer, which means our answer might be correct. I'm pretty sure it's not a 100% secure check with chances of false-positives, but I haven't come across one where the checksum failed on a correct answer, or where the checksum was correct but the answer was invalid.

The question: Can anyone explain how this works?

• Look up: digital roots – Robert Israel Jan 13 '17 at 7:24
• Number divided by 9 will have remainder equal to checksum. N*M divided by 9 will have remainder of the remainder of N divided by 9 times the remainder of M divided by nine checksummed. – fleablood Jan 13 '17 at 7:39
• So I've written a short little program to test this theory, and I can very easily create a lot of false negatives: bit.ly/2jez1BZ That 415 number you've got seems to be a tad bit magical. – fny Jan 13 '17 at 7:42
• When used as an arithmetic check, it's often known as casting out nines. It relies on something called modular arithmetic and, because we use a base-$10$ numeral system, the fact that $10 = 9 + 1$. – pjs36 Jan 13 '17 at 8:16
• 11 %False positives are fine as its just a test. There will be no "true negatives" (it will never fail if you have a right answer). If you got the right answer, the test will pass 100% of the time. If you got a wrong answer the test will pass 11% of the time. – fleablood Jan 13 '17 at 8:18

Let N = abcd = 1000a+100b+10c+d.

Divide by by 9 and take the remainder. You get

N = 9 (111a+11b+c) +(a+b+c+d)

So both N and the checksum have the same remainder.

So if checksum N = n and checksum M =m, that means N=9 x something + n, and M = 9 x thingsome + m. So NxM = 9 x athirdthing + nm. So checksum N xM = checksum nxm.

===

Example:

128 = 100 + 20 + 8=9 (11+2)+(1+2+8)

415= 400+10+5=9 (4×11+1)+(4+1+5)

So 128×415=

9[9×(11+2)(4×11+1)+(11+2)(4+1+5)+(4×11+1)(1+2+8)]+(1+2+8)×(4+1+5)

But if 128×415=abcdef then

128×415=

9×(a×11111+b×1111+c×111+d×11+e)+(a+b+c+d+e+f)

So (1+2+8)×(4+1+5) will have the same remainder as (a+b+c+d+e+f) when divided by 9.

You are actually re-doing the multiplication modulo $9$, using two properties:

1. the sum of the digits of a number is that number modulo $9$,

2. modulo "preserves the product"*, $(a\times b)\bmod9=(a\bmod 9)\times(b\bmod 9)$.

Indeed, let $a$ be written $\alpha\beta\gamma$. We have $$(100\alpha+10\beta+\gamma)\bmod9=(100\bmod9)\alpha+(10\bmod9)\beta+\gamma=\alpha+\beta+\gamma.$$

And

$$(a\bmod9)(b\bmod9)=(a-9a')(b-9b')=ab-9a'b-9ab'+81a'b'=ab\bmod9.$$

So if $c=a\times b$, the sum of the digits of $c$ must equal the product of the sum of the digits of $a$ and the sum of the digits of $b$.

$$(128\bmod9)\times(415\bmod9)=2\times1=2=53120\bmod9.$$

The converse is not necessarily true, equality can occur by accident. For instance if you swapped two digits while copying, or wrote a $9$ for a $0$, or made several errors.

$$(128\bmod9)\times(415\bmod9)=2\times1=2=5\color{red}{85}20\bmod9.$$

As the result of a modulo can take $9$ distinct values, one may estimate grossly that the check will detect errors in $8$ cases out of $9$, hence it improves the reliability of the computation by a factor $9$.

*For clarity of the notation, we left implicit that the sums of the digits and the products must be taken modulo $9$ themselves, to discard the possible carries.