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


*

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

*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.
