Any number written in our decimal system is made
up of powers of 10. For example,
65,321 = 6(10^4) + 5(10^3)+3(10^2) + 2(10)+1.
Now if you are interested in knowing whether 65,321 is a multiple of 5
(that is, is exactly divisible by 5), you can tell just by looking
at the last digit. That's because 10^4, 10^3, 10^2, and 10 are all
divisible by 5, so the last digit on the end gets to cast the deciding
vote.
Deciding whether 65,321 is a multiple of 3 is a little harder, since
no power of 10 is a multiple of 3. But
10 = 9 + 1 and
100 = 99 + 1 and
1000 = 999 + 1 etc.
So if you had 65,321,
65,321 = 6(10^4) + 5(10^3)+3(10^2) + 2(10)+1
you could rewrite this as
65,321 = 6(9999+1) + 5(999+1) + 3(99+1) + 2(9+1)+1
= 6(9999) + 5(999) + 3(99) + 2(9) + (6+5+3+2+1)
The first part is divisible by 3 so 65,321 is a multiple of 3 if the
sum of its digits 6+5+3+2+1 is divisible by 3.
Now we are ready for 11.
Eleven is a little more complicated, but look at this neat pattern:
99 is a multiple of 11
9999 is a multiple of 11
999999 is a multiple of 11
...
any even number of 9's makes a multiple of 11.
If you look at how we wrote
65,321 = 6(9999+1) + 5(999+1) + 3(99+1) + 2(9+1) + 1
you can see that every other term contains a number that is a multiple
of 11. Unfortunately, numbers with an odd number of 9's are not
divisible by 11.
But look at this pattern for these numbers:
10 = 10^1 = 11 - 1
1000 = 10^3 = 1001 - 1
100000 = 10^5 = 100001 - 1
10000000 = 10^7 = 10000001 - 1
and each of these numbers: 11, 1001, 100001, 10000001, ... is a
multiple of 11. (Check it out by long division . . . it's worthwhile
looking at the pattern you get when you divide 1 00 00 00 00 00 1 by
11!)
Back to the example of 65,321.
65,321 = 6(9999+1) + 5(1001 - 1) + 3(99+1) + 2(11-1) + 1
= 6(9999) + 6 + 5(1001) - 5 + 3(99) + 3 + 2(11) - 2 + 1
All the terms with parentheses are multiples of 11. So 65,321 will
be divisible by 11 if the remaining numbers are a multiple of 11.
That is,
6 - 5 + 3 - 2 + 1
Just the numbers you spoke about in your letter.
In general, the numbers in the odd positions are part of the "multiple
of 999...999 + 1" and so are added to the total. The numbers in the
even positions are part of the multples of "1000...0001 - 1" and so
get subtracted.
The final test is to look at this alternating sum and difference of
digits. But the result does not have to be 0 in order to be a multiple
of 11...ANY multiple of 11 will do. For example,
8030209 gives 8 + 3 + 2 + 9 = 22.
Since 22 is a multiple of 11, so is 8,030,209.