Why 99 & 1111 are special in divisibility tests using decimal digit sums? (casting out nines & elevens)

I don’t know if this is a well-known fact, but I have observed that every number, no matter how large, that is equally divided by 9, will equal 9 if you add all the numbers it is made from until there is 1 digit.

A quick example of what I mean:

999=891

8+9+1=18

1+8=9

This works even with really long numbers like 4376331

Why is that? This doesn’t work with any other number. Similarly for 11 and alternating digits sums.

Answer

Not quite right, since 9×0=0 and the digits don’t add up to 9; but otherwise correct.

The reason it works is that we write numbers in base 10, and when you divide 10 by 9, the remainder is 1. Take a number, say, 184631 (I just made it up). Remember what that really means:
184631=1+3×10+6×102+4×103+8×104+1×105.
The remainder when you divide any power of 10 by 9 is again just 1, so adding the digits gives you a number that has the same remainder when dividing by 9 as the original number does. Keep doing it until you get down to a single digit and you get the remainder of the original number when you divide by 9, except that you get 9 instead of 0 if the number is a nonzero multiple of 9.

But since every multiple of 9 is a multiple of 9, you will always get 9.

Note that you have a similar phenomenon with 3 (a divisor of 9), since adding the digits of a multiple of 3 will always result in one of the one-digit multiples of 3: 3, 6, or 9.

If we wrote in base 8, instead of base 10, then 7 would have the property: if you write a number in base 8 and add the digits (in base 8) until you get down to a single digit between 1 and 7, then the multiples of 7 will always end yielding 7, for precisely the same reason. And if we wrote in base 16, then 15 (or rather, F) would have the property. In general, if you write in base b, then b1 has the property.

This is a special case of casting out nines, which in turn is a special case of modular arithmetic. It is what is behind many divisibility tests (e.g., for 2, 3, 5, 9, and 11).

Coda. This reminds me of an anecdote a professor of mine used to relate: a student once came to him telling him he had discovered a very easy way to test divisibility of any number N by any number b: write N in base b, and see if the last digit is 0. I guess, equivalently, you could write N in base b+1, and add the digits to see if you get b at the end.

Attribution
Source : Link , Question Author : JD Isaacks , Answer Author : Arturo Magidin

Leave a Comment