# 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 $$99$$, will equal $$99$$ if you add all the numbers it is made from until there is $$11$$ digit.

A quick example of what I mean:

$$9∗99=8919*99 = 891$$

$$8+9+1=188+9+1 = 18$$

$$1+8=91+8 = 9$$

This works even with really long numbers like $$43763314376331$$

Why is that? This doesn’t work with any other number. Similarly for $$1111$$ and alternating digits sums.

Not quite right, since $9\times 0 = 0$ 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:

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 $b-1$ 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.