Casting Out Nines

When you first learned your 9 multiplication facts, your teacher may have taught you a helpful memory trick that the digits of the first few multiples of 9 add up to 9.  For example, 9 x 2 = 18 and 1 + 8 = 9; or 9 x 5 = 45 and 4 + 5 = 9.   This even works with higher multiples of 9 if you extend the pattern of adding digits.  For example, 9 x 317 = 2853 and 2 + 8 + 5 + 3 = 18 and 1 + 8 = 9.

In fact, you can quickly find the remainder of any number divided by 9 in a similar fashion.  For example, if you divide 24 by 9, the remainder is 6, because 9 x 2 = 18 and 24 – 18 = 6; but you can find the remainder or 24 divided by 9 even faster by using our new shortcut of adding digits: 2 + 4 = 6.  Using a bigger example, if you divide 223 by 9 the remainder is 7, because 9 x 24 = 216 and 223 – 216 = 7; but using the shortcut, 223 divided by 9 is 2 + 2 + 3 = 7.

This brings us to a nearly-forgotten but quick method for checking mathematical calculations called “casting out nines”.  In “casting out nines”, you find the remainders of each number in an equation divided by 9 by adding up the digits, and seeing if those remainders work in the same formula.  For example, let’s say that you have just calculated that 127 + 714 = 841.  For 127, 1 + 2 + 7 = 10 and 1 + 0 = 1.  For 714, 7 + 1 + 4 = 12 and 1 + 2 = 3.  For 841, 8 + 4 + 1 = 13 and 1 + 3 = 4.  The remainders of each of these numbers (1, 3, and 4) also add up: 1 + 3 = 4, and as long as we’re not a multiple of 9 off on our answer, we can assume that our calculation is correct.

127	(1 + 2 + 7 = 10, 1 + 0 = 1)	1
+ 714	(7 + 1 + 4 = 12, 1 + 2 = 3)	+ 3
841	(8 + 4 + 1 = 13, 1 + 3 = 4)	4

“Casting out nines” also works for subtraction:

627	(6 + 2 + 7 = 15, 1 + 5 = 6)	6
– 129	(1 + 2 + 9 = 12, 1 + 2 = 3)	– 3
498	(4 + 9 + 8 = 21, 2 + 1 = 3)	3

and even multiplication:

372	(3 + 7 + 2 = 12, 1 + 2 = 3)	3
x 29	(2 + 9 = 11, 1 + 1 = 2)	x 2
10788	(1 + 0 + 7 + 8 + 8 = 24, 2 + 4 = 6)	6

For addition and multiplication, it is sometimes necessary to “cast out a nine” one extra time on the remainder’s total or product, like in the calculation 248 + 627 = 875.  For 248, 2 + 4 + 8 = 14 and 1 + 4 = 5.  For 627, 6 + 2 + 7 = 15 and 1 + 5 = 6.  For 875, 8 + 7 + 5 = 20 and 2 + 0 = 2.  The remainders of each of these numbers (5, 6, and 2) don’t seem to add up at first glance, but 5 + 6 = 11 and 1 + 1 = 2, so they do.

248	(2 + 4 + 8 = 14, 1 + 4 = 5)	5
+ 627	(6 + 2 + 7 = 15, 1 + 5 = 6)	+ 6
875	(8 + 7 + 5 = 20, 2 + 0 = 2)	11	(1 + 1 = 2)

For subtraction, it is sometimes necessary to add nine to the difference if it is less than or equal to zero, like in the calculation 317 – 178 = 139.  For 317, 3 + 1 + 7 = 11 and 1 + 1 = 2.  For 178, 1 + 7 + 8 = 16 and 1 + 6 = 7.  For 139, 1 + 3 + 9 = 13 and 1 + 3 = 4.  The remainders of each of these numbers (2, 7, and 4) don’t seem to subtract properly at first glance, but 2 – 7 = -5, and -5 + 9 = 4.

317	(3 + 1 + 7 = 11, 1 + 1 = 2)	2
– 178	(1 + 7 + 8 = 16, 1 + 6 = 7)	– 7
139	(1 + 3 + 9 = 13, 1 + 3 = 4)	-5	(-5 + 9 = 4)

The reason why “casting out nines” works with the number 9 is because it is one less than the base 10 of our numbering system.  For example, viewing 223 as written in base 10, we can rewrite it as 2 x 100 + 2 x 10 + 3, or 2 x (99 + 1) + 2 x (9 + 1) + 3, or 2 x 99 + 2 + 2 x 9 + 2 + 3, and we see that 2 x 99 and 2 x 9 are divisible by 9, but 2, 2, and 3 (the three original digits) are not divisible by 9, which means 2 + 2 + 3 is a remainder. 

223	= 2 x 100 + 2 x 10 + 3 = 2 x (99 + 1) + 2 x (9 + 1) + 3 = 2 x 99 + 2 + 2 x 9 + 2 + 3 = 2 x 99 + 2 x 9 + 2 + 2 + 3

This same algorithm for finding the remainder when divided by 9 can be used for any number (although it may be necessary to “cast out nines” a few more times).  Then in an operation like addition, we can view all three numbers as multiples of 9 plus their remainder, so 9n₁ + r₁, 9n₂ + r₂, and 9n₃ + r₃ (where r₁ < 9, r₂ < 9, and r₃ < 9) in which 9n₁ + r₁ + 9n₂ + r₂ = 9(n₁ + n₂) + r₁ + r₂ = 9n₃ + r₃.  If r₁ + r₂ ≤ 9, then r₁ + r₂ = r₃, and if r₁ + r₂ > 9, then r₁ + r₂ – 9 = r₃, and we would obtain r₁ + r₂ = r₃ if we “cast out nines” one more time.  In mathematical jargon, r₁ + r₂ º r₃ (mod 9).  A similar argument for proves “casting out nines” for subtraction, and viewing multiplication as a series of addition proves “casting out nines” for multiplication.

As you can see, “casting out nines” is a quick method to check calculations.  It was once used frequently by accountants to check their work when calculations were done by pencils and paper.  If the remainders didn’t add up properly, it was a flag for an error.  It is a pity that this fascinating mathematical method isn’t well-known any more in our age of computers, where nearly everything is calculated for us.