Casting Out Nines - A Variation On The Explanation

A Variation On The Explanation

A nice trick for very young children to learn to add nine is to add ten to the digit and to count back one. Since we are adding 1 to the ten's digit and subtracting one from the unit's digit, the sum of the digits should remain the same. For example 9+2=11 with 1+1=2. When adding 9 to itself, we would thus expect the sum of the digits to be 9 as follows: 9+9=18 (1+8=9) and 9+9+9=27 (2+7=9). Let us look at a simple multiplication: 5×7=35 (3+5=8). Now consider (7+9)×5=16×5=80 (8+0=8) or 7×(9+5)=7×14=98 (9+8=17 1+7=8).

Any positive integer can be written as 9 × n + a where 'a' is a single digit 0 to 8 and 'n' is any positive integer. Thus, using the distributive rule (9 × n + a)×(9 × m + b)= 9 × 9 × n × m + 9 ×(am+bn) +ab. Since the first two factors are multiplied by 9, their sums will end up being 9 or 0, leaving us with 'ab'. In our example, 'a' was 7 and 'b' was 5. We would expect in any base system the number before that base would behave just like the nine.