Many of you know I am an elementary school teacher, third grade to be exact. Multiplication (and memorization of the facts) is the Big Math Deal in this grade. Push, push, push, memorize, memorize, drill and kill…
To help them learn and memorize facts, I teach them a finger trick for the 9s (I’m sure some of you know it). This is how it works: determine the factor you want to multiply 9 by. Then hold your hands in front of you, palms out, and count from left to right to that factor’s number (i.e., if you’re multiplying 9 x 3, you’d put down the 3rd–the left middle–finger). The answer is the number of fingers to the left of the down finger (in the tens place) and the number of fingers to the right of it (in the ones place). In the example of 9 x 3, two fingers are to the left, seven to the right: Wha-la, 27. This trick works for all factors up to 10.
Another way to look at the the trick is this. The sum of the digits will always equal 9 (until we go to three digits) because as we add one to the first digit we take one away from the second (Because 9 = 10 - 1). I assome you can figure out why the digits of 9 x 1 sum to nine.
Picking a finger one to the left (or right) you subtract one from one digit and add to the other.
Reading this over, it doesn’t sound very difinitive. Anyway, it’s still significantly slower than memorization.
If men had wings,
and bore black feathers,
few of them would be clever enough to be crows.
Slower than memorization, yes, but it is a way to help them remember the correct answer so I don’t get tests with “55” and “26” on them. Once they learn the right answers, they start to memorize them.
I should have been clearer that the sum of the digits of a multiple of nine will be nine until 9 x 21 = 189. The sum of the digits will always be a multiple of nine.
If you look research abacus use, you will find more similar tricks. These kinds of tricks allow master abacus users to simulate the speed of electronic adding machines.
Here are the rules (up through 16) to determine if a number is divisible:
1 – any number, of course.
2 – last digit divisible by 2 (i.e., the number is even. You’ll see later why I used this wording).
3 – sum of the digits is digits is divisible by 3
4 – last two digits divisible by four
5 – ends with 5 or 0
6 – rules for 2 and 3 both apply
7 – twice the ones digit minus the tens digit is divisible by 7 (e.g., 63 – 2 X 3 = 6 - 6 = 0, which is divisible by 7. Not a particulary useful test, of course).
8 – last three digits divisible by eight.
9 – sum of the digits is divisible by 9
10 – ends with 0
11 – sum of the even numbered digits = sum of the odd numbered digits (e.g., 121. First and third digits are 1 and 1 = 2, which = the second digit).
12 – rules for 4 and 3 both apply
13 – four times the units digit plus the number formed by the other digits is divisible by 13 (not very useful, but it works – 273: 3 x 4 =12 + 27 = 39)
14 – rules for 7 and 2 both apply
15 – rules for 5 and 3 both apply
16 – last four digits are divisible by 16
“East is east and west is west and if you take cranberries and stew them like applesauce they taste much more like prunes than rhubarb does.” – Marx
Essentially correct, but not completely (209, for example, is divisible by 11). A more complete way of saying it would be to check:
(Sum of the odd position digits) - (Sum of the even position digits),
which is divisible by 11 if and only if the original number is. (9-0+2 is divisible by eleven, so 209 is, too)
For 7 and 13 (and 11 as well) you can do a similar thing, only break up the digits into groups of 3. For example, 63123073 is divisible by 13 since 073 - 123 + 063 = 13, which is divisible by 13. (And 63123073 has remainder 6 when divided by 7, since 13 does, too).