eli@uw-june (Eli Messinger) (08/24/83)
There are actually quite a few "simple division properties" for base
10 arithmetic:
2: Last digit 0,2,4,6,8
3: Digits sum to a number divisible by 3
4: Last two digits (taken as a number twixt 00 and 99)
divisible by 4
5: Last digit 0 or 5
6: Divisible by 2 and 3
7: ?
8: ?
9: Digits sum to a number divisible by 9
10: Last digit 0
Does anybody have a proof of the rule for 3? I had it in the notes from
a number theory course I took -- which I can no longer find. (Send to me
by mail...)
... uw-june!elistan@clyde.UUCP (08/25/83)
If N is the number and r is the radix, (All the digits of N added together) mod (r-1) = N mod (r-1) Can anyone here supply a proof? P.S. This means it is easy to test binary numbers for divisibility by 3 (r-1=2**2-1), 7 (r-1=2**3-1), 15 (r-1=2**4-1), and so on. Especially helpful for hardware. Stan King phone: 201-386-7433 Bell Labs, Whippany, NJ Cornet: 8+232-7433 room 2A-111 uucp: clyde!stan
rlr@pyuxn.UUCP (08/25/83)
What hasn't been appropriately noted here is the recursion in the nine rule.
9: a number is divisible by 9 if the sum of its digits is divisible by 9
OK, so this rule can be applied recursively...
9877898798979877898798976 --> sum of digits is 198. Is it divisible by 9?
198 (1 + 9 + 8) --> sum of digits is 18. Is it divisible by 9?
18 (1 + 8) --> sum of digits is 9.
It always finally resolves to 9.
What about multiplication tricks like how to multiply two digit numbers by 11
(Add the two digits and place the result between them; if > 10 add resulting
ten's digit to hundreds place. 58 * 11 = {5 -- 13 -- 8} = 638). Any other
multiplication tricks? Richlaura@utcsstat.UUCP (Laura Creighton) (08/27/83)
another trick, or how I learned those darn multiplication tables (before then I added them up in my head. Good thing that I can add fast!). 6 times table -- 6*X, where X is even, is YX, where Y is one half of X. laura creighton utzoo!utcsstat!laura
ziegler@lzmi.UUCP (08/29/83)
My favorite base-10 numeric property is this: If you transpose any to adjacent digits in a number, the difference between the original number and the resulting number is always divisible by nine, and of the same order of magnitude as the rightmost transposed digit. The proof is left up to the interested student... (Whoops -- make that "any two adjacent digits") This one comes in very handy for bank tellers and accountants.
lake@alberta (08/30/83)
A simple trick for squaring a two-digit number ending in a 5 is to multiply the 10's digit by one plus itself and then append '25' to the end of the product. e.g. 75*75 can be calculated by multiplying (7+1)*7=56 and appending 25 to 56, so the product is 5625. Rob Lake University of Alberta alberta!lake Edmonton, Alberta
chongo@nsc.uucp (Curt Noll) (08/31/83)
>another trick, or how I learned those darn multiplication tables >(before then I added them up in my head. Good thing that I can >add fast!). > >6 times table -- 6*X, where X is even, is YX, where Y is one half >of X. > >laura creighton utzoo!utcsstat!laura i assume by YX you mean Y*10+X. i learned the 9*X (1<=X<=10) by: 9 * X = 10*(X-1) + (10-X). one can do this on ones fingers by holding down the digit X. the number of fingers to the left (i.e., twards 1) is the number of tens, and the number of fingers to the right is the number of ones. chongo /\99/\