intreasting fact?

Question

if you have to find if a 3digit number is divivisable by 3, i.e. 345 or 696 add the numbers up then divide by 3 i.e. 345 = 13 not divisable 696= 21 / 3 equals 7 so it can be devided by 3


haha no real use just crap

intreasting fact?
if you have to find if a 3digit number is divivisable by 3, i.e. 345 or 696 add the numbers up then divide by 3 i.e. 345 = 13 not divisable 696= 21 / 3 equals 7 so it can be devided by 3


haha no real use just crap



it is theres an number which doesnt work

Answer 1

345 =12 it is divisible. But thanks, never heard that little trick before.

Answer 2

It's a shame you are hardly numerate: it is "useless" to you because you are not very good at simple addition. 3+4+5 is 12, not 13, and 345 IS divisible by three. At least you got the OTHER example correct. It's a shame you think it only works with 3-digit numbers: it works with numbers of ANY size. Also note that once you have added the individual digits you can repeat the process until you get to a one-digit number. So with 696: 6+9+6 = 21, and 2+1 = 3. If you finally get to 3, 6 or 9, the original is divisible by 3.
Also if it adds to 3, 6 or 9 and the original number is EVEN, it is divisible by 6. If the result becomes 9, the number is divisible by 9 (as well as 3).
THis is often called "casting out nines", because you can do it by adding digits and dismissing them when they total 9. In the decimal system it ALWAYS words. If you use another number base, the same trick words, but it is based around another number: in octal (base 8) for example, you "cast out sevens". SO note:
(1) the method works with ANY size number -- not just three digits.
(2) you can repeat the process on the sum, rather than having to divide by 3.
(3) it can tell you if the number is divisible by 6 or 9 just as easily/

I'll give you another related fact that is VERY interesting: with a number ANY length, altrnately add and subtract the three-digit numbers which are parts of it. For example: 38,392,658 you calculate 38 - 392 + 658 = 304 Now with this result you can tell whether the original number is divisible by 7, 11 or 13. If IT is divisible by any of these, so is the big number. 304 is NOT divisible by 7 or 11 or 13, so neither is 38,392,858. Despite what you say, being able to factor a number properly is definitely of real use!

Answer 3

adding to Heinz M's list

7 is divisible if given the number ABCDEF(this is a 6 digit number but it can be any size)
ABCDE -2*F. You can take the new number and remove the last number. subtract double the number and if the new number is divisble by 7 the original number is disible by 7. You can repeat this until the number becomes 0 or 7(if the number becomes negative, just change it to positive)
903 becomes 90 -(3*2) =84
84 becomes 8-(4*2) = 0 and therefore 903 is divisible by 7


A number is divisble by 11 if the sum of every other number subtracted from the sum of the other numbers is divisible by 11.
5032016 becomes 5+3+0+6 -(0+2+1) =14-3 =11
and therefore 5032016 is divisble by 11

Answer 4

The same property holds for 9 also. The reason is that 10 divided by 3 (9) leaves a remainder of 1, and because no prime divides both 3 (9) and 10.

A proof of this property for 3 (it would be almost identical for 9).

Let a=(a0,a1,a2,a3,a4, . . ., an) be the number a0+a1•10+a2•10^2+ . . .+an•10^n where for every ai 0≤ai≤9 (if you considered the ais as digits, the number would look like (an)(an-1) . . .(a2)(a1)(a0)).

Note on number theory:
if a number n has a remainder of r when divided by m, we say n=r (mod m) or "n is congruent to r modulo m."
Some properties of this:
If s,t,r and n are numbers such that no prime divides s and n both, and s=r (mod n), then s^t=r^t (mod n).

Back to our problem:

a=a0+a1•10+a2•10^2+ . . .+an•10^n (mod 3) = a0+a1•(1)+a2•(1)^2+ . . . +an•(1)^n=a0+a1+a2+ . . . +an (mod 3). If 3 divides a, then 3 divides the sum of it's digits. Likewise, if 3 divides the sum of a's digits, 3 divides a.

Therefore this will occur for all such numbers, and there does not exists a number such that this doesn't happen.

A result from this is that the remainder of a number divided by 3 is the same as the remainder of the sum of it's digits divided by 3.

Answer 5

Math is helpful in many situations, and is anything but 'crap'.
The divisibility rules are (and this means divisible without remainder!)
1. Any even number is divisible by 2
2. Any number where its cross sum is divisible by 3, is also divisible by 3 (like 1,242 go 1+2+4+2= 9 which is divisible by 3, therefor 1,242 is also divisible by 3)
3. Any number, where the last 2 digits are divisible by 4, is divisible by 4.
4. Any number ending in either 5 or 0 is divisible by 5
5. Any number that is divisible by 3 AND by 2 (see 1. and 2. above) is divisible by 6.
6. 7 is the only number that has no easy rule
7. Any number, where the last 3 digits are divisible by 8, is divisible by eight
8. Any number where its cross sum is divisible by 9, is also divisible by 9 (like 1,242 go 1+2+4+2= 9 which is divisible by 9, therefor 1,242 is also divisible by 9)

Answer 6

the sum of the digits of any number divisible by three, is also dividible by three. In fact, that is one of the divisibility rules to check if a number is divisible by three.

For example, how do you know that 416748736284983724831 is divisible by three. You add up the integers and it gives you 105 and if you can't tell that 105 is divisible by three then you do the same thing to 105 which gives you 6 and since 6 is divisible by three, implies 105 is divisible by three, implies 416748736284983724831 is divisible by three.

It is a pretty cool rule because as you can see it converges pretty fast, giant numbers become smaller really fast. I was able to reduce 416748736284983724831 to a single digit number in three iterations.

The same is also true for any other (integral) power of nine (like for 9, 27, etc...)

Answer 7

if you have a really long number, you dont have to add them all up... just drop all the 3's, 6's, 9's, any 1's +2's etc cancell each other out... 5+7 cancels out, etc, any combination divisible by 3... cancel out till what you end up with... is it divisible by 3?
if you end up with 8 and 4... well you got it... =12... =divisible by thre

Answer 8

there ways of finding whether all numbers can divide into a number apart from 7 which is annoying

Answer 9

Its not limited to three digit numbers.

Answer 10

Its not a great trick man .one of my 2nd grade lesson