what you have is....

Originally posted by rubberjaw30
or, you can just say that if the sum of digits is divisible by three, then the number is divisible by three...

Swap 'three' for 'nine' and it's still true.

Originally posted by TheMaster37
[...] There is even one for multiples of 7, but that's a bit trickier.

The algorithms for testing divisibility by 7 I've come across are all horribly complicated, do you know one that's relatively simple? From the top of my head, these are the divisibility tests I know up to 25:

1: trivial
2: even number
3: sum of digits is divisible by 3
4: last two digits divisible by 4
5: ends in 0 or 5
6: divisible by 2 and 3
7: ???
8: last three digits divisible by 8
9: sum of digits equals 9
10: ends in 0
11: alternating sum of digits divisible by 11
12: divisible by 3 and 4
13: ???
14: divisible by 2 and 7
15: divisible by 3 and 5
16: last four digits divisible by 16
17: ???
18: divisible by 2 and 9
19: ???
20: divisible by 4 and 5 (more simply put: ends in 00, 20, 40, 60 or 80)
21: divisible by 3 and 7
22: divisible by 2 and 11
23: ???
24: divisible by 3 and 8
25: ends in 00, 25, 50 or 75.

Does anyone know a clever way to fill some of the blanks?

7 is a prime number so it would probably be just a difficult as any other. or i just might be saying this cuz its all i know. or maybe youve all gone through university and know more cuz im only gr.12. or maybe im writing just cuz!

The well known rule for divisibility by 7 is easier than it sounds:

1) Take the last digit of the number and double it.
2) Subtract the number obtained from 1) from the remaining number.
3) If this number is divisible by 7, so was the original number, otherwise it wasn't (zero is considered to be divisible by 7).

Here's an example:

Original number: 42357
Last digit is 7, doubling this makes 14, so we subtract 14 from 4235 (the original number without the last digit) making 4221.

To determine if 4221 if divisible by 7 we do the same thing - giving us the sum 422 - 2 = 420.

And again with 420: 42 - 0 = 42.

Just for fun we'll do the algorithm on this last number, 42: 4 - 4 = 0

So 42357 is divisible by 7.

The problem, of course, is that we have to do so many calculations to determine whether the original number was divisible by 7, probably more than if we had done a simple long division.

to find it easier than that if you look at 4221 if you split it into two parts 42 and 21 are both divisible by 7

Originally posted by tournymangr
to find it easier than that if you look at 4221 if you split it into two parts 42 and 21 are both divisible by 7

how does that work for three digits?

Originally posted by joe shmo
how does that work for three digits?

........i hate being proved wrong......

... or 4305 ?

new thing!! what fish weighs halve its weight plus 10 pounds??

doesn't 24 work?

Originally posted by tournymangr
new thing!! what fish weighs halve its weight plus 10 pounds??

a 20 pounder ... but I guess this is a trick question?????????

Originally posted by wolfgang59
a 20 pounder ... but I guess this is a trick question?????????

yeah its 20 but there was no trick. actually im just showing people the really easy puzzles my math class does b4 we start working very easy no??

Originally posted by TheMaster37
Same here! It's funny how numbers can behave so nicely 🙂

They may be the only logical thing left in this world

Originally posted by tournymangr
yeah its 20 but there was no trick. actually im just showing people the really easy puzzles my math class does b4 we start working very easy no??

your asking for a fish, not the wieght of a fish?

Originally posted by joe shmo
your asking for a fish, not the wieght of a fish?

I don't think he could possibly know what is the wieght of a fish.

Maybe this could help

http://www.softpile.com/Education/Science/Review_60001_index.html