This works with the number 3 and the number 9.
Is 987659387043185671335 divisible by 9? Add up each digit: 9+8+7+6+5+9+3+8+7+0+4+3+1+8+5+6+7+1+3+3+5 = 108 Add up each digit of that: 1+0+8 = 9.
9 is divisible by 9, so so is 987659387043185671335.If you can give me a reason why this works, you're cleverer than I am.
+1 :-)
I don't know about that. I expect pure mathematicians would say something along the lines of it's just the beauty and magic of maths.
In the menu-box on the right-hand side on the link I posted are some more interesting (?) facts about division:
A number is divisible by 3 if the sum of its digits is divisible by 3. (That one is the only one I knew)
A number is divisible by 4 if the last two digits of the number are divisible by 4 .
A number with at least 3 digits is divisible by 8 if its last three digits form a number divisible by 8.
A number is divisible by 11 if the alternating sum of its digits is divisible by 11. IOW for the number abcdefghij, if a-b+c-d+e-f+g-h+i-j is divisible by 11, then the number itself is divisible by 11. (This one is the most intriguing, and, like division by 9, difficult to see an intuitive explanation!).
A number is divisible by 25 if it ends with 00, 25, 50, or 75.
Well why does it matter in fact? Brian
The reason for the "four" rule is pretty obvious and, given the underlying pattern in multiples of 11, I can't help suspecting that it might become evident with a bit of algebra.
It all has to do with a decimal based system of representation
These two are kind of obvious since 100 is divisible by 4 and 1000 is divisible by 8.
If my memory was good enough to remember all those sorts of things, I guess I wouldn't need to use a calculator.
Because it's very odd.
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