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Unread 24 Jun 2004, 03:24   #1
'|'empest
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Maths people, enlighten me

There is a rule which states that a number diviedes by three if it's digits added together will also devide by three

for example, take the number 4572 is it divisable by 3?

4+5+7+2 = 18, so yes, the number is divisable by 3.

and according to the calculator 4572 devided by 3 is 1524


So, here's my question - how do you go about proving this rule mathematically?
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Unread 24 Jun 2004, 03:24   #2
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Re: Maths people, enlighten me

it works - end of story.
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Unread 24 Jun 2004, 03:40   #3
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Re: Maths people, enlighten me

abcde = a * 10000 + b * 1000 + c* 100 + d*10 + e
= a + 9999a + b + 999b + c + 99c + d + 9d + e
= (a + b + c + d + e) + (9999a + 999b + 99c + 9d)

the part on the right is obviously divisible by three, if the part on the left is divisible by three as well, the whole thing is
QED

(you woudl be more impressed if oyu knew how druknk i was)
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Unread 24 Jun 2004, 03:43   #4
'|'empest
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Re: Maths people, enlighten me

nice

did you generally know that or have to put thought into it?
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Unread 24 Jun 2004, 03:48   #5
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Re: Maths people, enlighten me

Quote:
Originally Posted by '|'empest
nice

did you generally know that or have to put thought into it?
thought about it.

wrote the first two lines in about a minute, stared at it while sipping beer for about five minutes, then added the third line with the exaplnantion.
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Unread 24 Jun 2004, 08:18   #6
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Re: Maths people, enlighten me

Quote:
Originally Posted by acropolis
abcde = a * 10000 + b * 1000 + c* 100 + d*10 + e
= a + 9999a + b + 999b + c + 99c + d + 9d + e
= (a + b + c + d + e) + (9999a + 999b + 99c + 9d)

the part on the right is obviously divisible by three, if the part on the left is divisible by three as well, the whole thing is
QED

(you woudl be more impressed if oyu knew how druknk i was)
man that formula is flawed as in it doesn't prove what waws oringially asked.

because the abcde can represent any number but not all numbers are divisable by 3

If number has four digits and each digit is represented by a(x),a(x+1),a(x+2,a(x+2)4...
if a(x) + a(x+1)/3 = even number
then the whole number is divisable by 3

oki bored will come back and fix
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Last edited by TheShadowMan; 24 Jun 2004 at 08:24. Reason: Gonna keep editing til i get proof
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Unread 24 Jun 2004, 08:21   #7
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Re: Maths people, enlighten me

Quote:
Originally Posted by TheShadowMan
man that formula is flawed as in it doesn't prove what waws oringially asked.

because the abcde can represent any number but not all numbers are divisable by 3
No it can't you buffoon.

The question is: 'If a+b+c+d+e is divisable by 3, then why is abcde divisible by 3?'

It's been proven that abcde is divisible by 3 if a+b+c+d+e, because its a given in the question

It's like proving that d e^x /dx = e^x, then you coming along and saying 'Ah, but d x^2/dx =! e^x'.

THAT'S NOT THE QUESTION
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Unread 24 Jun 2004, 08:27   #8
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Re: Maths people, enlighten me

yeh but you have prove that sum of digits are divisable by three to prove by the number is divisable by 3

else you can just say that the reason the number can be divided by 3 is because it is a multilpe 3
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Unread 24 Jun 2004, 08:35   #9
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Re: Maths people, enlighten me

Quote:
Originally Posted by TheShadowMan
yeh but you have prove that sum of digits are divisable by three to prove by the number is divisable by 3
Ok, let me put this in a simpler way.

IT'S NOT DIVISIBLE BY THREE IF THE SUM OF THE DIGITS AREN'T DIVISIBLE BY THREE, SO PROVING INDEPENDENT OF WHAT 'A+B+C+D+E' IS IS NOT ONLY POINTLESS, IT'S ALSO IMPOSSIBLE

Quote:
Originally Posted by TheShadowMan
else you can just say that the reason the number can be divided by 3 is because it is a multilpe 3
If it isn't a multiple of 3, it can't be divided integrally by 3. That's an obvious given to this little exercise dude.
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Unread 24 Jun 2004, 08:38   #10
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Re: Maths people, enlighten me

ah so all numbers that divisable by three, those digits of those numbers all when added together are divisable by three.

I understand now it just wasn't clear in the question
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Unread 24 Jun 2004, 08:43   #11
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Re: Maths people, enlighten me

Quote:
Originally Posted by TheShadowMan
I understand now it just wasn't clear in the question
Why not read the first line of the thread?
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Unread 24 Jun 2004, 08:50   #12
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Re: Maths people, enlighten me

every third integer starting at 3, divided by 3 is an integer.
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Unread 24 Jun 2004, 10:58   #13
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Re: Maths people, enlighten me

10.002
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Unread 24 Jun 2004, 11:24   #14
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Re: Maths people, enlighten me

Quote:
Originally Posted by Nodrog
10.002
true dat.
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Unread 24 Jun 2004, 14:32   #15
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Re: Maths people, enlighten me

a+b+c...+z=>a+b+c+...d=>9
(if you make a 1 digit of it by calcing it, like 54=>9 and 99=>18=>9)

then abcd...z is divisionable by 9
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Unread 24 Jun 2004, 14:38   #16
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Re: Maths people, enlighten me

"it's divisble by 9 iff the sum of the diigits is divisible by 9, assuming its an integer" would be an easier way of stating it.
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Unread 24 Jun 2004, 14:54   #17
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Re: Maths people, enlighten me

i suck at those things
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