The Maths Problem You Will Never Solve (Probably)!

Is this a probability problem?

For example, what is the probability that the OP will actually give us a problem to solve?

Reply 3

Pheylan

but there isn't a problem, so i have already solved it. :holmes:

Reply 4

milliondollarcorpse

It's even in the wrong forum...

Biggest fail thread of the day

Reply 5

andy12691

10

Then your question is 'prove that a number whose sum of all digits divisible by 3, is divisible by 3' ?

Reply 6

Guess Who =]

there was nothing to SOLVE weird person

Reply 7

Cities

The problem is to prove it

Reply 8

Pheylan

but there isn't a problem, so i have already solved it. :holmes:

ergosum

Here is the maths problem. To test whether a number is divisble by 3, you add the digits and check that the sum is divisible by 3 e.g. with 333, you add together the digits and check the sum is divisble by 3 (3+3+3 = 9 so 333 is divisble by 3).

A puzzling problem at first, it isn't as bad as it may look and hasn't been solved completely, but if you believe otherwise post in your proofs (or those of others who have solved this).

So are you asking us to prove it?
I would, but sadly I encountered a proof in a book as recently to ensure it has remained in my memory and a plagiarist I am not.

Reply 9

ibysaiyan

ergosum

Here is the maths problem. To test whether a number is divisble by 3, you add the digits and check that the sum is divisible by 3 e.g. with 333, you add together the digits and check the sum is divisble by 3 (3+3+3 = 9 so 333 is divisble by 3).

A puzzling problem at first, it isn't as bad as it may look and hasn't been solved completely, but if you believe otherwise post in your proofs (or those of others who have solved this).

Great.. and for once i wanted to plant in my mathematica seed.....

Reply 10

username227498

n1r4v

The problem is to prove it

But that's not solving, that's proofing? :confused:

Reply 11

username350927

14

I have solved it :smile:

as there is nothing to solve?

Reply 12

Rubgish

13

To be honest this thread is a total joke. Not only did I prove this in my camb interview, I also went ahead to prove that the same method works for 9 and a similar method works for 11. Methinks the OP is trying to get us to do his/her homework.

Reply 13

But a hint for anybody who is trying to prove it, but not much of a hint anyway because if I was asked to prove it I would do this instinctively, and I am under the impression others would do so as well.

Spoiler

Reply 14

Oh I Really Don't Care

17

You do realise this is not a problem. It has been "solved" and is relatively straightforward?

*sigh*..

OP

Alright people, I know its not a problem I just wanted to get attention for it. I've solved it, just wanted to see how other people have gone about it, and if there are different ways. I wasn't expecting this many posts so quickly I was going to delete it, but thanks anyway...although some of you guys seem quite annoyed...

Reply 17

stephen_king-evans

4

this is a simple way

Basically:

100a + 10b + 1c (HTU)

100+10+1=111

111/3 = 37 (so basically you can forget about place value)

if a+b+c is divisible by 3 then it will work

QED

Reply 18

Blu3j4yw4y

I get why it works, but I don't get how you would prove it with proper mathematical formula.

Essentially each digit is a measure of how far it is above a multiple of 3.

E.G. 10 is one unit away from a multiple of three (9).
20 is two units away from a multiple of three (18).
30 is three units away from a multiple of three (27).
40 is four units away from a multiple of three (36).
50 is five units away from a multiple of three (45).
60 is six units away from a multiple of three (54).

etc...

This works as you go up by tens too

100 is one unit away from a multiple of three (99).
200 is two units away from a multiple of three (198).
300 is three units away from a multiple of three (297).
etc..

1000 is one unit away from a multiple of three (999).
2000 is two units away from a multiple of three (1998).
3000 is three units away from a multiple of three (2997).

So, essentially. Each digital can be taken as a measure of how far the whole number is away from a multiple of three.

E.g. take 4285

(4 units away from a multiple of three) + (2 units away from a multiple of three) + (8 units away from a multiple of three) + (5 units away from a multiple of three)

=19

Therefore we can deduce that 4285 is 19 digits away from a multiple of three (4266, of course) therefore, as this 19 unit difference is not divisible by three, 4285 itself cannot be a multiple of three

Another example: 4575

(4 units away from a multiple of three) + (5 units away from a multiple of three) + (7 units away from a multiple of three) + (5 units away from a multiple of three)

= 21

This means 4575 is 21 units away from a multiple of three (4554) and as the difference itself is also dividible by three we know 4575 must be a multiple of three. (Adding two multiples of any given number will always give you another multiple).

I don't know how to write a mathematical proof, but that's why.

Sorry I stopped reading your proof halfway through, so I don't know if it is complete, but there is really easy way of doing it (not Unbounded's method).

For all:

1458=1(10^3)+4(10^2)+5(10^1)+8(10^0)

Reply 19