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The Maths Problem You Will Never Solve (Probably)! watch

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    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).
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    (Original post by 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).
    Erm...what is the problem?
    This is a known "trick" - I don't see how this is a problem we can 'solve'.
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    Is this a probability problem?

    For example, what is the probability that the OP will actually give us a problem to solve?
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    but there isn't a problem, so i have already solved it. :holmes:
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    It's even in the wrong forum...

    Biggest fail thread of the day
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    Then your question is 'prove that a number whose sum of all digits divisible by 3, is divisible by 3' ?
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    there was nothing to SOLVE weird person
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    The problem is to prove it
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    (Original post by Pheylan)
    but there isn't a problem, so i have already solved it. :holmes:
    :teehee:


    (Original post by 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.
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    (Original post by 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.....
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    (Original post by n1r4v)
    The problem is to prove it
    But that's not solving, that's proofing? :confused:
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    Oh dear.
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    Well, if you really want a proof...

    Note that for all natural numbers n:  10^n - 1 = (9)(10^{n-1} + 10^{n-2} + \cdots + 10+1) which is divisible by 3.

    Write a number N in its digits:  N = a_na_{n-1}...a_1 \iff N = \displaystyle\sum_{i=1}^n a_i10^{i-1}

    So let us consider N minus the sum of its digits:

     N - \displaystyle\sum_{i=1}^n a_i = \displaystyle\sum_{i=1}^n a_i(10^{i-1}-1) , which is divisible by 3. Hence N leaves the same remainder on division by 3 as the sum of its digits (as their difference is a multiple of 3), and so N is divisible by 3 if and only if the sum of its digits is divisible by 3.
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    I have solved it

    as there is nothing to solve?
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    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.
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    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.

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    Think about base 10.
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    You do realise this is not a problem. It has been "solved" and is relatively straightforward?
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    *sigh*..
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    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...
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    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
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    (Original post by 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(1  0^0)
 
 
 
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