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Thread: arithmetic problem

  1. #1
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    arithmetic problem

    hello i tried solving the following problem using fermat little theoreme but had difficulties advancing , perhaps i'm not aware of some necessary theorems :

    We write on a board all the natural numbers from 1 to 500 .We arbitrarily chose two , three, four or five numbers then we erase them and replace them by the remainder of the euclidian division of their sum by 13. In the following step , we do the same thing with a new list of natural numbers remaining on the board . we repeat this process many times.

    After a particular number of steps , only two numbers stay written on the board , one of them is 102. FIND THE OTHER NUMBER

    i couldn't solve this problem so if anyone got some advice or help i would be very grateful.
    Last edited by aray; Aug 23rd 2017 at 07:59 AM.
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  2. #2
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    Re: arithmetic problem

    10
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  3. #3
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    Re: arithmetic problem

    thank you for the answer but i didn't know how to find it sir
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  4. #4
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    Re: arithmetic problem

    13 is a prime and doesn't divide 102+a with a being the unknown so (102+a)^12≡1mod13 but didn't know what to do after
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  5. #5
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    Re: arithmetic problem

    yes i found 10 thank you sir
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  6. #6
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    Re: arithmetic problem

    i just didn't read the problem correctly so i didn't use the sum of all the numbers
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  7. #7
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    Re: arithmetic problem

    there was no need to use fermat; i saw that whenever you remove numbers and replace them by the remainder of their sum than the remainder of the sum of all the remaining numbers +the replacing remainder divided by 13 gives you 8 as a remainder .for example we remove 500 and 499 we get 11 and numbers from 1to 498 so the sum of numbers from 1 to 498 +11 divided by 13 gives you as a remainder 8 and in the final step we get 10 as the other number
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  8. #8
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    Re: arithmetic problem

    thank you for helping me reach the idea mr idea
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