# Thread: arithmetic problem

1. ## 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.

10

3. ## Re: arithmetic problem

thank you for the answer but i didn't know how to find it sir

4. ## 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

5. ## Re: arithmetic problem

yes i found 10 thank you sir

6. ## Re: arithmetic problem

i just didn't read the problem correctly so i didn't use the sum of all the numbers

7. ## 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

8. ## Re: arithmetic problem

thank you for helping me reach the idea mr idea