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Is 7^103 + 6^103 divisible by 13 if yes then why if no then why not??

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posted Jul 29, 2015 by anonymous

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1 Solution

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7^103 + 6^103 divisible by 13

using fermat's little theorem when P is a prime number ,,,,,,
any number ( let a ) has a power of p-1 i.e (a^p-1 ) when divided by p has a remainder 1,,,,

i.e 7^12 when divided by 13 has a remainder 1 and 6^12 when divided by 13 has a remainder 1

i.e 7^{(12*8)+7} ==== 7^7 and similarly 6^7

so the question reduces to find the remainder when 7^7 + 6^7 is divided by 13 ,,, now we can do some manupulations to get answer

remainder{ 7^7 + 6^7}/13 ==== remainder{ (-6)^7 + 6^7 } / 13 ;;; using negative remainder concept

remainder {0}/13 === remainder will be 0 , hence 13 will completely divide this number

solution Sep 28, 2015 by Ankit Kamboj
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