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If n^m leaves a remainder of 1 after division by 7 for all positive in
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02 Oct 2022, 09:08
Given that n^m leaves a remainder of 1 after division by 7 for all positive integers n that are not multiples of 7 and we need to find m could be equal to
Let's take each option choice and check which one satisfies the statement
(A) 2 => \(n^2\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7 Let's take n=2 we get \(2^2\) = 4 divided by 7 gives 4 remainder ≠ 1 => NOT POSSIBLE
(B) 3 => \(n^3\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7 Let's take n=2 we get \(2^3\) = 8 divided by 7 gives 1 remainder Let's take n=3 we get \(3^3\) = 27 divided by 7 gives 6 remainder ≠ 1 => NOT POSSIBLE
(C) 4 => \(n^4\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7 Let's take n=2 we get \(2^4\) = 16 divided by 7 gives 2 remainder ≠ 1 => NOT POSSIBLE
(D) 5 => \(n^5\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7 Let's take n=2 we get \(2^5\) = 32 divided by 7 gives 4 remainder ≠ 1 => NOT POSSIBLE
(E) 6 So, Answer will be E but let's check for 1 number => \(n^6\) gives 1 remainder when divided by 7 for all values of n which are not multiples of 7 Let's take n=2 we get \(2^6\) = 64 divided by 7 gives 1 remainder => POSSIBLE
So, Answer will be E Hope it helps!
Watch the following video to learn the Basics of Remainders