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Re: What is the remainder when 13^17 + 17^13 is divided by 10?
[#permalink]
16 Sep 2022, 08:41
We need to find what is the remainder when \(13^{17} + 17^{13}\) is divided by 10
Theory: Remainder of sum of two numbers = Sum of their individual remainders Remainder of any number by 10 = Unit's digit of that number
=> Remainder of \(13^{17} + 17^{13}\) by 10 = Remainder of \(13^{17}\) by 10 + Remainder of \(17^{13}\) by 10
Unit's digit of \(13^{17}\)
= Unit's digit of \(3^{17}\)
We can do this by finding the pattern / cycle of unit's digit of power of 3 and then generalizing it.
Unit's digit of \(3^1\) = 3 Unit's digit of \(3^2\) = 9 Unit's digit of \(3^3\) = 7 Unit's digit of \(3^4\) = 1 Unit's digit of \(3^5\) = 3
So, unit's digit of power of 3 repeats after every \(4^{th}\) number. => We need to divided 17 by 4 and check what is the remainder => 17 divided by 4 gives 1 remainder
=> \(3^{17}\) will have the same unit's digit as \(3^1\) = 3 => Unit's digits of \(13^{17}\) = 3
Unit's digit of \(17^{13}\)
= Unit's digit of \(7^{13}\)
We can do this by finding the pattern / cycle of unit's digit of power of 7 and then generalizing it.
Unit's digit of \(7^1\) = 7 Unit's digit of \(7^2\) = 9 Unit's digit of \(7^3\) = 3 Unit's digit of \(7^4\) = 1 Unit's digit of \(7^5\) = 7
So, unit's digit of power of 7 repeats after every \(4^{th}\) number. => We need to divided 13 by 4 and check what is the remainder => 13 divided by 4 gives 1 remainder
=> \(7^{13}\) will have the same unit's digit as \(7^1\) = 7 => Unit's digits of \(17^{13}\) = 7
=> Unit's digits of \(13^{17}\) + Unit's digits of \(17^{13}\) = 3 + 7 = 10
But remainder of \(13^{17} + 17^{13}\) by 10 cannot be more than or equal to 10 => Remainder = Remainder of 10 by 10 = 0
So, Answer will be 0 Hope it helps!
Watch the following video to learn the Basics of Remainders
gmatclubot
Re: What is the remainder when 13^17 + 17^13 is divided by 10? [#permalink]