Finding patterns is the key here.

17*1 = 1
17*17 = ..9
289*17 = ...3
...3*17 = ....1

Similarly, if we calculate the pattern for 13, we'll observe that it repeats in a similar way 13 does, after 4 multiplications.
So remainders will be (3+7)%10=0

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

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