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=>

\(3^1 = 3~3, 3^2 = 9~9, 3^3 = 27~7, 3^4 = 81~1, 3^5~3, …\)

So, the units digits of \(3^n\) have a period of \(4\).

They form the cycle \(3 -> 9 -> 7 -> 1.\)

Thus, \(3^n\) has the units digit of \(3\) if \(n\) has a remainder of \(1\) when it is divided by \(4\).

The remainder when \(2021\) is divided by \(4\) is \(1\), so the units digit of \(1993^{2021}\) is \(3\) and \(a = 3.\)

\(|a - 1| + |a - 2| + |a - 3| + |a - 4| + |a - 5|\)

\(= |3 - 1| + |3 - 2| + |3 - 3| + |3 - 4| + |3 - 5|\)

\(= 2 + 1 + 0 + 1 + 2\)

\(= 6\)

Therefore, E is the correct answer.
Answer: E
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If \(a\) is the remainder when \(1993^{2021}\) is divided by 10

Theory: Remainder of a number by 10 is same as the unit's digit of the number

(Watch this Video to Learn How to find Remainders of Numbers by 10)

Using Above theory Remainder of \(1993^{2021}\) by 10 = unit's digit of \(1993^{2021}\)

Unit's digit of \(1993^{2021}\) = Unit's digit of \(3^{2021}\)

Now to find the unit's digit of \(3^{2021}\), we need to find 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 2021 by 4 and check what is the remainder
=> 2021 divided by 4 gives 1 remainder

=> \(3^{2021}\) will have the same unit's digit as \(3^1\) = 3

=> a = Remainder of \(1993^{2021}\) by 10 = 3

=> \(|a - 1| + |a - 2| + |a - 3| + |a - 4| + |a - 5|\) = \(|3 - 1| + |3 - 2| + |3 - 3| + |3 - 4| + |3 - 5|\)
= \(|2| + |1| + |0| + |-1| + |-2|\) = 2 + 1 + 0 + 1 + 2 = 6

(Watch this video to Learn about the Basics of Absolute Values)


So, Answer will be E
Hope it helps!

MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

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