We need to find What is the remainder when \((2^{16})(3^{16})(7^{16})\) is divided by 10

\((2^{16})(3^{16})(7^{16})\) = \((2 * 3 * 7)^{16}\) = \(42^{16}\) = \((40 + 2)^{16}\)

Now, we have split 42 into two numbers, one (40) is a number closer to 42 and a multiple of 10 and other is a small number

Now, if we expand this using Binomial theorem then we will get all terms except the last term as a multiple of 40 => A multiple of 10

=> All terms except the last term will give us a remainder of 0 when divided by 10

=> Remainder of \((2^{16})(3^{16})(7^{16})\) by 10 is same as remainder of the last term = 16C16 * 2^16 * 40^0 = 2^16 by 10

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

Now, Let's find the unit's digit of \(2^{16}\) first.

We can do this by finding the pattern / cycle of unit's digit of power of 2 and then generalizing it.

Unit's digit of \(2^1\) = 2
Unit's digit of \(2^2\) = 4
Unit's digit of \(2^3\) = 8
Unit's digit of \(2^4\) = 6
Unit's digit of \(2^5\) = 2

So, unit's digit of power of 2 repeats after every \(4^{th}\) number.
=> We need to divided 16 by 4 and check what is the remainder
=> 16 divided by 4 gives 0 remainder

=> \(2^{16}\) will have the same unit's digit as \(2^4\) = 6
=> Unit's digits of \(2^{16}\) = 6

But remainder of \(2^{16}\) by 10 = 6

So, Answer will be D
Hope it helps!

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

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