What is the remainder when \((9^{2n + 3})(5^n)\) 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 \((9^{2n + 3})(5^n)\) by 10 = Unit's digit of \((9^{2n + 3})(5^n)\)

Now, Let's find the unit's digit of \((9^{2n + 3})\)

Now to find the unit's digit of \(9^{2n + 3}\), we need to find the pattern / cycle of unit's digit of power of 9 and then generalizing it.
Unit's digit of \(9^1\) = 9
Unit's digit of \(9^2\) = 1
Unit's digit of \(9^3\) = 9
Unit's digit of \(9^4\) = 1
=> Units' digit of Odd Power of 9 = 9
=> Units' digit of Even power of 9 = 1

2n + 3 = Even + Odd = Odd
=> Unit's digit of \(9^{2n + 3}\) = 9

Unit's digit of \(5^n\) = 5 (as all positive powers of 5 have units' digit of 5)

=> Unit's digit of \((9^{2n + 3})(5^n)\) = Unit's digit of 9 * 5 = 5

So, Answer will be D
Hope it helps!

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

BrushMyQuant very helpful explanation! This way is far more straightforward than my process; however, I'm hoping someone can explain why my method doesn't work.

I originally broke down the numbers into their primes so [(3^4n+6)(5^n)][/(2)(5)]
Since the single 5 in the denominator will always cancel leaving just the 2, I assumed the answer had to be 1.