Modular Arithmetic and Large Exponents

May 24, 2019

What are the last two digits of 2019²⁰¹⁹?

These types of questions are popular in tests like the AMC or MMPC, and often seem daunting, but modular arithmetic simplifies the problem down easily. Modular arithmetic can be described as using the remainder of a number when it’s divided by another number.

Before we solve the problem, let’s look at a simpler question: what is the last digit of 2019²⁰¹⁹?

Because we’re only concerned with the units digit, we only have to worry about 9²⁰¹⁹. If a value $(a+b), b \in \mathbb {Z^+} \leq 9$ were squared, then the last value would be b^2.

By the same logic, last digit of 2019ⁿ is equal to 9ⁿ regardless of what n is. Looking at the powers of 9, we get the following:

Powers of 9	Units digit of value
9¹	9
9²	1
9³	9

By the same logic, the last two digits must be only pertinent to powers of 19. Essentially, we will be finding the remainder of the value when divided by a hundred, which is what the last two digits would give us. This is where modular arithmetic comes in.

We can simplify the problem into a simple arithmetic equation:

$2019^2019 \equiv x (\mathsf{mod} 100)$

By simply inputting values into a calculator, we can arrive at the following conclusions.

$19^2 \equiv 61 (\mathsf{mod} 100)$

$61^2 \equiv 21 (\mathsf{mod} 100)$

$21^2 \equiv 41 (\mathsf{mod} 100)$

$41^2 \equiv 81 (\mathsf{mod} 100)$

$81^2 \equiv 61 (\mathsf{mod} 100)$

It’s obvious that the values now start to repeat. Furthermore, we know that the last nonrepetitive value, 41², is 19¹⁶. So lets look at $2019^{2019} \equiv x (\mathsf{mod} 16)$ in order to further simplify the problem. By just dividing, we know that $x=3$ . Thus, we arrive at the following equation: