The remainder when (2021)²⁰²³ is divided by 7 is :

Q: Wait, why do we simplify the base first? Can't we just calculate it directly?

Calculating $(2021)^{2023}$ directly is impossible because the number is astronomically large. Modular arithmetic allows us to work with the remainder of the base, which is $5$, because $(a \cdot b) \pmod n = (a \pmod n \cdot b \pmod n) \pmod n$. This property is the foundation of modular exponentiation.

Q: Why use $-2$ instead of $5$? Isn't $5$ the correct remainder?

Both are correct! $5 \equiv -2 \pmod 7$. However, using $-2$ is a strategic choice. Powers of $2$ are much smaller and easier to manage than powers of $5$. It simplifies the arithmetic significantly, reducing the chance of calculation errors.

Q: Why is the final answer $5$ and not $-2$?

In modular arithmetic, we typically express the final remainder as a positive integer between $0$ and $n-1$. Since we arrived at $-2 \pmod 7$, we simply add the divisor $7$ to convert it to a positive equivalent: $-2 + 7 = 5$.

Question

The remainder when $(2021)^{2023}$ is divided by 7 is :

Options:

(A) 1

(B) 2

(C) 5

(D) 6

Sigma Percentile · Accepted Answer

### The Clockwork Universe of Modular Arithmetic

My dear student, welcome to the fascinating world of modular arithmetic. Today, we are going to tackle a problem that looks like a monster: finding the remainder when $2021^{2023}$ is divided by $7$.

If you try to compute this directly, you will be here until the end of time. But in the realm of mathematics, we don't use brute force; we use elegance. We use the "Modulo Clock."

### Phase 1: Simplifying the Base

Imagine a clock with only $7$ numbers, from $0$ to $6$. When we divide by $7$, we are essentially moving around this clock.

The first step is to simplify our base, $2021$. We ask: where does $2021$ land on our $7$-hour clock? We perform the division:

$$2021 = 7 	imes 288 + 5$$

This tells us that $2021$ is equivalent to $5$ in the world of modulo $7$. So, our problem transforms from the terrifying $2021^{2023} \pmod 7$ into the much friendlier $5^{2023} \pmod 7$.

### Phase 2: The Negative Remainder Trick

Now, we have $5^{2023} \pmod 7$. While $5$ is correct, it is not the most efficient path.

Look at our clock again. If you are at $5$, you can either go forward $5$ steps or backward $2$ steps to reach the same position. Thus, $5 \equiv -2 \pmod 7$.

Why is this a breakthrough? Because working with $2$ is infinitely easier than working with $5$. Our expression becomes $(-2)^{2023} \pmod 7$.

Since $2023$ is an odd power, the negative sign persists: $-(2^{2023}) \pmod 7$. We have tamed the beast!

### Phase 3: The Cycle of Powers

Now, we need to understand how powers of $2$ behave modulo $7$. Let's trace them:

$$2^1 \equiv 2 \pmod 7$$
$$2^2 \equiv 4 \pmod 7$$
$$2^3 = 8 \equiv 1 \pmod 7$$

Do you see the magic? $2^3$ gives us a remainder of $1$. This is the holy grail of modular arithmetic!

Because $1$ raised to any power is just $1$, the powers of $2$ will repeat in a cycle of $3$. We just need to see how many cycles of $3$ fit into our exponent, $2023$.

### Phase 4: The Final Synthesis

We divide the exponent $2023$ by $3$:

$$2023 = 3 	imes 674 + 1$$

This means $2^{2023} = (2^3)^{674} 	imes 2^1$. Substituting our cycle value, we get:

$$(1)^{674} 	imes 2^1 = 1 	imes 2 = 2$$

But wait! Remember the negative sign we parked outside in Phase 2? We must bring it back. So, our result is $-2 \pmod 7$.

Finally, to convert this to a positive remainder, we add the divisor: $-2 + 7 = 5$. And there you have it—the remainder is **$5$**. You have successfully navigated the clockwork universe of modular arithmetic!

Animated Solution for Mathematics - Binomial Theorem: The remainder when $(2021)^{2023}$ is divided by 7 is :

Select Answer:

Visualized Solution

The Modulo Objective

Simplifying the Base

Locating on the Mod Clock

The Negative Remainder Trick

Rewriting the Expression

Finding the Cycle of Powers

Breaking Down the Exponent

Computing the Final Remainder

Converting to a Positive Remainder

The Sigma Insight: Binomial Expansion for Positive Integral Index

The Clockwork Universe of Modular Arithmetic

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The remainder when (2021)2023 is divided by 7 is :