The remainder when 428^2024 is divided by 21 is__________

Q: Wait, why can we just replace $428$ with $8$? Isn't that changing the number?

It is not changing the number; it is changing the perspective! Modular arithmetic is about equivalence classes. If $a \equiv b \pmod{n}$, then $a^k \equiv b^k \pmod{n}$. We are not interested in the value of $428^{2024}$ itself, but only in its remainder when divided by $21$. By replacing $428$ with $8$, we are simply working with a smaller, equivalent representative of the same remainder class.

Q: How do I know when to stop looking for a cycle?

You stop when you hit $1$ or $-1$. These are the 'holy grails' of modular arithmetic. Since $1^k$ is always $1$, and $(-1)^k$ is either $1$ or $-1$ depending on whether $k$ is even or odd, these values make calculating massive exponents trivial. In this problem, hitting $1$ at $8^2$ was our golden ticket.

Q: What if the exponent was odd, like $2025$?

That is a fantastic question! If the exponent were $2025$, you would simply use the laws of exponents: $8^{2025} = 8^{2024} \times 8^1$. Since we already know $8^{2024} \equiv 1 \pmod{21}$, the expression becomes $1 \times 8^1 = 8$. The remainder would simply be $8$.

Question

The remainder when $428^{2024}$ is divided by 21 is__________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians! Today, we are standing before a mathematical mountain. Look at the expression $428^{2024}$.

If you were to try and calculate this directly, even the most powerful supercomputer would start to sweat. But in the world of JEE Advanced, we do not use brute force; we use elegance. We use the power of modular arithmetic to tame the beast.

### Shrinking the Giant

Our first step is to simplify the base. We are looking for the remainder of $428^{2024}$ when divided by $21$.

In modular arithmetic, we only care about the remainder. So, let us divide $428$ by $21$. We find that:

$$428 = 21 	imes 20 + 8$$

This tells us that $428 \equiv 8 \pmod{21}$. Now, thanks to the beautiful property of modular arithmetic which states that if $a \equiv b \pmod{n}$, then $a^k \equiv b^k \pmod{n}$, we can replace our massive base of $428$ with the much friendlier $8$.

Our problem has now transformed into finding the remainder of $8^{2024} \pmod{21}$. Much better, right?

### The Hunt for the Cycle

Even $8^{2024}$ is still too large to calculate. We need to find a pattern, a cycle that repeats. Let us map the powers of $8$ modulo $21$:

$$8^1 \equiv 8 \pmod{21}$$

Now, let us look at $8^2$. We calculate:

$$8^2 = 64$$

If we divide $64$ by $21$, we get $64 = 21 	imes 3 + 1$. This is the breakthrough! We have found that $8^2 \equiv 1 \pmod{21}$.

This is the 'holy grail' of modular arithmetic. Once you hit a remainder of $1$, you have found the cycle. Any power of $8$ that is a multiple of $2$ will now result in a remainder of $1$.

### The Final Victory

Now, we just need to bridge the gap to our original exponent, $2024$. Since we know $8^2 \equiv 1 \pmod{21}$, we can rewrite $8^{2024}$ using the laws of exponents:

$$8^{2024} = (8^2)^{1012}$$

Substituting our identity, we get $(1)^{1012} \pmod{21}$. And as we all know, $1$ raised to any power is simply $1$.

The remainder is **$1$**. Just like that, the mountain has been climbed. We did not need to calculate the massive number; we only needed to understand its structure. Keep this mindset, and no problem in the JEE exam will ever be too big for you.

Animated Solution for Mathematics - Binomial Theorem: The remainder when $42 8^{2024}$ is divided by 21 is__________

Enter Numerical Value:

Visualized Solution

The Problem Statement

Simplifying the Base

Applying Power Property

Finding a Cycle

The Magic of $8^{2}$

Rewriting the Exponent

Final Substitution and Result

The Sigma Insight: Binomial Theorem for Positive Integral Index

Analyzing the Setup

Shrinking the Giant

The Hunt for the Cycle

The Final Victory

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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 4282024 is divided by 21 is__________