Finding Remainders using Binomial Theorem | JEE Math Tricks

Why This Technique is a Game-Changer

The Binomial Theorem provides an elegant way to find remainders of large powers without tedious calculations. Instead of computing $7^{97}$ directly, we can find its remainder when divided by 64 in just 2-3 steps!

🎯 Core Insight

When finding remainder of $a^n$ modulo $m$, we try to express $a$ as $(m \pm k)$ or $(m \cdot r \pm 1)$. Then using binomial expansion, most terms become multiples of $m$, leaving only a simple remainder.

$$(m \pm k)^n = \sum_{r=0}^n \binom{n}{r} m^{n-r} (\pm k)^r$$

All terms with $m^{n-r}$ where $n-r \geq 1$ are divisible by $m$, so remainder depends only on the last term(s).

The Binomial Theorem Refresher

Binomial Theorem Formula

$$(a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r$$

Where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ are binomial coefficients

Case 1: (mk ± 1)^n

When $a = mk \pm 1$, then:

$$(mk \pm 1)^n = (mk)^n \pm n(mk)^{n-1} + \cdots \pm n(mk) + 1$$

All terms except last are divisible by $m$

Case 2: (m ± k)^n

When $a = m \pm k$, then:

$$(m \pm k)^n = m^n \pm n m^{n-1}k + \cdots \pm n m k^{n-1} + k^n$$

All terms except last are divisible by $m$

Classic Example Medium

Problem 1: The Classic 7^97 Problem

Find the remainder when $7^{97}$ is divided by 64.

Step-by-Step Solution:

Step 1: Express 7 in terms of 64

We notice that $7 = 8 - 1$, but 8 doesn't directly relate to 64. Better approach:

$7^2 = 49 = 64 - 15$ ❌ (not helpful)

Instead, think: $7 = 2^3 - 1$? No, let's try $7 = 8 - 1$ and $8 = 2^3$

Actually, the key insight: $7 = 8 - 1$ and we need modulo 64 = $8^2$

Step 2: Write 7 as (8 - 1) and apply binomial theorem

$7^{97} = (8 - 1)^{97}$

Using binomial expansion:

$(8 - 1)^{97} = \sum_{r=0}^{97} \binom{97}{r} 8^{97-r} (-1)^r$

Step 3: Identify terms divisible by 64

Since we're dividing by 64 = $8^2$, any term with $8^{97-r}$ where $97-r \geq 2$ (i.e., $r \leq 95$) will be divisible by 64.

So only the last two terms ($r = 96$ and $r = 97$) might not be divisible by 64.

Step 4: Calculate the last two terms

Last two terms:

$T_{96} = \binom{97}{96} 8^1 (-1)^{96} = 97 \times 8 \times 1 = 776$

$T_{97} = \binom{97}{97} 8^0 (-1)^{97} = 1 \times 1 \times (-1) = -1$

Sum of last two terms: $776 - 1 = 775$

Step 5: Find the remainder

Since all other terms are divisible by 64, remainder is the remainder when 775 is divided by 64:

$775 ÷ 64 = 12$ remainder $775 - 12 \times 64 = 775 - 768 = 7$

✅ Final Answer: Remainder = 7

💡 Key Insight

When finding $(a-1)^n$ modulo $a^2$, only the last two terms matter: $na - 1$

$(a-1)^n \equiv (-1)^n + n \cdot a \cdot (-1)^{n-1} \pmod{a^2}$

JEE Main 2019 Easy

Problem 2: Simple (a+1)^n Case

Find the remainder when $9^{99}$ is divided by 8.

Smart Solution:

Step 1: Express 9 as (8 + 1)

$9^{99} = (8 + 1)^{99}$

Step 2: Apply binomial theorem

$(8 + 1)^{99} = \sum_{r=0}^{99} \binom{99}{r} 8^{99-r} 1^r$

All terms except the last ($r = 99$) contain at least one factor of 8.

Step 3: Identify the remainder

Only the term with $r = 99$ doesn't have factor of 8:

$T_{99} = \binom{99}{99} 8^0 1^{99} = 1$

So remainder is 1.

✅ Final Answer: Remainder = 1

💡 General Rule

For $(m + 1)^n$ modulo $m$, remainder is always 1, since all terms except last contain $m$.

JEE Advanced 2017 Hard

Problem 3: Double Binomial Application

Find the remainder when $17^{100}$ is divided by 18.

Advanced Solution:

Step 1: Express 17 as (18 - 1)

$17^{100} = (18 - 1)^{100}$

Step 2: Apply binomial theorem

$(18 - 1)^{100} = \sum_{r=0}^{100} \binom{100}{r} 18^{100-r} (-1)^r$

Step 3: Analyze terms modulo 18

All terms with $r \leq 99$ contain at least one factor of 18, so they are divisible by 18.

Only the last term ($r = 100$) remains:

$T_{100} = \binom{100}{100} 18^0 (-1)^{100} = 1$

✅ Final Answer: Remainder = 1

💡 Advanced Insight

When divisor is $m$ and number is $(m ± k)^n$, if $m$ is prime to $k$, we might need to consider more terms depending on the power relationships.

🚀 Problem-Solving Framework

Step 1: Identify the Pattern

Look for numbers close to the divisor
Try to express as $(m ± k)$ or $(mk ± 1)$
Check if $k$ is small (preferably 1)
Higher powers of 2 often work well

Step 2: Apply Binomial Theorem

Write the binomial expansion
Identify which terms vanish modulo $m$
Focus on the last 1-3 terms
Consider the sign pattern carefully

Common Patterns to Recognize

$(m+1)^n \equiv 1$

$(m-1)^n \equiv ±1$

$(2^a-1)^n$ mod $2^b$

$(mk±1)^n$ mod $m$

Problems 4-6 Available in Full Version

Includes 3 more challenging JEE problems with advanced binomial applications

📝 Test Your Understanding

Try these problems using the binomial theorem approach:

1. Find remainder when $5^{100}$ is divided by 24

Hint: $5 = 4 + 1$ or $5 = 6 - 1$? Which is better?

2. Find remainder when $3^{50}$ is divided by 8

Hint: $3 = 4 - 1$ or $3^2 = 9 = 8 + 1$?

3. Find remainder when $6^{99}$ is divided by 7

Hint: $6 = 7 - 1$

📋 Quick Reference Formulas

For modulo m:

$(m+1)^n \equiv 1 \pmod{m}$
$(m-1)^n \equiv (-1)^n \pmod{m}$
$(mk+1)^n \equiv 1 \pmod{m}$
$(mk-1)^n \equiv (-1)^n \pmod{m}$

For modulo m²:

$(m+1)^n \equiv 1 + nm \pmod{m^2}$
$(m-1)^n \equiv (-1)^n(1 - nm) \pmod{m^2}$
Last two terms matter for m²

Mastered Binomial Remainders?

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