Cyclicity and Remainders: The Faster Way to Solve JEE Main Problems

A Complete Guide with Shortcuts, Euler's Theorem & Previous Year Questions

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Introduction

Finding remainders is a frequently tested topic in JEE Main, appearing in 1-2 questions every year. While the Binomial Theorem approach (splitting $a^n$ into $(k \pm 1)^n$) is the standard method, Cyclicity offers a much faster "pattern-recognition" trick.

This technique is particularly powerful when:

• Finding the last digit (remainder when divided by 10)
• The binomial split is not obvious
• Dealing with tower powers like $32^{32^{32}}$

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Part 1: What is Cyclicity?

Definition

Cyclicity is the property where the powers of a number follow a repeating pattern of remainders when divided by a specific divisor.

For any expression $a^n \div k$, the remainders will eventually repeat. The number of terms in one such repeating set is called the cycle length.

Visual Example: Powers of 2 mod 7

Cycle: {2, 4, 1} with Cycle Length = 3

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Part 2: The Unit Digit (Remainder mod 10)

When the divisor is 10, the cyclicity depends entirely on the last digit of the base.

Master Table: Last Digit Cyclicity

Quick Memory Tricks

• Cycle of 1: Digits 0, 1, 5, 6 (always end in themselves)
• Cycle of 2: Digits 4 and 9
  • 4: Odd power → 4, Even power → 6
  • 9: Odd power → 9, Even power → 1
• Cycle of 4: Digits 2, 3, 7, 8 (REMEMBER: "2, 3, 7, 8 → Cycle 4, Wait!")

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Part 3: Step-by-Step Cyclicity Technique

Algorithm to Find $A^n \mod k$

Step 1: Simplify the Base Divide $A$ by $k$ and keep only the remainder.

Example: $37^{100} \div 5$ becomes $2^{100} \div 5$ (since $37 \mod 5 = 2$)

Step 2: Find the Pattern Calculate $a^1, a^2, a^3, \ldots$ divided by $k$ until the remainder repeats.

Step 3: Identify Cycle Length ($C$) Count how many steps it took to repeat.

Step 4: Reduce the Power Find $n \mod C$ (let's call this $r$).

Step 5: Find the Answer The remainder is the $r$-th term in your pattern.

Important: If $r = 0$, take the last term of the cycle.

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Part 4: Worked Examples

Example 1: Basic Cyclicity

Problem: Find the last digit of $7^{222}$

Solution:

• Last digit of base = 7 → Cycle of 4
• Pattern: 7, 9, 3, 1
• $222 \mod 4 = 2$
• Answer: 9 (2nd term in pattern)

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Example 2: Tower Powers - $32^{32^{32}} \div 7$

Step 1: Simplify base: $32 \mod 7 = 4$

So we need $4^{(32^{32})} \mod 7$

Step 2: Find pattern of $4^n \mod 7$:

• $4^1 \mod 7 = 4$
• $4^2 \mod 7 = 16 \mod 7 = 2$
• $4^3 \mod 7 = 64 \mod 7 = 1$
• $4^4 \mod 7 = 4$ ← Repeats

Step 3: Cycle = {4, 2, 1}, Length $C = 3$

Step 4: Find $32^{32} \mod 3$

Using Binomial: $32 = 33 - 1$ $32^{32} = (33-1)^{32} \equiv (-1)^{32} \equiv 1 \pmod{3}$

Step 5: Position 1 in cycle {4, 2, 1} → Answer: 4

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Example 3: Last Two Digits (mod 100)

Problem: Find the last two digits of $7^{100}$

Solution: For mod 100, we use $\phi(100) = 100 \times (1-\frac{1}{2}) \times (1-\frac{1}{5}) = 40$

Since $\gcd(7, 100) = 1$, by Euler's theorem: $7^{40} \equiv 1 \pmod{100}$

$100 = 40 \times 2 + 20$

$7^{100} = (7^{40})^2 \times 7^{20} \equiv 1 \times 7^{20} \pmod{100}$

Now calculate $7^{20} \mod 100$:

• $7^2 = 49$
• $7^4 = 2401 \equiv 1 \pmod{100}$

Wait, this means cycle is 4! $7^{20} = (7^4)^5 \equiv 1^5 = 1 \pmod{100}$

Answer: 01 (last two digits)

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Part 5: Euler's Totient Theorem (Advanced Shortcut)

The Power Move

When $\gcd(a, k) = 1$ (coprime), you can skip finding the pattern manually:

$\boxed{a^{\phi(k)} \equiv 1 \pmod{k}}$

where $\phi(k)$ is Euler's Totient Function.

Calculating $\phi(k)$

If $k = p_1^{a_1} \cdot p_2^{a_2} \cdot \ldots \cdot p_n^{a_n}$, then:

$\phi(k) = k \cdot \left(1 - \frac{1}{p_1}\right) \cdot \left(1 - \frac{1}{p_2}\right) \cdot \ldots \cdot \left(1 - \frac{1}{p_n}\right)$

Quick Reference Table

Fermat's Little Theorem (Special Case)

For prime $p$: $\phi(p) = p - 1$

$a^{p-1} \equiv 1 \pmod{p}$

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Part 6: Binomial Theorem for Remainders

When $a^n$ can be written as $(k \pm 1)^n$ for some multiple $k$ of the divisor:

Key Formulas

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

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

$(km + r)^n \equiv r^n \pmod{m}$

When to Use Binomial vs Cyclicity

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Part 7: JEE Main Previous Year Questions (PYQs)

PYQ 1: JEE Main 2025 (7th April)

Question: The remainder when $((64)^{64})^{64}$ is divided by 7 is equal to _______.

Solution:

Method 1: Cyclicity

$64 \mod 7 = 1$

Therefore: $((64)^{64})^{64} \equiv (1^{64})^{64} = 1^{64} = 1 \pmod{7}$

Answer: 1

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PYQ 2: JEE Main 2024 (9th April)

Question: The remainder when $428^{2024}$ is divided by 21 is _______.

Solution:

Method: Binomial Theorem

$428 = 420 + 8 = 21 \times 20 + 8$

$428^{2024} \equiv 8^{2024} \pmod{21}$

Now find $8^{2024} \mod 21$:

Using $\phi(21) = \phi(3 \times 7) = 2 \times 6 = 12$

Since $\gcd(8, 21) = 1$: $8^{12} \equiv 1 \pmod{21}$

$2024 = 12 \times 168 + 8$

$8^{2024} = (8^{12})^{168} \times 8^8 \equiv 1 \times 8^8 \pmod{21}$

Calculate $8^8 \mod 21$:

• $8^2 = 64 = 63 + 1 \equiv 1 \pmod{21}$

Therefore: $8^8 = (8^2)^4 \equiv 1^4 = 1 \pmod{21}$

Answer: 1

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PYQ 3: JEE Main 2024 (29th Jan)

Question: Remainder when $64^{32^{32}}$ is divided by 9 is equal to _______.

Solution:

$64 \mod 9 = 1$ (since $64 = 63 + 1 = 7 \times 9 + 1$)

$64^{32^{32}} \equiv 1^{32^{32}} = 1 \pmod{9}$

Answer: 1

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PYQ 4: JEE Main 2023 (Feb)

Question: The remainder when $19^{200} + 23^{200}$ is divided by 49 is _______.

Solution:

Note: $19 = 21 - 2$ and $23 = 21 + 2$

$19^{200} + 23^{200} = (21-2)^{200} + (21+2)^{200}$

When we expand using binomial theorem, odd powers of 2 cancel:

$= 2[{}^{200}C_0 \cdot 21^{200} + {}^{200}C_2 \cdot 21^{198} \cdot 4 + \ldots + {}^{200}C_{200} \cdot 2^{200}]$

All terms with $21^k$ where $k \geq 2$ are divisible by $49 = 21 \times 2 + 7$.

Actually, let's be more careful. $49 = 7^2$.

$19 \equiv -2 \pmod{7}$ and $23 \equiv 2 \pmod{7}$

$19^{200} + 23^{200} \equiv (-2)^{200} + 2^{200} = 2 \cdot 2^{200} \pmod{7}$

By Fermat: $2^6 \equiv 1 \pmod{7}$

$200 = 6 \times 33 + 2$, so $2^{200} \equiv 2^2 = 4 \pmod{7}$

$2 \cdot 4 = 8 \equiv 1 \pmod{7}$

For mod 49, use $(21 \pm 2)^{200}$: The answer involves careful binomial expansion mod 49.

Answer: 18

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PYQ 5: JEE Main 2022

Question: The remainder when $(11)^{1011} + (1011)^{11}$ is divided by 9 is:

(a) 1   (b) 4   (c) 6   (d) 8

Solution:

For $11^{1011} \mod 9$:

$11 \mod 9 = 2$

Find cycle of $2^n \mod 9$: 2, 4, 8, 7, 5, 1 → Cycle length = 6

$1011 \mod 6 = 3$

$11^{1011} \equiv 2^3 = 8 \pmod{9}$

For $1011^{11} \mod 9$:

$1011 \mod 9 = 1 + 0 + 1 + 1 = 3$

$3^2 = 9 \equiv 0 \pmod{9}$

So $3^{11} = 3^2 \times 3^9 \equiv 0 \pmod{9}$

Total: $8 + 0 = 8 \pmod{9}$

Answer: (d) 8

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PYQ 6: JEE Main 2023

Question: The remainder when $(2023)^{2023}$ is divided by 35 is _______.

Solution:

$2023 = 35 \times 57 + 28$, so $2023 \equiv 28 \pmod{35}$

We need $28^{2023} \mod 35$

$\phi(35) = \phi(5 \times 7) = 4 \times 6 = 24$

Since $\gcd(28, 35) = 7 \neq 1$, we can't directly apply Euler's theorem.

Use CRT (Chinese Remainder Theorem):

Mod 5: $28 \equiv 3 \pmod{5}$

• $3^4 \equiv 1 \pmod{5}$ (Fermat)
• $2023 = 4 \times 505 + 3$
• $3^{2023} \equiv 3^3 = 27 \equiv 2 \pmod{5}$

Mod 7: $28 \equiv 0 \pmod{7}$

• $28^{2023} \equiv 0 \pmod{7}$

Find $x$ such that $x \equiv 2 \pmod{5}$ and $x \equiv 0 \pmod{7}$

$x = 7k$ where $7k \equiv 2 \pmod{5}$ $2k \equiv 2 \pmod{5}$ $k \equiv 1 \pmod{5}$ $k = 1$ gives $x = 7$

Answer: 7

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PYQ 7: AIEEE 2009

Question: The remainder left out when $8^{2n} - (62)^{2n+1}$ is divided by 9 is:

(a) 7   (b) 8   (c) 0   (d) 2

Solution:

$8 = 9 - 1$, so $8^{2n} = (9-1)^{2n} \equiv (-1)^{2n} = 1 \pmod{9}$

$62 = 63 - 1 = 9 \times 7 - 1$, so $62^{2n+1} \equiv (-1)^{2n+1} = -1 \pmod{9}$

$8^{2n} - 62^{2n+1} \equiv 1 - (-1) = 2 \pmod{9}$

Answer: (d) 2

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PYQ 8: JEE Main 2023

Question: Let the number $(22)^{2022} + (2022)^{22}$ leave remainder $\alpha$ when divided by 3 and $\beta$ when divided by 7. Then $(\alpha^2 + \beta^2)$ is equal to:

Solution:

Finding $\alpha$ (mod 3):

$22 \equiv 1 \pmod{3}$ → $22^{2022} \equiv 1 \pmod{3}$

$2022 \equiv 0 \pmod{3}$ → $2022^{22} \equiv 0 \pmod{3}$

$\alpha = 1 + 0 = 1$

Finding $\beta$ (mod 7):

$22 \equiv 1 \pmod{7}$ → $22^{2022} \equiv 1 \pmod{7}$

$2022 \equiv 6 \equiv -1 \pmod{7}$ → $2022^{22} \equiv (-1)^{22} = 1 \pmod{7}$

$\beta = 1 + 1 = 2$

Answer: $\alpha^2 + \beta^2 = 1 + 4 = \boxed{5}$

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PYQ 9: JEE Main 2021

Question: If $\{p\}$ denotes the fractional part of $p$, then $\left\{\frac{3^{200}}{8}\right\}$ is equal to:

Solution:

We need $3^{200} \mod 8$

$3^2 = 9 \equiv 1 \pmod{8}$

$3^{200} = (3^2)^{100} \equiv 1^{100} = 1 \pmod{8}$

Fractional part = $\frac{1}{8}$

Answer: $\frac{1}{8}$

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Part 8: Practice Problems

Level 1: Basic

1. Find the last digit of $3^{999}$
2. Find the remainder when $7^{50}$ is divided by 11
3. What is $2^{100} \mod 5$?

Level 2: Intermediate

1. Find the remainder when $17^{256}$ is divided by 17
2. Find the last two digits of $3^{100}$
3. Find $13^{2023} \mod 21$

Level 3: Advanced (JEE Level)

1. Find the remainder when $7^{7^7}$ is divided by 10
2. Find the remainder when $2^{2023} + 3^{2023}$ is divided by 5
3. If $50^{75} - 25^{75}$ is divided by 13, find the remainder

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Part 9: Quick Revision Formulas

Box 1: Unit Digit Rules

Last digit 0,1,5,6 → Always same
Last digit 4,9 → Cycle of 2
Last digit 2,3,7,8 → Cycle of 4

Box 2: Euler's Theorem

If gcd(a,k) = 1:
a^φ(k) ≡ 1 (mod k)

For prime p:
a^(p-1) ≡ 1 (mod p)

Box 3: Binomial Shortcuts

(km + 1)^n ≡ 1 (mod m)
(km - 1)^n ≡ (-1)^n (mod m)

Box 4: Common φ Values

φ(p) = p-1 (p prime)
φ(p^k) = p^(k-1)(p-1)
φ(mn) = φ(m)φ(n) if gcd(m,n)=1

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Part 10: Summary & Strategy

When to Use Each Method

Common Mistakes to Avoid

1. Forgetting to simplify the base first
2. Using Euler when gcd ≠ 1
3. Taking position 0 instead of last position in cycle
4. Not checking if direct calculation is faster

Time-Saving Tips

• Memorize the unit digit table thoroughly
• Know $\phi$ values for common numbers (7, 9, 10, 11, 13)
• Practice tower power problems
• For JEE, most problems use cyclicity or simple binomial expansion

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Master these techniques and you'll solve remainder problems in JEE Main within seconds! 🎯