Remainder Theorem – Complete Guide
Master the Remainder Theorem for SSC, UPSC, UPPSC, BPSC, Railway & Banking exams. Clear concept, worked examples, shortcut tricks, and previous year questions all in one place!
๐ What is the Remainder Theorem?
The Remainder Theorem (Polynomial Division Theorem)
When a polynomial f(x) is divided by a linear divisor (x − a), the remainder is the value of the polynomial at x = a, i.e., simply substitute a into f(x) to get the remainder directly — no long division needed!
Remainder in Number Division
In competitive exams, "Remainder Theorem" often refers to finding the remainder when a number (or expression) is divided by another number. Both forms — polynomial and numeric — are tested heavily.
๐ Key Concepts You Must Know
1. Polynomial Remainder Theorem
If f(x) ÷ (x − a), the remainder = f(a). Substitute a directly into polynomial.
Example: f(x) = x³ − 4x + 5, divide by (x − 2)
Remainder = f(2) = 8 − 8 + 5 = 5
2. Factor Theorem (Special Case)
If the remainder is zero, then (x − a) is a factor of f(x). This is the Factor Theorem — a special case of Remainder Theorem.
Use this to check divisibility of polynomials without long division.
3. Cyclicity Method
Powers of numbers follow a repeating pattern (cycle) when divided. Key for finding remainders of large powers like 7⁵⁰ ÷ 10.
Cyclicity of 2: 2,4,8,6 (cycle=4). Cyclicity of 3: 3,9,7,1 (cycle=4).
4. Fermat's Little Theorem
If p is prime and gcd(a, p) = 1, then the remainder of aแต⁻¹ ÷ p is always 1.
Very useful for finding remainders of huge power expressions in SSC/UPSC.
5. Chinese Remainder Theorem
Finds a number that gives specific remainders when divided by multiple divisors. Advanced concept asked in UPSC and Banking exams.
Useful when a number leaves remainder 2 on ÷3 and remainder 3 on ÷5 type questions.
6. Modular Arithmetic Basics
a ≡ r (mod d) means a leaves remainder r when divided by d. Used to simplify large calculations.
Key rule: (a × b) mod d = [(a mod d) × (b mod d)] mod d
๐ Important Rules & Divisibility Shortcuts
Div by 2
Remainder when divided by 2 = last digit mod 2
Div by 3 or 9
Sum all digits → divide by 3 or 9 → that is the remainder
Div by 10
The units digit directly gives the remainder when divided by 10
Div by 5
Only the last digit matters. Units digit 0 or 5 → R=0
Power Rule
Any number of form (kd ± 1)โฟ gives remainder ±1 when divided by d
aโฟ − bโฟ Rule
Also: aโฟ + bโฟ is divisible by (a+b) when n is odd
Negative Remainder
If remainder comes out negative, add divisor to make it positive
✏️ Fully Solved Examples (Step-by-Step)
๐ Example 1 – Polynomial Remainder
⭐ Easy- Identify a: Divisor is (x − 2), so a = 2
- Apply Remainder Theorem: Remainder = f(a) = f(2)
Substitute x = 2 into f(x) - Calculate:
f(2) = (2)³ − 3(2)² + 2(2) − 5= 8 − 3(4) + 4 − 5= 8 − 12 + 4 − 5 = −5
๐ Example 2 – Large Power Remainder (Cyclicity)
⭐⭐ Medium- Find cyclicity of 7: Powers of 7 repeat their units digit every 4 steps
7¹ = 7, 7² = 49, 7³ = 343, 7⁴ = 2401 → Units digits: 7, 9, 3, 1 → Cycle = 4 - Divide power by cycle length:
50 ÷ 4 = 12 remainder 2 - Remainder 2 means same as 7²:
7² has units digit = 9 - Therefore: 7⁵⁰ ÷ 10 gives units digit = 9, so remainder = 9
๐ Example 3 – Number Remainder (Digit Sum Rule)
⭐ Easy- Rule: Remainder when divided by 9 = sum of digits mod 9
- Sum the digits of 5765:
5 + 7 + 6 + 5 = 23 - Digit sum 23 → sum again:
2 + 3 = 5 - Therefore: 5765 ÷ 9 → remainder = 5
๐ Example 4 – Fermat's Little Theorem
⭐⭐⭐ Hard- Apply Fermat's Little Theorem: Since 7 is prime and gcd(2,7)=1
2^(7−1) = 2^6 ≡ 1 (mod 7) - Express 100 in terms of 6:
100 = 6 × 16 + 4, so 2¹⁰⁰ = (2⁶)¹⁶ × 2⁴ - Simplify:
(2⁶)¹⁶ ≡ 1¹⁶ = 1 (mod 7)2⁴ = 16 = 2×7 + 2, so 2⁴ ≡ 2 (mod 7) - Combine:
2¹⁰⁰ ≡ 1 × 2 = 2 (mod 7)
๐ Example 5 – Factor Theorem Application
⭐⭐ Medium- Part 1 – Apply Factor Theorem: Check f(3)
f(3) = (3)³ − 7(3) + 6 = 27 − 21 + 6 = 12 ≠ 0Hmm! Let's check f(−3):f(−3) = −27 + 21 + 6 = 0 ✅So (x + 3) is the factor, not (x − 3). This illustrates careful sign reading. - Part 2 – Find k: (x − 1) is a factor, so f(1) = 0
f(1) = (1)² + k(1) − 2 = 01 + k − 2 = 0 → k = 1
⚡ Shortcut Tricks for Competitive Exams
⚡ Trick 1 – (N+1)โฟ ÷ N
When any number of the form (aN + 1) is raised to any power and divided by N:
(aN + 1)โฟ ÷ N → Remainder = 1
Example: 101⁵⁰ ÷ 100 → 101 = 100+1, so Remainder = 1
⚡ Trick 2 – (N−1)โฟ ÷ N
When a number of the form (aN − 1) is raised to a power n:
If n is even → R = 1; if n is odd → R = N−1
Example: 99⁵⁰ ÷ 100 → 99=100−1, 50 is even → R = 1
99⁵¹ ÷ 100 → 51 is odd → R = 99
⚡ Trick 3 – Cyclicity of Powers
Units digit (= remainder ÷ 10) follows cycles:
Cycle of 2,3,7,8 = 4; of 4,9 = 2; of 0,1,5,6 = 1
Divide power by cycle → use remainder to pick position in cycle
⚡ Trick 4 – Digit Sum for ÷9 or ÷3
Instead of dividing huge numbers by 9 or 3:
Remainder = (Sum of digits) mod 9 or mod 3
Example: 987654 ÷ 9 → 9+8+7+6+5+4=39 → 3+9=12 → 1+2=3 → R=3
⚡ Trick 5 – Negative Remainder
Using a negative form of the dividend can simplify calculations:
If N = dq + r, use r = r − d (negative form)
Example: 19 ÷ 7 → 19 = 21−2 → use remainder −2 → actual R = 7−2 = 5
⚡ Trick 6 – Remainder of Sum/Product
For complex expressions split and find each remainder:
R(a×b ÷ d) = R(a÷d) × R(b÷d), then mod d
Example: (34 × 27) ÷ 5 → R(34÷5)=4, R(27÷5)=2 → 4×2=8 → R=3
๐ Previous Year Questions (PYQ) – Exam-Wise
| Question | Exam | Answer | Concept Used |
|---|---|---|---|
| What is the remainder when x³ + 3x² − 2x + 1 is divided by (x − 1)? | SSC CGL | 3 | Polynomial R.T. |
| Find remainder: 17²⁵ ÷ 18 | UPPSC | 17 | (N−1)โฟ odd → N−1 |
| What is units digit of 3⁴⁵? | RRB NTPC | 3 | Cyclicity of 3 |
| Remainder of 2⁵⁶ ÷ 7 | SBI PO | 1 | Fermat's Theorem |
| If (x − 2) is a factor of x² + kx − 4, find k | BPSC | k = 0 | Factor Theorem |
| What is remainder when 5765432 is divided by 9? | RRB Group D | 5 | Digit Sum Rule |
| Find remainder: (13 × 17 × 19) ÷ 7 | SSC CHSL | 1 | Modular Arithmetic |
| Remainder of 99⁵⁰ ÷ 100 | UPPSC PCS | 1 | (N−1)โฟ even → 1 |
| Find k if x³ − 2x² + kx + 3 gives remainder 7 when divided by (x − 2) | IBPS PO | k = 3 | Polynomial R.T. |
| What is the remainder when 4³⁰ is divided by 5? | SSC MTS | 1 | (N−1)โฟ even → 1 |
Digit sum, simple substitution
Cyclicity, factor theorem, (N±1) rule
Fermat's, Chinese Remainder, combined
๐ฏ Exam Strategy & Top Tips
- Always try the (N±1) rule and cyclicity first — they solve 60% of remainder questions fastest.
- For polynomial f(x) ÷ (x − a), just substitute x = a. Never do long division in exams.
- For (ax − b), rewrite as a(x − b/a) and use x = b/a as substitution value for f(x).
- When the answer is negative, add divisor to convert to positive remainder.
- Divisibility by 9: digit sum method. Divisibility by 11: alternating digit sum method.
- Memorise the cyclicity table — it will save 2–3 minutes per exam.
- Practice at least 20 PYQs on remainder theorem before any exam.
