Remainder Theorem and Division Algorithm
Remainder Theorem and Division Algorithm
Remainder questions in NTPC look heavy (7²⁹ + 4, 25²⁵, products of three numbers) but every one of them collapses once you replace each number by its remainder, or by a negative remainder. That single habit is the whole topic.
1. Division Algorithm
Dividend = Divisor × Quotient + Remainder, with 0 ≤ Remainder < Divisor
If the divisor is d, the possible remainders are 0, 1, ..., d − 1. A question stating remainder 15 with divisor 12 is impossible; options built on such data are traps.
2. Remainder Arithmetic Rules
| Rule | Statement | Example |
|---|---|---|
| Sum rule | Rem(a + b) = Rem(Rem a + Rem b) | (47 + 59) ÷ 6: 5 + 5 = 10 → 4 |
| Product rule | Rem(a × b) = Rem(Rem a × Rem b) | (47 × 59) ÷ 6: 5 × 5 = 25 → 1 |
| Power rule | Rem(aⁿ) = Rem((Rem a)ⁿ) | 26⁵ ÷ 5: 1⁵ → 1 |
| Negative remainder | Use a − d when it is easier | 34 ÷ 35 → −1 |
| (d + 1)ⁿ ÷ d | Always 1 | 8²⁰ ÷ 7 → 1 |
| (d − 1)ⁿ ÷ d | 1 if n even, d − 1 if n odd | 34³¹ ÷ 35 → 34 |
| Factor divisor | If N leaves r on division by d, and k is a factor of d, then N ÷ k leaves Rem(r ÷ k) | N ÷ 56 leaves 29 → N ÷ 8 leaves 5 |
| Cycle of remainders | Powers repeat in a cycle; find the cycle length and reduce the exponent | 2ⁿ ÷ 7: 2, 4, 1, 2, 4, 1 (cycle 3) |
✗ N ÷ 8 leaves 5, so N ÷ 56 leaves 5 | ✓ The reverse direction does not work; only a FACTOR of the original divisor gives a fixed remainder
3. Successive Division
If N divided by d₁ gives quotient q₁ and remainder r₁, and q₁ divided by d₂ gives remainder r₂, and so on, build N from the inside out:
Smallest N = r₁ + d₁ × (r₂ + d₂ × r₃) and the general N adds multiples of d₁ × d₂ × d₃
4. Fermat's Little Theorem and the Polynomial Remainder Theorem
If p is prime and a is not a multiple of p, then a^(p − 1) leaves remainder 1 on division by p
f(x) ÷ (x − a) leaves remainder f(a); f(x) ÷ (ax − b) leaves remainder f(b/a)
हिंदी नोट: शेषफल निकालते समय बड़ी संख्या को उसके शेषफल से बदल दीजिए। जैसे 34 को 35 से भाग देने पर शेषफल −1 मान लेना सबसे तेज़ तरीका है। विषम घात पर −1 का मान −1 रहता है, सम घात पर 1 हो जाता है।
Exam Pointer: The 2020-21 NTPC shifts carried remainders of (power + constant), divisor from quotient and remainder, a number leaving a known remainder and its square, and rod-cutting style division. The CBT-2 Level block below covers polynomial remainder and Fermat-type questions, one step above CBT-1. Distractor options are usually the remainder before the final reduction (like 25 instead of 4 for divisor 7).
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Missing term in the division algorithm
[PYQ: NTPC 2020-21 cycle]
EXAM LEVEL
Q. In a division the divisor is 47, the quotient is 128 and the remainder is 39. Find the dividend.
Dividend = 47 × 128 + 39 = 6,016 + 39 = 6,055.
Answer: 6,055
EXAMATLAS LEVEL
Q. In a division sum the divisor is 4 times the quotient and 3 times the remainder. If the remainder is 12, find the dividend.
Divisor = 3 × 12 = 36. Divisor is also 4 times the quotient, so quotient = 9. Dividend = 36 × 9 + 12 = 336. Check that the remainder 12 is less than the divisor 36, so the data is consistent.
Answer: 336
Pattern 2: Remainder of a power plus a constant
[PYQ: NTPC CBT-1 5-Apr-2021 Shift-2]
EXAM LEVEL
Q. Find the remainder when 7⁴³ + 3 is divided by 6.
7 leaves 1 on division by 6, so 7⁴³ leaves 1⁴³ = 1. Add 3: remainder 4.
Answer: 4
EXAMATLAS LEVEL
Q. Find the remainder when 3⁵³ + 5²⁸ is divided by 8.
3² = 9 leaves 1 on division by 8, so 3⁵² = (3²)²⁶ leaves 1 and 3⁵³ leaves 3. Similarly 5² = 25 leaves 1, so 5²⁸ leaves 1. Sum leaves 3 + 1 = 4.
Answer: 4
Pattern 3: Same number, a different divisor or a function of the number
[PYQ: NTPC CBT-1 13-Jan-2021 Shift-1 | NTPC CBT-1 4-Jan-2021 Shift-2]
EXAM LEVEL
Q. A number leaves remainder 29 when divided by 56. What is the remainder when the same number is divided by 8?
8 is a factor of 56, so N = 56k + 29 and 56k is a multiple of 8. Remainder = 29 ÷ 8 → 5.
Answer: 5
EXAMATLAS LEVEL
Q. A number leaves remainder 5 when divided by 7. What is the remainder when three times the square of the number is divided by 7?
Replace N by 5. N² leaves 25 → 4. Three times that is 12 → 5. Note that the answer coincidentally equals the original remainder; do not skip the working because of it.
Answer: 5
Pattern 4: Successive division
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. A number on successive division by 4 and 5 leaves remainders 3 and 2 respectively. Find the smallest such number.
Build from the inside: the second quotient can be 0, so the first quotient = 2. N = 4 × 2 + 3 = 11. Check: 11 ÷ 4 = 2 remainder 3, and 2 ÷ 5 = 0 remainder 2. The general form is 20k + 11; if a key insists that the last quotient is at least 1, its answer is 31, so check the options.
Answer: 11
EXAMATLAS LEVEL
Q. A number on successive division by 3, 5 and 7 leaves remainders 2, 4 and 1. What remainder does it leave when divided by 105?
Let the last quotient be k. Then the second quotient = 7k + 1, the first quotient = 5(7k + 1) + 4 = 35k + 9, and N = 3(35k + 9) + 2 = 105k + 29. Remainder on dividing by 105 = 29.
Answer: 29
Pattern 5: Division in word problems (pieces, packets, leftovers)
[PYQ: NTPC 2020-21 cycle]
EXAM LEVEL
Q. How many complete pieces of 23 cm can be cut from a rod 10 m long, and what length is left over?
10 m = 1,000 cm. 1,000 ÷ 23 = 43 remainder 11 (23 × 43 = 989). So 43 pieces and 11 cm left.
Answer: 43 pieces, 11 cm left
EXAMATLAS LEVEL
Q. A shopkeeper packs 1,000 sweets in boxes of 18. What is the least number of extra sweets he must buy so that no sweet is left unpacked, and how many boxes will he then fill?
1,000 ÷ 18 leaves remainder 10 (18 × 55 = 990). He needs 18 − 10 = 8 more sweets to complete one more box. Total 1,008 sweets = 56 boxes. Answering 10 (the remainder) is the trap.
Answer: 8 sweets, 56 boxes
Pattern 6: Negative remainder and products
[PYQ: NTPC 2020-21 cycle]
EXAM LEVEL
Q. Find the remainder when 34³¹ is divided by 35.
34 = 35 − 1 leaves −1. (−1)³¹ = −1, and −1 + 35 = 34.
Answer: 34
EXAMATLAS LEVEL
Q. Find the remainder when 17 × 23 × 29 is divided by 12.
17 leaves 5, 23 leaves 11 = −1, 29 leaves 5. Product: 5 × (−1) × 5 = −25. Add 36 (a multiple of 12) to make it positive: 11. Check: 17 × 23 × 29 = 11,339 and 12 × 944 = 11,328, difference 11.
Answer: 11
CBT-2 LEVEL
Pattern 7: Polynomial remainder theorem and Fermat's theorem
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. When x³ − 4x² + 2x + k is divided by (x − 2), the remainder is 5. Find k.
Remainder = f(2) = 8 − 16 + 4 + k = k − 4. Set k − 4 = 5, so k = 9.
Answer: 9
EXAMATLAS LEVEL
Q. Find the remainder when 3¹⁰⁰ is divided by 7.
7 is prime, so by Fermat 3⁶ leaves 1. 100 = 6 × 16 + 4, so 3¹⁰⁰ leaves the same as 3⁴ = 81, and 81 = 7 × 11 + 4. Remainder 4. Without Fermat, list 3ⁿ ÷ 7: 3, 2, 6, 4, 5, 1 and the cycle of 6 gives the same answer.
Answer: 4
60-Second Revision
- Dividend = divisor × quotient + remainder; remainder is always less than divisor.
- Replace every number by its remainder (or negative remainder) before multiplying or powering.
- (d + 1)ⁿ leaves 1; (d − 1)ⁿ leaves 1 for even n and d − 1 for odd n.
- Known remainder for d gives a fixed remainder only for factors of d.
- Successive division: build N from the innermost quotient outward.
- CBT-2: f(x) ÷ (x − a) leaves f(a); a^(p − 1) leaves 1 for prime p.