Division Algorithm & Successive Division
The division algorithm expresses a number in terms of its divisor, quotient and remainder. Successive division extends this idea by dividing a number, then dividing the obtained quotient again. These relations are useful for reconstructing numbers, finding missing values and solving many competitive-exam number-system problems without performing lengthy calculations.
1. Basic Idea of Division
When one positive integer is divided by another, the result consists of:
- Dividend — the number being divided.
- Divisor — the number by which we divide.
- Quotient — the number of complete times the divisor fits into the dividend.
- Remainder — the amount left after complete groups have been removed.
2. Division Algorithm
For a positive integer dividend N and positive divisor d:
N = dq + r
where:
- N = Dividend
- d = Divisor
- q = Quotient
- r = Remainder
3. Essential Condition on the Remainder
The remainder always satisfies:
0 ≤ r < d
Very Important: A remainder can never be equal to or greater than the divisor.
4. Example of the Division Algorithm
Divide 47 by 6.
We have:
47 = 6 × 7 + 5
Therefore:
- Dividend = 47
- Divisor = 6
- Quotient = 7
- Remainder = 5
5. Fundamental Formula
Dividend = Divisor × Quotient + Remainder
This single formula forms the basis of most elementary division-algorithm questions.
6. Finding the Dividend
If divisor, quotient and remainder are known:
Dividend = Divisor × Quotient + Remainder
Example: Divisor = 13, quotient = 24 and remainder = 7.
Dividend = 13 × 24 + 7 = 319
7. Finding the Quotient
From:
N = dq + r
we get:
q = (N − r)/d
Example: A number 157 leaves remainder 7 when divided by 15.
q = (157 − 7)/15 = 150/15 = 10
8. Finding the Divisor
If dividend N, quotient q and remainder r are known:
d = (N − r)/q
Example: Dividend = 183, quotient = 11 and remainder = 7.
d = (183 − 7)/11 = 176/11 = 16
9. Finding the Remainder
If dividend, divisor and quotient are known:
r = N − dq
Example:
N = 236, d = 15, q = 15
Then:
r = 236 − 15 × 15 = 11
10. Exact Division
A division is called exact when remainder = 0.
Thus:
N = dq
and d divides N exactly.
Example:
84 = 7 × 12 + 0
11. Non-Exact Division
If remainder is non-zero, the division is not exact.
Example:
86 = 7 × 12 + 2
12. Maximum Possible Remainder
Since:
0 ≤ r < d
the maximum possible remainder when dividing by d is:
d − 1
Example: When a number is divided by 17, the maximum possible remainder is 16.
13. Minimum Possible Remainder
The minimum possible remainder is:
0
This occurs when the dividend is exactly divisible by the divisor.
14. Divisor Must Be Greater than the Remainder
If a number leaves remainder r, then:
Divisor > r
Example: If remainder is 11, divisor must be at least 12.
15. When Divisor is Greater than Dividend
If 0 ≤ N < d, then:
N = d × 0 + N
Therefore:
- Quotient = 0
- Remainder = N
Example:
7 ÷ 12 → Quotient 0, Remainder 7
16. Division by 1
Every integer N divided by 1 gives:
N = 1 × N + 0
Therefore quotient = N and remainder = 0.
17. Reconstructing a Number
Many examination questions give the divisor, quotient and remainder and ask for the original number.
Always use:
N = dq + r
18. Solved Example: Reconstruct the Dividend
A number when divided by 23 gives quotient 17 and remainder 11. Find the number.
N = 23 × 17 + 11
= 391 + 11 = 402
Therefore, the number is 402.
19. Difference Between Dividend and Remainder
From:
N = dq + r
we get:
N − r = dq
Therefore, N − r is always divisible by d.
20. Useful Divisibility Interpretation
If N leaves remainder r when divided by d:
d divides N − r
This provides a useful way to identify possible divisors.
21. Example: Possible Divisor
A number 83 leaves remainder 5 when divided by d.
Then:
d divides 83 − 5 = 78
Also d > 5.
Thus any possible divisor must be a divisor of 78 greater than 5.
Possible values include:
6, 13, 26, 39, 78
22. Quotient and Remainder are Unique
For a fixed positive divisor d and dividend N, there is exactly one pair of integers q and r satisfying:
N = dq + r, 0 ≤ r < d
Important Property: For a fixed dividend and positive divisor, quotient and remainder are unique.
23. Successive Division
In successive division, we first divide a number by one divisor and then divide the quotient obtained by another divisor.
This is different from independently dividing the original number by both divisors.
Important: In successive division, the second divisor acts on the first quotient, not on the original dividend.
24. Two-Step Successive Division
Suppose N is first divided by d1:
N = d1q1 + r1
and the quotient q1 is then divided by d2:
q1 = d2q2 + r2
25. Combining Two Successive Divisions
Substitute:
q1 = d2q2 + r2
into:
N = d1q1 + r1
to obtain:
N = d1d2q2 + d1r2 + r1
26. Combined Remainder in Two-Step Division
Therefore, when N is directly divided by d1d2, the remainder is:
R = d1r2 + r1
Because:
0 ≤ r1 < d1, 0 ≤ r2 < d2
the combined remainder automatically satisfies:
0 ≤ R < d1d2
27. Example: Two Successive Divisions
A number is divided by 5 and leaves remainder 3. The quotient is then divided by 7 and leaves remainder 4. Find the remainder when the original number is divided by 35.
Here:
- d1 = 5
- r1 = 3
- d2 = 7
- r2 = 4
Therefore:
R = 5 × 4 + 3
= 23
Hence the original number leaves remainder 23 when divided by 35.
28. Why We Cannot Simply Add the Remainders
In the previous example, the successive remainders are 3 and 4.
The combined remainder is not:
3 + 4 = 7
Instead:
5 × 4 + 3 = 23
Common Trap: In successive division, do not simply add intermediate remainders.
29. Reverse Successive Division
If the final quotient and intermediate remainders are given, reconstruct the original number by working backwards.
For two steps:
q1 = d2q2 + r2
then:
N = d1q1 + r1
30. Example: Reverse Successive Division
A number is divided by 4 and leaves remainder 3. The quotient is divided by 6 and leaves remainder 2. The final quotient is 5. Find the original number.
First reconstruct the first quotient:
q1 = 6 × 5 + 2 = 32
Now reconstruct the original number:
N = 4 × 32 + 3 = 131
Therefore, the original number is 131.
31. Direct Formula for Two-Step Reconstruction
From:
N = d1d2q2 + d1r2 + r1
we can directly reconstruct N if d1, d2, q2, r1 and r2 are known.
32. Same Example Using the Direct Formula
For d1 = 4, d2 = 6, q2 = 5, r1 = 3 and r2 = 2:
N = 4 × 6 × 5 + 4 × 2 + 3
= 120 + 8 + 3 = 131
33. Three-Step Successive Division
Suppose:
N = d1q1 + r1
q1 = d2q2 + r2
q2 = d3q3 + r3
34. Combined Formula for Three Successive Divisions
Substituting successively gives:
N = d1d2d3q3 + d1d2r3 + d1r2 + r1
35. Combined Remainder for Three Divisors
When N is directly divided by:
d1d2d3
the remainder is:
R = d1d2r3 + d1r2 + r1
36. Example: Three-Step Combined Remainder
A number is divided successively by 3, 4 and 5. The respective remainders are 2, 1 and 3. Find the remainder when the original number is divided by 60.
Here:
R = 3 × 4 × 3 + 3 × 1 + 2
= 36 + 3 + 2
= 41
Therefore, the required remainder is 41.
37. Reconstructing After Three Successive Divisions
If final quotient q3 is also given:
N = d1d2d3q3 + d1d2r3 + d1r2 + r1
38. Example: Three-Step Reconstruction
A number is divided successively by 2, 3 and 5. The remainders are respectively 1, 2 and 4, and the final quotient is 6. Find the original number.
Use:
N = 2 × 3 × 5 × 6 + 2 × 3 × 4 + 2 × 2 + 1
= 180 + 24 + 4 + 1
= 209
39. Checking the Reconstruction
Verify 209:
- 209 ÷ 2 → quotient 104, remainder 1.
- 104 ÷ 3 → quotient 34, remainder 2.
- 34 ÷ 5 → quotient 6, remainder 4.
Hence the reconstruction is correct.
40. Repeated Division by the Same Divisor
If the same divisor d is used repeatedly:
N = dq1 + r1
q1 = dq2 + r2
q2 = dq3 + r3
41. Formula After Repeated Division by d
After three divisions:
N = d3q3 + d2r3 + dr2 + r1
42. General Repeated-Division Pattern
After k successive divisions by the same divisor d:
N = dkqk + rkdk−1 + rk−1dk−2 + ... + r2d + r1
Advanced Observation: The successive remainders behave like positional digits in base d, read in reverse order of their appearance.
43. Example: Repeated Division by 10
Suppose a number is repeatedly divided by 10.
The successive remainders give the decimal digits from right to left.
Example: 538:
- 538 ÷ 10 → remainder 8
- 53 ÷ 10 → remainder 3
- 5 ÷ 10 → remainder 5
Reading the remainders in reverse order gives:
538
44. Successive Division and Place Value
Repeated division by 10 explains why:
538 = 5 × 102 + 3 × 10 + 8
This is a direct consequence of the successive-division formula.
45. Order of Divisors Matters
In successive division, changing the order of divisors changes the weights attached to the intermediate remainders.
For divisors d1 and d2:
R = d1r2 + r1
If their order is reversed, a different pair of successive remainders may occur.
Important: Although d1d2 = d2d1, successive-division remainders depend on the order in which division is performed.
46. Direct Division versus Successive Division
Suppose N is first divided by 4 and the obtained quotient is divided by 5.
This is not the same procedure as independently dividing N by 4 and N by 5.
The correct successive relation is:
N = 4(5q2 + r2) + r1
47. Combined Quotient
In two-step successive division:
N = d1d2q2 + R
where:
R = d1r2 + r1
Thus q2 is also the quotient when N is directly divided by d1d2.
48. Example: Combined Quotient and Remainder
A number is divided by 6, giving quotient 17 and remainder 5. The quotient 17 is divided by 4, giving quotient 4 and remainder 1.
Original number:
N = 6 × 17 + 5 = 107
Direct division by 24:
107 = 24 × 4 + 11
Combined remainder formula:
6 × 1 + 5 = 11
So direct quotient = 4 and remainder = 11.
49. Successive Division with Zero Remainder
An intermediate remainder may be zero.
Example:
A number is divided by 5 with remainder 0. The quotient is divided by 3 with remainder 2.
Combined remainder modulo 15 is:
R = 5 × 2 + 0 = 10
50. Successive Division where the Second Remainder is Zero
If r2 = 0:
R = d1 × 0 + r1 = r1
Thus the combined remainder equals the first remainder.
51. Solved Example: Find the Original Number
A number is divided by 7 and leaves remainder 4. The quotient is divided by 5 and leaves remainder 3. If the final quotient is 8, find the original number.
N = 7 × 5 × 8 + 7 × 3 + 4
= 280 + 21 + 4
= 305
52. Solved Example: Find the Combined Remainder
A number leaves remainder 5 when divided by 8. The obtained quotient leaves remainder 6 when divided by 9. What remainder will the original number leave when divided by 72?
R = 8 × 6 + 5
= 53
53. Solved Example: Three Successive Remainders
A number is successively divided by 4, 5 and 6 and leaves remainders 3, 2 and 1 respectively. Find its remainder when divided by 120.
R = 4 × 5 × 1 + 4 × 2 + 3
= 20 + 8 + 3
= 31
54. Solved Example: Missing Divisor
When 263 is divided by d, quotient is 12 and remainder is 11. Find d.
Using:
263 = 12d + 11
we get:
252 = 12d
d = 21
Check: 11 < 21, so the remainder condition is satisfied.
55. Solved Example: Check Whether Data is Possible
Can a number divided by 13 have remainder 15?
No.
Since:
0 ≤ r < 13
the maximum possible remainder is 12.
Therefore remainder 15 is impossible.
56. Relationship with Remainder Properties
The division algorithm provides the basic representation:
N = dq + r
while properties of sums, products and powers of remainders were covered separately in Topic 1.12 : Remainders & Remainder Properties.
This topic focuses mainly on division structure and successive division rather than repeating general modular-arithmetic rules.
57. Relationship with HCF
Repeated division also forms the foundation of Euclid's method for finding HCF.
However, detailed HCF algorithms belong to:
Chapter 3 : HCF & LCM
and are not duplicated here.
58. Common Exam Traps
Trap 1: Always use Dividend = Divisor × Quotient + Remainder.
Trap 2: Remainder must satisfy 0 ≤ r < divisor.
Trap 3: The maximum remainder for divisor d is d−1.
Trap 4: If divisor is greater than the dividend, quotient is 0 and remainder is the dividend.
Trap 5: In successive division, divide the obtained quotient, not the original dividend again.
Trap 6: Do not simply add successive remainders.
Trap 7: For two divisions, combined remainder = d1r2 + r1.
Trap 8: For three divisions, combined remainder = d1d2r3 + d1r2 + r1.
Trap 9: The order of divisors matters in successive division.
Trap 10: When reconstructing a number, work backwards from the final quotient or use the combined formula carefully.
59. Quick Revision
- Division Algorithm: N = dq + r.
- Remainder condition: 0 ≤ r < d.
- Dividend = Divisor × Quotient + Remainder.
- Quotient = (Dividend − Remainder)/Divisor.
- Divisor = (Dividend − Remainder)/Quotient.
- Remainder = Dividend − Divisor × Quotient.
- Exact division means remainder = 0.
- Maximum remainder for divisor d is d−1.
- If dividend < divisor, quotient = 0 and remainder = dividend.
- If N leaves remainder r on division by d, then d divides N−r.
- For fixed N and positive d, quotient and remainder are unique.
- Successive division means dividing the obtained quotient again.
- Two-step formula: N = d1d2q2 + d1r2 + r1.
- Two-step combined remainder = d1r2 + r1.
- Three-step combined remainder = d1d2r3 + d1r2 + r1.
- Reverse successive division reconstructs the original number from the final quotient.
- Repeated division by d produces powers d, d2, d3, etc. in reconstruction.
- Repeated division by 10 reveals decimal digits from right to left.
- Successive remainders must not simply be added.
- The order of successive divisors matters.
- Detailed remainder algebra belongs to Topic 1.12.
- Detailed Euclidean HCF method belongs to Chapter 3.
Previous Year Questions (PYQs)
Practice these genuine previous-year questions on the division algorithm, dividend-divisor-quotient-remainder relation, successive division, combined remainder and reverse reconstruction.
Jharkhand Police SI PYQ2017 · Official Paper
Q1. A number is divided by 125. The quotient is 85 and the remainder is 22. Find the number.
A. 2665
B. 10603
C. 2835
D. 10647
Correct Answer: D. 10647
Explanation:
Use:
Dividend = Divisor × Quotient + Remainder
= 125 × 85 + 22
= 10,625 + 22
= 10,647.
Therefore, the required number is 10,647.
RRB NTPC PYQ31 July 2021 · CBT-I · Shift II
Q2. When 12401 is divided by a certain number, the quotient is 76 and the remainder is 13. Find the divisor.
A. 136
B. 947
C. 163
D. 948
Correct Answer: C. 163
Explanation:
Using:
Dividend = Divisor × Quotient + Remainder
12401 = d × 76 + 13
d × 76 = 12401 − 13 = 12388
d = 12388/76 = 163.
Hence the divisor is 163.
SSC CGL PYQ30 November 2016 · Tier-II Quantitative Abilities
Q3. In a division, the divisor is four times the quotient and twice the remainder. If the remainder is 80, find the dividend.
A. 6480
B. 9680
C. 8460
D. 4680
Correct Answer: A. 6480
Explanation:
Remainder = 80.
The divisor is twice the remainder:
Divisor = 2 × 80 = 160.
The divisor is four times the quotient:
160 = 4 × Quotient
Quotient = 40.
Therefore:
Dividend = 160 × 40 + 80
= 6400 + 80
= 6480.
SSC CPO PYQ24 November 2020 · Shift II
Q4. A number is successively divided by 3, 4 and 7, giving remainders 2, 3 and 5 respectively. What remainder will it leave when divided by 84?
A. 30
B. 71
C. 53
D. 48
Correct Answer: B. 71
Explanation:
For three successive divisions:
Combined remainder = d1d2r3 + d1r2 + r1.
Therefore:
R = 3 × 4 × 5 + 3 × 3 + 2
= 60 + 9 + 2
= 71.
Since 3 × 4 × 7 = 84, the required remainder is 71.
SSC CHSL PYQ1 July 2024 · Tier-I · Shift II
Q5. A number is successively divided by 3, 5 and 7. The respective remainders are 2, 1 and 3, and the final quotient is 3. Find the original number.
A. 367
B. 360
C. 365
D. 362
Correct Answer: C. 365
Explanation:
Work backwards from the final quotient.
Last division by 7:
7 × 3 + 3 = 24.
Previous division by 5:
5 × 24 + 1 = 121.
First division by 3:
3 × 121 + 2 = 365.
Therefore, the original number is 365.
Odisha Police SI PYQ6 July 2022 · Paper II
Q6. A number leaves remainders 1 and 4 when divided successively by 4 and 5. If the order of divisors is changed to 5 and 4, what will the successive remainders be?
A. 1, 2
B. 2, 3
C. 3, 2
D. 4, 1
Correct Answer: B. 2, 3
Explanation:
From the first successive division:
N = 4(5q + 4) + 1
= 20q + 17.
Now divide N first by 5:
N = 5(4q + 3) + 2.
So the first new remainder is 2 and the quotient is 4q + 3.
Now divide this quotient by 4:
4q + 3 leaves remainder 3.
Therefore the new successive remainders are 2, 3.
Practice MCQs
Practice these exam-oriented questions on the division algorithm, missing values, maximum remainder, successive division, combined remainder and reverse reconstruction.
Practice MCQ
Q1. A division has divisor 37, quotient 24 and remainder 19. What is the dividend?
A. 888
B. 897
C. 907
D. 917
Correct Answer: C. 907
Explanation:
Dividend = Divisor × Quotient + Remainder
= 37 × 24 + 19
= 888 + 19
= 907.
Practice MCQ
Q2. When 1259 is divided by a number, the quotient is 17 and the remainder is 1. Find the divisor.
A. 72
B. 73
C. 74
D. 76
Correct Answer: C. 74
Explanation:
Divisor = (Dividend − Remainder)/Quotient
= (1259 − 1)/17
= 1258/17
= 74.
Practice MCQ
Q3. What is the greatest possible remainder when a positive integer is divided by 29?
A. 27
B. 28
C. 29
D. 30
Correct Answer: B. 28
Explanation:
For divisor d, the remainder satisfies:
0 ≤ r < d.
Therefore the maximum possible remainder is d − 1.
29 − 1 = 28.
Practice MCQ
Q4. A number is divided successively by 4 and 6 and leaves remainders 3 and 5 respectively. What remainder will it leave when divided by 24?
A. 17
B. 20
C. 21
D. 23
Correct Answer: D. 23
Explanation:
For two successive divisions:
R = d1r2 + r1.
Therefore:
R = 4 × 5 + 3
= 23.
Practice MCQ
Q5. A number is divided successively by 5, 7 and 3 and leaves remainders 4, 2 and 1 respectively. What remainder will it leave when divided by 105?
A. 39
B. 44
C. 49
D. 54
Correct Answer: C. 49
Explanation:
For three successive divisions:
R = d1d2r3 + d1r2 + r1.
R = 5 × 7 × 1 + 5 × 2 + 4
= 35 + 10 + 4
= 49.
Practice MCQ
Q6. A number is successively divided by 3, 5 and 7. The remainders are 2, 4 and 6 respectively, and the final quotient is 2. Find the original number.
A. 304
B. 309
C. 314
D. 319
Correct Answer: C. 314
Explanation:
Work backwards:
After division by 7:
7 × 2 + 6 = 20.
Before division by 5:
5 × 20 + 4 = 104.
Original number:
3 × 104 + 2 = 314.
Practice MCQ
Q7. A number is divided three times successively by 8. The remainders are 5, 3 and 6 respectively, and the final quotient is 2. Find the number.
A. 1397
B. 1413
C. 1437
D. 1453
Correct Answer: C. 1437
Explanation:
For repeated division by 8:
N = 83q3 + 82r3 + 8r2 + r1.
= 83 × 2 + 82 × 6 + 8 × 3 + 5
= 1024 + 384 + 24 + 5
= 1437.
Practice MCQ
Q8. The number 151 leaves remainder 7 when divided by d. Which of the following cannot be the value of d?
A. 9
B. 12
C. 16
D. 10
Correct Answer: D. 10
Explanation:
If 151 leaves remainder 7 when divided by d, then:
d divides 151 − 7 = 144.
Also d > 7.
9, 12 and 16 all divide 144, but 10 does not.
Therefore 10 cannot be the divisor.
Practice MCQ
Q9. A number is first divided by 6 and then the obtained quotient is divided by 5. The respective remainders are 2 and 4, and the final quotient is 7. What are the quotient and remainder when the original number is divided directly by 30?
A. Quotient 7, Remainder 26
B. Quotient 7, Remainder 24
C. Quotient 8, Remainder 26
D. Quotient 8, Remainder 24
Correct Answer: A. Quotient 7, Remainder 26
Explanation:
The direct divisor is:
6 × 5 = 30.
The final quotient remains 7.
Combined remainder:
R = 6 × 4 + 2
= 26.
Hence the direct division gives quotient 7 and remainder 26.
Practice MCQ
Q10. A number is divided by 12 and leaves remainder 7. The quotient is then divided by 5 and leaves remainder 4. If the final quotient is 6, find the original number.
A. 403
B. 409
C. 415
D. 427
Correct Answer: C. 415
Explanation:
Reconstruct the first quotient:
q1 = 5 × 6 + 4 = 34.
Now reconstruct the original number:
N = 12 × 34 + 7
= 408 + 7
= 415.
विभाजन एल्गोरिथ्म एवं क्रमिक विभाजन
Division Algorithm किसी संख्या को divisor, quotient और remainder के संबंध में व्यक्त करता है। Successive Division में पहले किसी संख्या को divide किया जाता है और फिर प्राप्त quotient को अगली संख्या से divide किया जाता है। इन concepts की सहायता से original number की reconstruction, missing values, combined remainder तथा competitive examinations के अनेक number-system questions बहुत तेजी से हल किए जा सकते हैं।
1. Division का मूल विचार
जब किसी positive integer को किसी अन्य positive integer से divide किया जाता है, तो चार मुख्य quantities होती हैं:
- Dividend (भाज्य) — वह संख्या जिसे divide किया जा रहा है।
- Divisor (भाजक) — वह संख्या जिससे divide किया जा रहा है।
- Quotient (भागफल) — divisor, dividend में पूर्ण रूप से जितनी बार जाता है।
- Remainder (शेषफल) — पूर्ण division के बाद बची हुई मात्रा।
2. Division Algorithm
यदि dividend N और positive divisor d है, तो:
N = dq + r
जहाँ:
- N = Dividend
- d = Divisor
- q = Quotient
- r = Remainder
3. Remainder की आवश्यक Condition
Remainder हमेशा:
0 ≤ r < d
को satisfy करता है।
अत्यंत महत्वपूर्ण: Remainder कभी भी divisor के बराबर या उससे बड़ा नहीं हो सकता।
4. Division Algorithm का उदाहरण
47 को 6 से divide करें।
हम लिख सकते हैं:
47 = 6 × 7 + 5
अतः:
- Dividend = 47
- Divisor = 6
- Quotient = 7
- Remainder = 5
5. Fundamental Formula
Dividend = Divisor × Quotient + Remainder
Elementary division questions का सबसे महत्वपूर्ण आधार यही formula है।
6. Dividend ज्ञात करना
यदि divisor, quotient और remainder दिए गए हों, तो:
Dividend = Divisor × Quotient + Remainder
उदाहरण: Divisor = 13, quotient = 24 और remainder = 7।
Dividend = 13 × 24 + 7
= 312 + 7 = 319
7. Quotient ज्ञात करना
हम जानते हैं:
N = dq + r
इसलिए:
q = (N − r)/d
उदाहरण: 157 को 15 से divide करने पर remainder 7 मिलता है।
q = (157 − 7)/15
= 150/15 = 10
8. Divisor ज्ञात करना
यदि dividend N, quotient q और remainder r दिए गए हों:
d = (N − r)/q
उदाहरण: Dividend = 183, quotient = 11 और remainder = 7।
d = (183 − 7)/11
= 176/11 = 16
9. Remainder ज्ञात करना
यदि dividend, divisor और quotient ज्ञात हों:
r = N − dq
उदाहरण:
N = 236, d = 15, q = 15
अतः:
r = 236 − 15 × 15
= 236 − 225 = 11
10. Exact Division
यदि remainder = 0 हो, तो division को exact division कहते हैं।
इस स्थिति में:
N = dq
और divisor, dividend को पूरी तरह divide करता है।
उदाहरण:
84 = 7 × 12 + 0
11. Non-Exact Division
यदि remainder non-zero हो, तो division exact नहीं होता।
उदाहरण:
86 = 7 × 12 + 2
12. Maximum Possible Remainder
चूँकि:
0 ≤ r < d
इसलिए divisor d होने पर maximum possible remainder:
d − 1
होगा।
उदाहरण: किसी संख्या को 17 से divide करने पर maximum remainder 16 हो सकता है।
13. Minimum Possible Remainder
Minimum possible remainder:
0
होता है। यह तब होता है जब dividend, divisor से exactly divisible हो।
14. Divisor हमेशा Remainder से बड़ा होगा
यदि किसी division में remainder r है, तो:
Divisor > r
उदाहरण: यदि remainder 11 है, तो divisor कम-से-कम 12 होना चाहिए।
15. जब Divisor, Dividend से बड़ा हो
यदि:
0 ≤ N < d
तो:
N = d × 0 + N
अतः:
- Quotient = 0
- Remainder = N
उदाहरण:
7 ÷ 12 → Quotient = 0, Remainder = 7
16. 1 से Division
किसी भी integer N को 1 से divide करने पर:
N = 1 × N + 0
अतः quotient = N और remainder = 0।
17. किसी Number की Reconstruction
कई examination questions में divisor, quotient और remainder देकर original number पूछा जाता है।
ऐसे questions में सीधे:
N = dq + r
का उपयोग करें।
18. हल किया गया उदाहरण: Dividend ज्ञात करें
किसी संख्या को 23 से divide करने पर quotient 17 और remainder 11 प्राप्त होता है। संख्या ज्ञात करें।
N = 23 × 17 + 11
= 391 + 11 = 402
अतः required number = 402।
19. Dividend और Remainder का Difference
चूँकि:
N = dq + r
इसलिए:
N − r = dq
अर्थात:
N − r हमेशा divisor d से divisible होगा।
20. Useful Divisibility Interpretation
यदि N को d से divide करने पर remainder r मिलता है, तो:
d divides N − r
यह property possible divisor ज्ञात करने वाले questions में बहुत उपयोगी है।
21. उदाहरण: Possible Divisor ज्ञात करना
83 को d से divide करने पर remainder 5 मिलता है।
तो:
d divides 83 − 5 = 78
साथ ही:
d > 5
इसलिए d, 78 का ऐसा factor होगा जो 5 से बड़ा हो।
Possible values:
6, 13, 26, 39, 78
22. Quotient और Remainder की Uniqueness
किसी fixed dividend N और positive divisor d के लिए integers q और r की केवल एक ही pair होती है जो:
N = dq + r, 0 ≤ r < d
को satisfy करती है।
महत्वपूर्ण गुण: Fixed dividend और positive divisor के लिए quotient तथा remainder unique होते हैं।
23. Successive Division क्या है?
Successive Division में पहले original number को किसी divisor से divide किया जाता है। उसके बाद प्राप्त quotient को अगली संख्या से divide किया जाता है।
यह original number को अलग-अलग divisors से independently divide करने से अलग concept है।
महत्वपूर्ण: Successive division में दूसरी division पहले प्राप्त quotient पर की जाती है, original dividend पर नहीं।
24. Two-Step Successive Division
मान लें N को पहले d1 से divide किया गया:
N = d1q1 + r1
अब quotient q1 को d2 से divide किया गया:
q1 = d2q2 + r2
25. दो Successive Divisions को Combine करना
दूसरी equation:
q1 = d2q2 + r2
को पहली equation में substitute करें:
N = d1q1 + r1
हमें मिलता है:
N = d1d2q2 + d1r2 + r1
26. Two-Step Division का Combined Remainder
जब original number N को directly:
d1d2
से divide किया जाता है, तो remainder होगा:
R = d1r2 + r1
क्योंकि:
0 ≤ r1 < d1
और:
0 ≤ r2 < d2
इसलिए combined remainder automatically:
0 ≤ R < d1d2
को satisfy करता है।
27. उदाहरण: Two Successive Divisions
किसी संख्या को 5 से divide करने पर remainder 3 मिलता है। प्राप्त quotient को 7 से divide करने पर remainder 4 मिलता है। Original number को 35 से divide करने पर remainder क्या होगा?
यहाँ:
- d1 = 5
- r1 = 3
- d2 = 7
- r2 = 4
अतः:
R = 5 × 4 + 3
= 23
इसलिए original number को 35 से divide करने पर remainder 23 मिलेगा।
28. Remainders को सीधे Add क्यों नहीं कर सकते?
ऊपर के example में successive remainders 3 और 4 हैं।
Combined remainder:
3 + 4 = 7
नहीं होगा।
Correct calculation है:
5 × 4 + 3 = 23
Common Trap: Successive division में intermediate remainders को simply add नहीं करना चाहिए।
29. Reverse Successive Division
यदि final quotient और intermediate remainders दिए हों, तो original number ज्ञात करने के लिए backwards work करें।
Two-step case में:
q1 = d2q2 + r2
और फिर:
N = d1q1 + r1
30. उदाहरण: Reverse Successive Division
किसी number को 4 से divide करने पर remainder 3 मिलता है। प्राप्त quotient को 6 से divide करने पर remainder 2 मिलता है। Final quotient 5 है। Original number ज्ञात करें।
पहले first quotient reconstruct करें:
q1 = 6 × 5 + 2 = 32
अब original number:
N = 4 × 32 + 3
= 131
31. Two-Step Reconstruction का Direct Formula
हमने पाया:
N = d1d2q2 + d1r2 + r1
यदि d1, d2, q2, r1 और r2 दिए हों, तो इसी formula से original number सीधे निकाला जा सकता है।
32. वही उदाहरण Direct Formula से
यहाँ:
d1 = 4, d2 = 6, q2 = 5, r1 = 3, r2 = 2
अतः:
N = 4 × 6 × 5 + 4 × 2 + 3
= 120 + 8 + 3
= 131
33. Three-Step Successive Division
मान लें:
N = d1q1 + r1
q1 = d2q2 + r2
q2 = d3q3 + r3
34. Three Successive Divisions का Combined Formula
Successive substitution करने पर:
N = d1d2d3q3 + d1d2r3 + d1r2 + r1
35. Three Divisors का Combined Remainder
जब N को directly:
d1d2d3
से divide किया जाता है, तो remainder होगा:
R = d1d2r3 + d1r2 + r1
36. उदाहरण: Three-Step Combined Remainder
किसी संख्या को क्रमशः 3, 4 और 5 से divide किया जाता है। क्रमशः remainders 2, 1 और 3 हैं। Original number को 60 से divide करने पर remainder ज्ञात करें।
Formula:
R = 3 × 4 × 3 + 3 × 1 + 2
= 36 + 3 + 2
= 41
अतः required remainder = 41।
37. Three Successive Divisions के बाद Original Number
यदि final quotient q3 भी दिया गया हो, तो:
N = d1d2d3q3 + d1d2r3 + d1r2 + r1
38. उदाहरण: Three-Step Reconstruction
किसी संख्या को क्रमशः 2, 3 और 5 से divide किया जाता है। Remainders क्रमशः 1, 2 और 4 हैं तथा final quotient 6 है। Original number ज्ञात करें।
Formula:
N = 2 × 3 × 5 × 6 + 2 × 3 × 4 + 2 × 2 + 1
= 180 + 24 + 4 + 1
= 209
39. Reconstruction को Verify करना
209 को check करें:
- 209 ÷ 2 → quotient 104, remainder 1
- 104 ÷ 3 → quotient 34, remainder 2
- 34 ÷ 5 → quotient 6, remainder 4
सभी conditions satisfy होती हैं। इसलिए original number 209 सही है।
40. Same Divisor से Repeated Division
यदि किसी number को बार-बार same divisor d से divide किया जाए:
N = dq1 + r1
q1 = dq2 + r2
q2 = dq3 + r3
41. तीन बार Same Divisor से Division का Formula
तीन successive divisions के बाद:
N = d3q3 + d2r3 + dr2 + r1
42. Repeated Division का General Pattern
यदि same divisor d से k successive divisions हों, तो:
N = dkqk + rkdk−1 + rk−1dk−2 + ... + r2d + r1
Advanced Observation: Same divisor d से successive division में प्राप्त remainders base-d representation के positional digits की तरह कार्य करते हैं। इन्हें प्राप्त होने के reverse order में पढ़ा जाता है।
43. उदाहरण: 10 से Repeated Division
यदि किसी decimal number को repeatedly 10 से divide किया जाए, तो successive remainders उसके digits को right से left क्रम में देते हैं।
उदाहरण: 538
- 538 ÷ 10 → quotient 53, remainder 8
- 53 ÷ 10 → quotient 5, remainder 3
- 5 ÷ 10 → quotient 0, remainder 5
Remainders मिले:
8, 3, 5
इन्हें reverse order में पढ़ने पर:
538
44. Successive Division और Place Value
Repeated division by 10 यह भी समझाता है कि:
538 = 5 × 102 + 3 × 10 + 8
यह repeated-division formula का ही decimal place-value रूप है।
45. Divisors का Order महत्वपूर्ण है
Successive division में divisors का order बदलने पर intermediate remainders के weights बदल जाते हैं।
यदि पहले d1 और फिर d2 से divide किया जाए:
R = d1r2 + r1
यदि divisors का order बदल दिया जाए, तो successive remainders की pair भी बदल सकती है।
महत्वपूर्ण: यद्यपि d1d2 = d2d1, successive-division remainders division के order पर निर्भर करते हैं।
46. Direct Division और Successive Division में अंतर
मान लें N को पहले 4 से divide किया जाता है और प्राप्त quotient को 5 से divide किया जाता है।
यह N को separately 4 और 5 से divide करने के समान नहीं है।
Correct successive relation:
N = 4(5q2 + r2) + r1
47. Combined Quotient
Two-step successive division में:
N = d1d2q2 + R
जहाँ:
R = d1r2 + r1
और क्योंकि R < d1d2, इसलिए q2 वही quotient है जो N को directly d1d2 से divide करने पर प्राप्त होगा।
48. उदाहरण: Combined Quotient और Remainder
किसी number को 6 से divide करने पर quotient 17 और remainder 5 मिलता है। Quotient 17 को 4 से divide करने पर quotient 4 और remainder 1 मिलता है।
Original number:
N = 6 × 17 + 5 = 107
अब directly 24 से:
107 = 24 × 4 + 11
Combined remainder formula:
R = 6 × 1 + 5 = 11
अतः:
- Direct quotient = 4
- Direct remainder = 11
49. Successive Division में Zero Remainder
Intermediate remainder zero भी हो सकता है।
उदाहरण: किसी number को 5 से divide करने पर remainder 0 मिलता है। प्राप्त quotient को 3 से divide करने पर remainder 2 मिलता है।
15 से direct division का combined remainder:
R = 5 × 2 + 0
= 10
50. यदि Second Remainder Zero हो
यदि:
r2 = 0
तो:
R = d1 × 0 + r1
अतः:
R = r1
इस स्थिति में combined remainder first remainder के बराबर होगा।
51. हल किया गया उदाहरण: Original Number ज्ञात करें
किसी number को 7 से divide करने पर remainder 4 मिलता है। प्राप्त quotient को 5 से divide करने पर remainder 3 मिलता है। Final quotient 8 है। Original number ज्ञात करें।
Formula:
N = 7 × 5 × 8 + 7 × 3 + 4
= 280 + 21 + 4
= 305
52. हल किया गया उदाहरण: Combined Remainder ज्ञात करें
किसी number को 8 से divide करने पर remainder 5 मिलता है। प्राप्त quotient को 9 से divide करने पर remainder 6 मिलता है। Original number को 72 से divide करने पर remainder क्या होगा?
R = 8 × 6 + 5
= 53
53. हल किया गया उदाहरण: Three Successive Remainders
किसी number को क्रमशः 4, 5 और 6 से divide करने पर remainders क्रमशः 3, 2 और 1 मिलते हैं। Number को 120 से divide करने पर remainder ज्ञात करें।
R = 4 × 5 × 1 + 4 × 2 + 3
= 20 + 8 + 3
= 31
54. हल किया गया उदाहरण: Missing Divisor
263 को d से divide करने पर quotient 12 और remainder 11 मिलता है। d ज्ञात करें।
Division algorithm:
263 = 12d + 11
इसलिए:
263 − 11 = 12d
252 = 12d
d = 21
Check:
11 < 21
अतः remainder condition भी satisfy होती है।
55. हल किया गया उदाहरण: दिए गए Data की Validity
क्या किसी संख्या को 13 से divide करने पर remainder 15 हो सकता है?
नहीं।
क्योंकि:
0 ≤ r < 13
Maximum possible remainder:
12
इसलिए remainder 15 impossible है।
56. Remainder Properties से संबंध
Division Algorithm का basic representation है:
N = dq + r
जबकि sums, differences, products और powers के remainder rules को हमने अलग से:
Topic 1.12 : Remainders & Remainder Properties
में cover किया है।
इस topic का मुख्य focus division structure, number reconstruction और successive division है; इसलिए general modular-arithmetic properties को यहाँ unnecessarily repeat नहीं किया गया है।
57. HCF से संबंध
Repeated division का concept आगे चलकर Euclid's method द्वारा HCF ज्ञात करने का आधार भी बनता है।
लेकिन detailed HCF algorithm को:
Chapter 3 : HCF & LCM
में पढ़ाया जाएगा। इसलिए यहाँ उसकी unnecessary duplication नहीं की गई है।
58. परीक्षा में होने वाली सामान्य गलतियाँ
गलती 1: हमेशा Dividend = Divisor × Quotient + Remainder का उपयोग करें।
गलती 2: Remainder के लिए 0 ≤ r < divisor होना अनिवार्य है।
गलती 3: Divisor d के लिए maximum remainder d−1 होता है।
गलती 4: यदि divisor dividend से बड़ा है, तो quotient 0 और remainder dividend होगा।
गलती 5: Successive division में अगली division प्राप्त quotient पर होती है, original dividend पर नहीं।
गलती 6: Successive remainders को simply add न करें।
गलती 7: Two-step combined remainder = d1r2 + r1।
गलती 8: Three-step combined remainder = d1d2r3 + d1r2 + r1।
गलती 9: Successive division में divisors का order महत्वपूर्ण है।
गलती 10: Original number reconstruct करते समय final quotient से reverse work करें या combined formula सावधानी से लगाएँ।
59. त्वरित पुनरावृत्ति
- Division Algorithm: N = dq + r।
- Remainder condition: 0 ≤ r < d।
- Dividend = Divisor × Quotient + Remainder।
- Quotient = (Dividend − Remainder)/Divisor।
- Divisor = (Dividend − Remainder)/Quotient।
- Remainder = Dividend − Divisor × Quotient।
- Exact division में remainder = 0 होता है।
- Divisor d के लिए maximum remainder = d−1।
- यदि dividend < divisor, तो quotient = 0 और remainder = dividend।
- यदि N को d से divide करने पर remainder r हो, तो d, N−r को divide करता है।
- Fixed N और positive d के लिए quotient तथा remainder unique होते हैं।
- Successive division में प्राप्त quotient को अगली संख्या से divide किया जाता है।
- Two-step formula: N = d1d2q2 + d1r2 + r1।
- Two-step combined remainder = d1r2 + r1।
- Three-step combined remainder = d1d2r3 + d1r2 + r1।
- Reverse successive division से final quotient से original number reconstruct किया जा सकता है।
- Same divisor d से repeated division में d, d2, d3 आदि powers प्राप्त होती हैं।
- Repeated division by 10 decimal digits को right से left reveal करती है।
- Successive remainders को सीधे add नहीं करना चाहिए।
- Successive divisors का order remainders को प्रभावित करता है।
- Detailed remainder algebra Topic 1.12 में covered है।
- Detailed Euclidean HCF method Chapter 3 में covered होगा।
पिछले वर्षों में पूछे गए प्रश्न (PYQs)
Division Algorithm, dividend-divisor-quotient-remainder relation, successive division, combined remainder और reverse reconstruction पर आधारित इन पिछले वर्षों के प्रश्नों का अभ्यास करें।
Jharkhand Police SI PYQ2017 · Official Paper
प्रश्न 1. किसी संख्या को 125 से divide करने पर quotient 85 और remainder 22 प्राप्त होता है। संख्या ज्ञात करें।
A. 2665
B. 10603
C. 2835
D. 10647
सही उत्तर: D. 10647
व्याख्या:
Formula:
Dividend = Divisor × Quotient + Remainder
= 125 × 85 + 22
= 10,625 + 22
= 10,647।
अतः required number = 10,647।
RRB NTPC PYQ31 जुलाई 2021 · CBT-I · Shift II
प्रश्न 2. 12401 को किसी संख्या से divide करने पर quotient 76 और remainder 13 प्राप्त होता है। Divisor ज्ञात करें।
A. 136
B. 947
C. 163
D. 948
सही उत्तर: C. 163
व्याख्या:
Division Algorithm से:
12401 = d × 76 + 13
d × 76 = 12401 − 13
= 12388
d = 12388/76
= 163।
अतः divisor = 163।
SSC CGL PYQ30 नवंबर 2016 · Tier-II Quantitative Abilities
प्रश्न 3. एक division में divisor, quotient का चार गुना तथा remainder का दो गुना है। यदि remainder 80 है, तो dividend ज्ञात करें।
A. 6480
B. 9680
C. 8460
D. 4680
सही उत्तर: A. 6480
व्याख्या:
Remainder = 80।
Divisor, remainder का दो गुना है:
Divisor = 2 × 80 = 160।
Divisor, quotient का चार गुना है:
160 = 4 × Quotient
Quotient = 40।
अब:
Dividend = 160 × 40 + 80
= 6400 + 80
= 6480।
SSC CPO PYQ24 नवंबर 2020 · Shift II
प्रश्न 4. किसी संख्या को क्रमशः 3, 4 और 7 से divide करने पर remainders क्रमशः 2, 3 और 5 प्राप्त होते हैं। उस संख्या को 84 से divide करने पर remainder क्या होगा?
A. 30
B. 71
C. 53
D. 48
सही उत्तर: B. 71
व्याख्या:
Three successive divisions के लिए:
Combined remainder = d1d2r3 + d1r2 + r1।
अतः:
R = 3 × 4 × 5 + 3 × 3 + 2
= 60 + 9 + 2
= 71।
और 3 × 4 × 7 = 84। इसलिए required remainder = 71।
SSC CHSL PYQ1 जुलाई 2024 · Tier-I · Shift II
प्रश्न 5. किसी संख्या को क्रमशः 3, 5 और 7 से divide किया जाता है। Remainders क्रमशः 2, 1 और 3 हैं तथा final quotient 3 है। Original number ज्ञात करें।
A. 367
B. 360
C. 365
D. 362
सही उत्तर: C. 365
व्याख्या:
Final quotient से reverse calculation करें।
7 से division से पहले quotient:
7 × 3 + 3 = 24।
5 से division से पहले quotient:
5 × 24 + 1 = 121।
Original number:
3 × 121 + 2
= 365।
अतः original number = 365।
Odisha Police SI PYQ6 जुलाई 2022 · Paper II
प्रश्न 6. किसी संख्या को क्रमशः 4 और 5 से divide करने पर remainders 1 और 4 प्राप्त होते हैं। यदि divisors का क्रम बदलकर पहले 5 और फिर 4 कर दिया जाए, तो नए successive remainders क्या होंगे?
A. 1, 2
B. 2, 3
C. 3, 2
D. 4, 1
सही उत्तर: B. 2, 3
व्याख्या:
Original successive division से:
N = 4(5q + 4) + 1
= 20q + 17।
अब N को पहले 5 से divide करें:
N = 5(4q + 3) + 2।
इसलिए पहला नया remainder = 2 और quotient = 4q + 3।
अब इस quotient को 4 से divide करने पर remainder = 3।
अतः नए successive remainders = 2, 3।
अभ्यास प्रश्न (Practice MCQs)
Division Algorithm, missing values, maximum remainder, successive division, combined remainder तथा reverse reconstruction पर आधारित इन परीक्षा-उपयोगी प्रश्नों का अभ्यास करें।
Practice MCQ
प्रश्न 1. किसी division में divisor 37, quotient 24 और remainder 19 है। Dividend क्या होगा?
A. 888
B. 897
C. 907
D. 917
सही उत्तर: C. 907
व्याख्या:
Dividend = Divisor × Quotient + Remainder
= 37 × 24 + 19
= 888 + 19
= 907।
Practice MCQ
प्रश्न 2. 1259 को किसी संख्या से divide करने पर quotient 17 और remainder 1 मिलता है। Divisor ज्ञात करें।
A. 72
B. 73
C. 74
D. 76
सही उत्तर: C. 74
व्याख्या:
Divisor = (Dividend − Remainder)/Quotient
= (1259 − 1)/17
= 1258/17
= 74।
Practice MCQ
प्रश्न 3. किसी positive integer को 29 से divide करने पर greatest possible remainder क्या होगा?
A. 27
B. 28
C. 29
D. 30
सही उत्तर: B. 28
व्याख्या:
Divisor d के लिए:
0 ≤ r < d।
इसलिए maximum possible remainder = d − 1।
29 − 1 = 28।
Practice MCQ
प्रश्न 4. किसी संख्या को क्रमशः 4 और 6 से divide करने पर remainders 3 और 5 मिलते हैं। संख्या को 24 से divide करने पर remainder क्या होगा?
A. 17
B. 20
C. 21
D. 23
सही उत्तर: D. 23
व्याख्या:
Two successive divisions के लिए:
R = d1r2 + r1।
इसलिए:
R = 4 × 5 + 3
= 23।
Practice MCQ
प्रश्न 5. किसी संख्या को क्रमशः 5, 7 और 3 से divide करने पर remainders 4, 2 और 1 प्राप्त होते हैं। उस संख्या को 105 से divide करने पर remainder क्या होगा?
A. 39
B. 44
C. 49
D. 54
सही उत्तर: C. 49
व्याख्या:
Three successive divisions के लिए:
R = d1d2r3 + d1r2 + r1।
R = 5 × 7 × 1 + 5 × 2 + 4
= 35 + 10 + 4
= 49।
Practice MCQ
प्रश्न 6. किसी संख्या को क्रमशः 3, 5 और 7 से divide किया जाता है। Remainders क्रमशः 2, 4 और 6 हैं तथा final quotient 2 है। Original number ज्ञात करें।
A. 304
B. 309
C. 314
D. 319
सही उत्तर: C. 314
व्याख्या:
Final quotient से reverse work करें।
7 से division से पहले:
7 × 2 + 6 = 20।
5 से division से पहले:
5 × 20 + 4 = 104।
Original number:
3 × 104 + 2
= 314।
Practice MCQ
प्रश्न 7. किसी संख्या को 8 से लगातार तीन बार divide किया जाता है। Remainders क्रमशः 5, 3 और 6 हैं तथा final quotient 2 है। संख्या ज्ञात करें।
A. 1397
B. 1413
C. 1437
D. 1453
सही उत्तर: C. 1437
व्याख्या:
Repeated division by 8 के लिए:
N = 83q3 + 82r3 + 8r2 + r1।
= 83 × 2 + 82 × 6 + 8 × 3 + 5
= 1024 + 384 + 24 + 5
= 1437।
Practice MCQ
प्रश्न 8. 151 को d से divide करने पर remainder 7 मिलता है। निम्न में से कौन-सा d का मान नहीं हो सकता?
A. 9
B. 12
C. 16
D. 10
सही उत्तर: D. 10
व्याख्या:
यदि 151 को d से divide करने पर remainder 7 मिलता है, तो:
d, 151 − 7 = 144 को divide करेगा।
साथ ही d > 7 होना चाहिए।
9, 12 और 16 सभी 144 के factors हैं, लेकिन 10 नहीं है।
इसलिए 10 divisor नहीं हो सकता।
Practice MCQ
प्रश्न 9. किसी संख्या को पहले 6 से divide किया जाता है और फिर प्राप्त quotient को 5 से divide किया जाता है। Remainders क्रमशः 2 और 4 हैं तथा final quotient 7 है। यदि original number को directly 30 से divide किया जाए, तो quotient और remainder क्या होंगे?
A. Quotient 7, Remainder 26
B. Quotient 7, Remainder 24
C. Quotient 8, Remainder 26
D. Quotient 8, Remainder 24
सही उत्तर: A. Quotient 7, Remainder 26
व्याख्या:
Direct divisor:
6 × 5 = 30।
Final quotient 7 ही direct quotient होगा।
Combined remainder:
R = 6 × 4 + 2
= 26।
अतः quotient = 7 और remainder = 26।
Practice MCQ
प्रश्न 10. किसी संख्या को 12 से divide करने पर remainder 7 मिलता है। प्राप्त quotient को 5 से divide करने पर remainder 4 मिलता है। यदि final quotient 6 है, तो original number ज्ञात करें।
A. 403
B. 409
C. 415
D. 427
सही उत्तर: C. 415
व्याख्या:
पहले intermediate quotient reconstruct करें:
q1 = 5 × 6 + 4
= 34।
अब original number:
N = 12 × 34 + 7
= 408 + 7
= 415।