Chapter 1 Revision : Number System Quick Revision, PYQs & Practice Questions अध्याय 1 पुनरावृत्ति : संख्या पद्धति त्वरित पुनरावृत्ति, PYQs एवं अभ्यास प्रश्न

Revise the complete Number System chapter for JSSC, JPSC, SSC, Railway and other competitive examinations with a concise chapter-wise quick revision, important formulas and properties, verified previous-year questions and mixed practice MCQs. Covers all major Chapter 1 concepts from numbers and digits to divisibility, factors, remainders, unit digits, factorials, number of digits, division algorithm and important number properties with detailed answers and explanations. JSSC, JPSC, SSC, Railway एवं अन्य प्रतियोगी परीक्षाओं के लिए पूरे Number System chapter की त्वरित पुनरावृत्ति करें। इसमें Chapter 1 के सभी प्रमुख concepts—numbers and digits, number classification, prime-composite numbers, factors-multiples, divisibility, factors की संख्या, remainders, unit digits, factorials, number of digits, division algorithm तथा महत्वपूर्ण number properties—के concise revision notes, महत्वपूर्ण formulas, verified PYQs और mixed practice MCQs विस्तृत उत्तर एवं व्याख्या सहित शामिल हैं।

Chapter 1 : Number System अध्याय 1 : संख्या पद्धति

Chapter 1 Revision: Number System

Use this chapter-end revision sheet to quickly recall the most important concepts, formulas, properties and shortcuts from the complete Number System chapter before attempting the verified PYQs and mixed Practice MCQs. It covers all 17 topics of Chapter 1 without repeating the full theory already explained in the individual study materials.

1. Number System Classification

Number TypeMeaning / ExamplesImportant Point
Natural Numbers1, 2, 3, 4, ...Counting numbers; smallest natural number = 1.
Whole Numbers0, 1, 2, 3, ...Natural numbers together with 0.
Integers..., −3, −2, −1, 0, 1, 2, 3, ...Includes negative integers, zero and positive integers.
Rational Numbersp/q, where q ≠ 0Decimal expansion terminates or repeats.
Irrational Numbers√2, √3, π, ...Decimal expansion is non-terminating and non-repeating.
Real NumbersAll rational and irrational numbersRepresented on the real number line.
Key Inclusion: Natural Numbers ⊂ Whole Numbers ⊂ Integers ⊂ Rational Numbers ⊂ Real Numbers.

2. Numbers and Digits — Essential Facts

  • The ten decimal digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
  • A digit is a symbol; a number may contain one or more digits.
  • Smallest positive integer = 1.
  • There is no greatest integer.
  • 0 has one digit.
  • A minus sign is not counted as a digit.
  • Leading zeros do not increase the number of digits.

3. Greatest and Smallest n-Digit Numbers

For n ≥ 1:

Smallest n-digit positive number = 10n−1

Greatest n-digit number = 10n − 1

Number of n-digit positive integers = 9 × 10n−1

Example: For 4-digit numbers:

  • Smallest = 1000
  • Greatest = 9999
  • Total = 9000

4. Positive and Negative Numbers

  • Positive number: greater than 0.
  • Negative number: less than 0.
  • 0 is neither positive nor negative.
  • For a positive number a, −a is its additive inverse.
  • The absolute value |a| represents distance from zero and is never negative.

5. Even and Odd Numbers

Every integer is either:

2k — even

2k+1 — odd

OperationResult
Even + EvenEven
Odd + OddEven
Even + OddOdd
Even − EvenEven
Odd − OddEven
Even − OddOdd
Even × Any IntegerEven
Odd × OddOdd
  • 0 is even.
  • Even number squared → even.
  • Odd number squared → odd.
  • Every odd square leaves remainder 1 when divided by 8.

6. Prime and Composite Numbers

  • A prime number has exactly two positive factors: 1 and itself.
  • A composite number has more than two positive factors.
  • 1 is neither prime nor composite.
  • 2 is the smallest prime number.
  • 2 is the only even prime number.
  • 4 is the smallest composite number.

7. Co-prime Numbers

Two or more integers are co-prime when their HCF is 1.

  • Co-prime numbers need not themselves be prime.
  • Example: 8 and 15 are co-prime.
  • Any two consecutive positive integers are co-prime.
  • Distinct prime numbers are always co-prime.
Exam Trap: “Co-prime” does not mean that both numbers must be prime.

8. Twin Prime Numbers

Two prime numbers differing by 2 are called twin primes.

Examples:

(3,5), (5,7), (11,13), (17,19), (29,31)

Except for the pair (3,5), twin primes greater than 3 are of the form:

6n−1 and 6n+1

9. Factors and Multiples

If a divides b exactly, then:

  • a is a factor of b.
  • b is a multiple of a.

Important facts:

  • 1 is a factor of every positive integer.
  • Every positive integer is a factor of itself.
  • A positive integer has finitely many factors but infinitely many multiples.

10. Prime Factorisation

Every integer greater than 1 can be expressed uniquely, apart from order, as a product of primes.

General form:

N = p1a1p2a2...pkak

Example:

360 = 23 × 32 × 5

11. Divisibility Rules — Essential Table

DivisorRule
2Last digit is even: 0, 2, 4, 6 or 8.
3Sum of digits is divisible by 3.
4Number formed by the last two digits is divisible by 4.
5Last digit is 0 or 5.
6Number is divisible by both 2 and 3.
8Number formed by the last three digits is divisible by 8.
9Sum of digits is divisible by 9.
10Last digit is 0.
11Difference between sums of alternate digits is 0 or a multiple of 11.
12Number is divisible by both 3 and 4.
15Number is divisible by both 3 and 5.
25Last two digits are 00, 25, 50 or 75.

12. Number of Factors

If:

N = paqbrc

then number of positive factors:

d(N) = (a+1)(b+1)(c+1)

Example:

72 = 23 × 32

d(72) = (3+1)(2+1) = 12

13. Odd and Even Factors

If:

N = 2apbqc...

then:

Number of odd factors = (b+1)(c+1)...

Number of even factors:

Total factors − Odd factors

14. Perfect Square and Number of Factors

Important Theorem: A positive integer has an odd number of positive factors if and only if it is a perfect square.

This happens because factors normally occur in pairs d and N/d, except when d = √N.

15. Sum of Factors

If:

N = paqb

then:

σ(N) = (1+p+p2+...+pa)(1+q+q2+...+qb)

Using geometric progression:

1+p+...+pa = (pa+1−1)/(p−1)

16. Product of All Positive Factors

If N has d positive factors, then:

Product of all positive factors = Nd/2

For a perfect square, this formula remains valid even though d is odd.

17. Division Algorithm

For positive divisor d:

N = dq+r

with:

0 ≤ r < d

  • Dividend = Divisor × Quotient + Remainder.
  • Maximum possible remainder = d−1.
  • If N
  • If remainder = 0, division is exact.

18. Basic Remainder Properties

If:

a ≡ r1 (mod m),   b ≡ r2 (mod m)

then:

  • a+b ≡ r1+r2 (mod m)
  • a−b ≡ r1−r2 (mod m)
  • ab ≡ r1r2 (mod m)

Reduce the result again modulo m if necessary.

19. Same-Remainder Property

If two numbers A and B leave the same remainder when divided by d, then:

d divides A−B

If several numbers leave the same remainder, a possible common divisor must divide their pairwise differences.

20. Unit Digit Cycles

Last Digit of BaseUnit-Digit CycleLength
001
111
22, 4, 8, 64
33, 9, 7, 14
44, 62
551
661
77, 9, 3, 14
88, 4, 2, 64
99, 12
Exam Shortcut: Divide the exponent by the cycle length. If the remainder is 0, use the final element of that cycle.

21. Unit Digit of Products and Sums

For products, only the unit digits of individual factors matter.

Example:

23 × 47 × 16

Use:

3 × 7 × 6 = 126

Required unit digit = 6.

For sums and differences, first determine the unit digit of each term and then combine them modulo 10.

22. Factorial — Core Facts

For a positive integer n:

n! = n(n−1)(n−2)...3×2×1

  • 0! = 1
  • 1! = 1
  • n! = n × (n−1)!
  • For n ≥ 5, n! always ends in at least one zero.

23. Highest Power of a Prime in n!

The exponent of a prime p in n! is:

vp(n!) = ⌊n/p⌋ + ⌊n/p2⌋ + ⌊n/p3⌋ + ...

Continue until the denominator exceeds n.

Example: Highest power of 2 dividing 10!:

⌊10/2⌋ + ⌊10/4⌋ + ⌊10/8⌋ = 5+2+1 = 8

Therefore:

28 divides 10!

24. Trailing Zeros in a Factorial

Each trailing zero requires one factor 10 = 2×5.

Since factorials contain more factors 2 than factors 5, count the factors 5:

Z(n!) = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...

Example: Trailing zeros in 100!:

⌊100/5⌋ + ⌊100/25⌋ = 20+4 = 24

25. Factorial Product and Quotient Caution

For factorial products and quotients, prime exponents are often safer than directly manipulating trailing-zero counts.

For an expression X:

Trailing zeros = min(v2(X), v5(X))

Exam Trap: For factorial quotients, do not blindly subtract the trailing-zero counts of numerator and denominator. Compare the remaining powers of 2 and 5.

26. Number of Digits — Basic Formula

For a positive integer N:

Number of digits = ⌊log10N⌋ + 1

Important boundaries:

  • 10n has n+1 digits.
  • 10n−1 has n digits.

27. Number of Digits in a Power

For an:

Digits in an = ⌊n log10a⌋ + 1

Useful common logarithms:

ValueApproximate log10
20.30103
30.47712
50.69897
70.84510

28. Useful 2 and 5 Shortcut

Because:

2n × 5n = 10n

pair equal powers of 2 and 5 before using logarithms.

Example:

220 × 515 = 25 × 1015 = 32 × 1015

Hence total digits = 17.

29. Digits in Products

If A has m digits and B has n digits, then AB can have:

m+n−1 or m+n digits

This is useful when an exact product is unnecessary.

30. Digits in Factorials

For n!:

Digits in n! = ⌊log10(n!)⌋ + 1

and:

log(n!) = log 1 + log 2 + ... + log n

For very large n, Stirling's approximation may be used, but exact log-sum is preferable when an exact digit count is required.

31. Successive Division — Two Steps

If:

N = d1q1 + r1

and:

q1 = d2q2 + r2

then:

N = d1d2q2 + d1r2 + r1

Combined remainder when dividing N directly by d1d2:

R = d1r2 + r1

32. Successive Division — Three Steps

For divisors d1, d2, d3 and successive remainders r1, r2, r3:

R = d1d2r3 + d1r2 + r1

Remember: In successive division, the second divisor acts on the quotient obtained from the first division, not on the original number again.

33. Reverse Successive Division

If the final quotient is given, reconstruct the original number backwards.

Example: A number is divided by 4 and leaves remainder 3. The quotient is divided by 6 and leaves remainder 2. Final quotient = 5.

First quotient = 6×5+2 = 32

Original number = 4×32+3 = 131

34. Greatest and Smallest Number Formation

  • Greatest number from given digits → arrange digits in descending order.
  • Smallest number when zero is absent → arrange digits in ascending order.
  • If zero is present → place the smallest non-zero digit first, then zero(s), then remaining digits in ascending order.
  • For distinct digits, never repeat a digit unless repetition is explicitly permitted.

Example: Using 0, 2, 5, 8 once each:

  • Greatest = 8520
  • Smallest = 2058

35. Greatest and Smallest Multiples

If N leaves remainder r on division by d:

Greatest multiple of d not exceeding N = N−r

If r ≠ 0:

Smallest multiple of d greater than N = N+(d−r)

If r = 0, N itself is already a multiple.

36. Greatest/Smallest Number with a Required Remainder

To find a number satisfying:

N ≡ r (mod d)

  • For a greatest-number problem, start from the upper boundary and move downward to the required residue.
  • For a smallest-number problem, start from the lower boundary and move upward to the required residue.

Example: Greatest three-digit number leaving remainder 5 on division by 17:

991 = 17×58+5

37. Counting Multiples in a Range

Number of positive multiples of d not exceeding N:

⌊N/d⌋

Number of multiples of d from A to B inclusive:

⌊B/d⌋ − ⌊(A−1)/d⌋

38. Two-Digit Number and its Reverse

If a is the tens digit and b is the units digit:

Original number = 10a+b

Reverse = 10b+a

Difference:

9(a−b)

Sum:

11(a+b)

Therefore: Difference is divisible by 9 and sum is divisible by 11.

39. Three-Digit Number and its Reverse

For:

100a+10b+c

reverse is:

100c+10b+a

Difference:

99(a−c)

Hence the difference is always divisible by 99.

40. Appending Digits and Zeros

If digit d is appended to integer N:

New number = 10N+d

If k zeros are appended:

New number = N×10k

If a digit d is prefixed to a k-digit number N:

New number = d×10k+N

41. Repeated-Digit Numbers

The n-digit number consisting entirely of 1s is:

111...111 = (10n−1)/9

If digit d is repeated n times:

d(10n−1)/9

42. Repeating a Number Block

If a two-digit number N is written twice:

NN = 101N

If a three-digit number N is written twice:

NN = 1001N

Since:

1001 = 7×11×13

Important: Every six-digit number of the form abcabc is divisible by 7, 11 and 13.

43. Consecutive Integers

Consecutive integers are represented as:

n, n+1, n+2, ...

  • Any two consecutive positive integers are co-prime.
  • Among two consecutive integers, one is even.
  • Among three consecutive integers, one is divisible by 3 and at least one is even.

44. Product of Consecutive Integers

Product of any k consecutive integers is divisible by:

k!

Special cases:

  • 2 consecutive integers → divisible by 2
  • 3 consecutive integers → divisible by 6
  • 4 consecutive integers → divisible by 24
  • 5 consecutive integers → divisible by 120

45. Standard Number Sums

SeriesFormula
1+2+3+...+nn(n+1)/2
1+3+5+...+(2n−1)n2
2+4+6+...+2nn(n+1)
12+22+...+n2n(n+1)(2n+1)/6
13+23+...+n3[n(n+1)/2]2

46. Consecutive Squares

Difference between consecutive squares:

(n+1)2−n2

= 2n+1

Hence the difference between two consecutive perfect squares is always odd.

47. Important Number Forms

Every integer can be represented in one of the forms:

2k or 2k+1

Every integer can also be represented uniquely as one of:

3k, 3k+1, 3k+2

Similarly, modulo m every integer belongs to exactly one residue class:

mk, mk+1, ..., mk+(m−1)

48. High-Value Number System Facts

  • 0 is even.
  • 0 is neither positive nor negative.
  • 1 is neither prime nor composite.
  • 2 is the smallest and only even prime.
  • 4 is the smallest composite number.
  • Any two consecutive positive integers are co-prime.
  • Every odd square is congruent to 1 modulo 8.
  • A positive integer has an odd number of positive factors iff it is a perfect square.
  • A difference of a two-digit number and its reverse is divisible by 9.
  • A sum of a two-digit number and its reverse is divisible by 11.

49. Chapter 1 — One-Minute Formula Sheet

  • Smallest n-digit number = 10n−1
  • Greatest n-digit number = 10n−1
  • Number of n-digit positive integers = 9×10n−1
  • N = dq+r, with 0≤r
  • Maximum remainder for divisor d = d−1
  • If N=paqb..., number of factors = (a+1)(b+1)...
  • Product of all factors = Nd(N)/2
  • Digits in N = ⌊log10N⌋+1
  • Digits in an = ⌊nlog10a⌋+1
  • vp(n!) = ⌊n/p⌋+⌊n/p2⌋+...
  • Z(n!) = ⌊n/5⌋+⌊n/25⌋+...
  • Two-step successive remainder = d1r2+r1
  • Two-digit number = 10a+b
  • Reverse difference = 9(a−b)
  • Reverse sum = 11(a+b)
  • abcabc = 1001×abc
  • 1001 = 7×11×13
  • Multiples of d in [A,B] = ⌊B/d⌋−⌊(A−1)/d⌋
  • Product of k consecutive integers is divisible by k!
  • 1+2+...+n = n(n+1)/2
  • First n odd numbers sum = n2
  • First n even numbers sum = n(n+1)
  • Sum of first n squares = n(n+1)(2n+1)/6
  • Sum of first n cubes = [n(n+1)/2]2

50. Chapter 1 — Most Important Exam Traps

Trap 1: 1 is neither prime nor composite.
Trap 2: 0 is even but neither positive nor negative.
Trap 3: Co-prime numbers need not themselves be prime.
Trap 4: Remainder must always be smaller than the divisor.
Trap 5: In a unit-digit cycle, exponent remainder 0 means use the last term of the cycle.
Trap 6: Trailing zeros in n! are counted through powers of 5, not merely by dividing n by 10.
Trap 7: 10n has n+1 digits.
Trap 8: Do not round logarithms too early in digit-count questions.
Trap 9: In successive division, each new divisor acts on the previous quotient.
Trap 10: Successive remainders are not simply added.
Trap 11: A leading zero cannot be used to create a smaller multi-digit number.
Trap 12: Distinct-digit questions do not allow repetition unless explicitly stated.
Trap 13: Product of k consecutive integers is divisible by k!, but need not equal k!.
Trap 14: Same remainder on division implies divisibility of differences, not necessarily divisibility of the original numbers.
Trap 15: For factorial quotients, trailing-zero counts should not always be directly subtracted.

Ready for Chapter 1 Practice

You have now revised the complete Number System chapter. The next section should test these concepts through verified previous-year questions first, followed by mixed chapter-level Practice MCQs.

Chapter 1 Previous Year Questions (PYQs)

Test your complete Number System preparation through these mixed previous-year questions. Each question includes the concept being tested so that you can identify weak areas after attempting it.

SSC CGL PYQ Divisibility Rules 3 March 2020 · Shift III

Q1. Which digit should replace * in 94*2357 so that the resulting number is divisible by 11?

A. 8
B. 7
C. 3
D. 1
Correct Answer: C. 3
Explanation:
For divisibility by 11, the difference between the sums of alternate digits must be 0 or a multiple of 11.

For 94x2357:
9 − 4 + x − 2 + 3 − 5 + 7 = 8 + x.

Since x is a digit, the possible multiple of 11 is 11.
8 + x = 11
x = 3.
RRB NTPC PYQ Number of Factors 31 January 2021 · CBT-I · Shift I

Q2. How many factors of 27 × 33 × 54 × 7 are even?

A. 320
B. 40
C. 280
D. 84
Correct Answer: C. 280
Explanation:
Total factors:
(7+1)(3+1)(4+1)(1+1)
= 8×4×5×2
= 320.

Odd factors contain no factor 2:
(3+1)(4+1)(1+1)
= 4×5×2
= 40.

Even factors = 320 − 40 = 280.
SSC CPO PYQ Sum of Factors 11 November 2022 · Tier-I · Shift I

Q3. Find the sum of all odd positive divisors of 216.

A. 16
B. 14
C. 40
D. 600
Correct Answer: C. 40
Explanation:
216 = 23 × 33.

An odd divisor cannot contain any factor 2.
Therefore odd divisors come only from powers of 3:

1 + 3 + 32 + 33
= 1 + 3 + 9 + 27
= 40.
RRC Group D PYQ Same Remainder Property 16 October 2018 · Shift III

Q4. What is the greatest number by which 76, 151 and 226 can be divided so that the same remainder is obtained in every case?

A. 60
B. 70
C. 75
D. 69
Correct Answer: C. 75
Explanation:
If several numbers leave the same remainder on division by d, then d divides their differences.

151 − 76 = 75
226 − 151 = 75
226 − 76 = 150

Required greatest divisor:
HCF(75, 75, 150) = 75.
SSC CHSL PYQ Unit Digit & Cyclicity 19 March 2018 · Shift I

Q5. Find the unit digit of 153144 − 115123 − 111510 + 21625.

A. 1
B. 5
C. 6
D. 3
Correct Answer: A. 1
Explanation:
Unit digit of 153144 depends on 3144.
The cycle of 3 is 3, 9, 7, 1. Since 144 is divisible by 4, unit digit = 1.

115123 → 5
111510 → 1
21625 → 6

Required unit digit:
1 − 5 − 1 + 6 = 1.
RPF SI PYQ Factorial & Trailing Zeros 19 December 2018 · Shift I

Q6. How many trailing zeros are there in 76!?

A. 18
B. 16
C. 20
D. 14
Correct Answer: A. 18
Explanation:
Trailing zeros in n! are determined by the number of factors 5.

Z(76!) = ⌊76/5⌋ + ⌊76/25⌋ + ⌊76/125⌋
= 15 + 3 + 0
= 18.
CDS PYQ Number of Digits 5 February 2017 · Mathematics

Q7. How many digits are present in 240, given log102 = 0.301?

A. 14
B. 13
C. 12
D. 11
Correct Answer: B. 13
Explanation:
For N = 240:

log N = 40 log 2
= 40×0.301
= 12.04.

Number of digits = ⌊12.04⌋ + 1
= 12 + 1
= 13.
Jharkhand Police SI PYQ Division Algorithm 2017 · Official Paper

Q8. A number is divided by 125. If the quotient is 85 and the remainder is 22, what is the number?

A. 2665
B. 10603
C. 2835
D. 10647
Correct Answer: D. 10647
Explanation:
Dividend = Divisor × Quotient + Remainder.

N = 125×85 + 22
= 10,625 + 22
= 10,647.
SSC CGL PYQ Number of Factors 9 March 2018 · Tier-II

Q9. If N = 411 + 412 + 413 + 414, how many positive factors does N have?

A. 92
B. 48
C. 50
D. 51
Correct Answer: A. 92
Explanation:
Factor out 411:

N = 411(1+4+16+64)
= 411×85
= 222×5×17.

Therefore the number of positive factors is:
(22+1)(1+1)(1+1)
= 23×2×2
= 92.
KVS PRT PYQ Greatest/Smallest Numbers 22 February 2023 · Shift II

Q10. Find the difference between the greatest and smallest five-digit numbers in which no digit is repeated.

A. 89,999
B. 88,531
C. 88,888
D. 89,998
Correct Answer: B. 88,531
Explanation:
Greatest five-digit number with distinct digits:
98,765.

Smallest five-digit number with distinct digits:
10,234.

Difference:
98,765 − 10,234
= 88,531.
UPSC CSE PYQ Repeated Number Property 2023 · Preliminary Examination · CSAT

Q11. A six-digit number is formed as XYZXYZ, where XYZ represents any three-digit block. Such a number is always divisible by which of the following?

A. 7 and 11 only
B. 11 and 13 only
C. 7 and 13 only
D. 7, 11 and 13
Correct Answer: D. 7, 11 and 13
Explanation:
Let XYZ represent the number N.

XYZXYZ = 1000N + N
= 1001N.

Now:
1001 = 7×11×13.

Hence every number of the form XYZXYZ is divisible by 7, 11 and 13.
SSC MTS PYQ Standard Number Sums 16 September 2017 · Shift III

Q12. What is the sum of the first 15 odd natural numbers?

A. 255
B. 225
C. 235
D. 215
Correct Answer: B. 225
Explanation:
The sum of the first n odd natural numbers is:

1+3+5+...+(2n−1) = n2.

For n = 15:
152 = 225.
SSC MTS PYQ Prime & Composite Numbers 30 October 2017 · Shift I

Q13. Which of the following numbers is neither prime nor composite?

A. 2
B. 1
C. 3
D. 5
Correct Answer: B. 1
Explanation:
A prime number has exactly two positive factors, while a composite number has more than two positive factors.

The number 1 has only one positive factor — 1 itself.

Therefore, 1 is neither prime nor composite.
RRC Group D PYQ Twin Prime Numbers 17 September 2018 · Shift I

Q14. Which of the following pairs consists of twin prime numbers?

A. (37, 41)
B. (3, 7)
C. (43, 47)
D. (71, 73)
Correct Answer: D. (71, 73)
Explanation:
Twin primes are two prime numbers whose difference is exactly 2.

73 − 71 = 2, and both 71 and 73 are prime.

The differences in the other given pairs are 4.

Hence (71, 73) is the twin-prime pair.
RRB Group D PYQ Co-prime Numbers 25 August 2022 · Shift II

Q15. Which of the following pairs of numbers is co-prime?

A. (17, 23)
B. (14, 21)
C. (12, 24)
D. (15, 25)
Correct Answer: A. (17, 23)
Explanation:
Two numbers are co-prime if their HCF is 1.

17 and 23 are distinct prime numbers, so their only common positive factor is 1.

The other pairs have common factors greater than 1.

Hence (17, 23) is the co-prime pair.
SSC CHSL PYQ Rational Numbers 15 November 2025 · Tier-I · Shift I

Q16. The decimal expansion of a rational number is:

A. Non-terminating and non-repeating
B. Terminating or repeating
C. Terminating only
D. Repeating only
Correct Answer: B. Terminating or repeating
Explanation:
Every rational number can be written as p/q, where p and q are integers and q ≠ 0.

Its decimal expansion is either:
• terminating, or
• non-terminating but repeating/recurring.

A non-terminating, non-repeating decimal represents an irrational number.
RRC Group D PYQ Perfect-Square Factors 18 September 2018 · Shift II

Q17. How many factors of 1296 are perfect squares?

A. 8
B. 9
C. 12
D. 10
Correct Answer: B. 9
Explanation:
Prime factorisation:

1296 = 24 × 34.

For a factor to be a perfect square, the exponents of both primes must be even.

Possible exponent of 2: 0, 2, 4 → 3 choices.
Possible exponent of 3: 0, 2, 4 → 3 choices.

Total perfect-square factors:
3 × 3 = 9.
SSC MTS PYQ Same Remainder 26 July 2022 · Shift III

Q18. Find the smallest natural number x that must be subtracted from 1800 so that 1800 − x leaves remainder 5 when divided by each of 7, 11 and 23.

A. 24
B. 25
C. 26
D. 20
Correct Answer: A. 24
Explanation:
LCM(7, 11, 23) = 7×11×23 = 1771.

1800 = 1771 + 29.

At present the remainder relative to a common multiple is 29. We require remainder 5.

Therefore:
x = 29 − 5 = 24.

Indeed:
1800 − 24 = 1776 = 1771 + 5.
SSC CPO PYQ Successive Division 24 November 2020 · Shift II

Q19. A number is successively divided by 3, 4 and 7, giving remainders 2, 3 and 5 respectively. What remainder will the original number leave when divided by 84?

A. 30
B. 71
C. 53
D. 48
Correct Answer: B. 71
Explanation:
For three successive divisions:

R = d1d2r3 + d1r2 + r1.

Therefore:
R = 3×4×5 + 3×3 + 2
= 60 + 9 + 2
= 71.

Since 3×4×7 = 84, the number leaves remainder 71 when divided by 84.
SSC CHSL PYQ Reverse Successive Division 1 July 2024 · Tier-I · Shift II

Q20. 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.

Before division by 7:
7×3 + 3 = 24.

Before division by 5:
5×24 + 1 = 121.

Original number:
3×121 + 2 = 365.
CDS PYQ Number of Digits 3 September 2023 · Elementary Mathematics

Q21. How many digits are there in the expansion of 125100? Given log102 = 0.301.

A. 69
B. 70
C. 209
D. 210
Correct Answer: D. 210
Explanation:
125 = 53, so:

125100 = 5300.

log 5 = log(10/2)
= 1 − 0.301
= 0.699.

Therefore:
log(5300) = 300×0.699 = 209.7.

Number of digits = ⌊209.7⌋ + 1
= 210.
SSC GD PYQ Greatest Number / Divisibility 7 March 2024 · Shift IV

Q22. What is the greatest four-digit number exactly divisible by 15, 20, 25 and 30?

A. 9900
B. 9300
C. 9700
D. 9930
Correct Answer: A. 9900
Explanation:
The required number must be a common multiple of 15, 20, 25 and 30.

LCM(15,20,25,30) = 300.

The greatest four-digit multiple of 300 is:
300×33 = 9900.

300×34 = 10,200, which is a five-digit number.
SSC MTS PYQ Greatest Number with Remainder 9 May 2023 · Shift I

Q23. What is the greatest four-digit number which leaves remainder 2 when divided by both 3 and 4?

A. 9994
B. 9996
C. 9998
D. 9995
Correct Answer: C. 9998
Explanation:
The required number has the form:

LCM(3,4)×k + 2
= 12k + 2.

The greatest four-digit multiple of 12 is 9996.

Therefore:
9996 + 2 = 9998.

It leaves remainder 2 on division by both 3 and 4.
SSC GD PYQ Prime Numbers 19 May 2026 · Shift III

Q24. How many prime numbers are there between 70 and 80?

A. 3
B. 4
C. 5
D. 2
Correct Answer: A. 3
Explanation:
The odd numbers between 70 and 80 are:

71, 73, 75, 77, 79.

75 = 3×25, so it is composite.
77 = 7×11, so it is composite.

The prime numbers are:
71, 73 and 79.

Hence the required number of primes is 3.
RRB NTPC PYQ Number of Factors 23 July 2021 · CBT-I · Shift I

Q25. How many positive factors does the number 12,288 have?

A. 22
B. 26
C. 24
D. 28
Correct Answer: B. 26
Explanation:
Prime factorisation:

12,288 = 212 × 3.

Therefore, number of positive factors:

(12+1)(1+1)
= 13×2
= 26.
SSC Selection Post PYQ Number of Factors 30 July 2025 · Graduate Level · Shift III

Q26. What is the total number of positive factors of 3600?

A. 45
B. 30
C. 35
D. 20
Correct Answer: A. 45
Explanation:
Prime factorisation:

3600 = 36×100
= (22×32)(22×52)
= 24×32×52.

Hence number of factors:

(4+1)(2+1)(2+1)
= 5×3×3
= 45.
SSC CGL PYQ Remainders & Powers 14 July 2023 · Tier-I · Shift I

Q27. What is the remainder when 2654081 + 9 is divided by 266?

A. 8
B. 6
C. 1
D. 9
Correct Answer: A. 8
Explanation:
Since:

265 ≡ −1 (mod 266),

2654081 ≡ (−1)4081 ≡ −1 (mod 266).

Therefore:

2654081 + 9 ≡ −1 + 9
≡ 8 (mod 266).

Hence the remainder is 8.
SSC CGL PYQ Remainder Properties 26 September 2024 · Tier-I · Shift I

Q28. Find the remainder when 920 + 2 is divided by 4.

A. 1
B. 3
C. 0
D. 2
Correct Answer: B. 3
Explanation:
9 ≡ 1 (mod 4).

Therefore:
920 ≡ 120 ≡ 1 (mod 4).

Hence:
920 + 2 ≡ 1+2
≡ 3 (mod 4).

Required remainder = 3.
SSC CGL PYQ Remainders & Powers 9 December 2022 · Tier-I · Shift IV

Q29. What is the remainder when 1919 + 20 is divided by 18?

A. 3
B. 2
C. 1
D. 0
Correct Answer: A. 3
Explanation:
19 ≡ 1 (mod 18).

Therefore:
1919 ≡ 1 (mod 18).

Also:
20 ≡ 2 (mod 18).

Thus:
1919 + 20 ≡ 1+2
≡ 3 (mod 18).
SSC CGL PYQ Algebraic Remainder Property 25 July 2023 · Tier-I · Shift III

Q30. What is the remainder when x17 + 1 is divided by x + 1?

A. x
B. x − 1
C. 0
D. 1
Correct Answer: C. 0
Explanation:
By the Remainder Theorem, substitute x = −1.

Remainder = (−1)17 + 1
= −1 + 1
= 0.

Equivalently, an + bn is divisible by a+b when n is odd.
SSC Selection Post PYQ Unit Digit & Cyclicity 3 February 2022 · Matric Level · Shift II

Q31. If N = 30738 + 52420, what is the unit digit of N?

A. 6
B. 5
C. 3
D. 4
Correct Answer: B. 5
Explanation:
For 30738, consider 738.

Unit-digit cycle of 7:
7, 9, 3, 1.

38 mod 4 = 2, so the unit digit is 9.

For 52420, consider 420.
An even power of 4 has unit digit 6.

Therefore:
9+6 = 15.

Required unit digit = 5.
UP Police SI PYQ Last Two Digits 14 December 2017 · Shift II

Q32. What are the last two digits of 56283 × 141283 × 125254?

A. 25
B. 10
C. 01
D. 00
Correct Answer: D. 00
Explanation:
We only need to determine whether the product contains at least two factors of 10.

56 contains several factors of 2, while 125 = 53 supplies a large number of factors of 5.

Hence the complete product contains far more than two pairs of 2×5 = 10.

Therefore it is divisible by 100 and its last two digits are 00.
RPF SI PYQ Factorial & Trailing Zeros 16 January 2019 · Shift I

Q33. Find the number of trailing zeros in 735!.

A. 162
B. 192
C. 172
D. 182
Correct Answer: D. 182
Explanation:
Trailing zeros in n! are counted using powers of 5:

Z(735!) = ⌊735/5⌋ + ⌊735/25⌋ + ⌊735/125⌋ + ⌊735/625⌋

= 147 + 29 + 5 + 1
= 182.
UPSC CAPF PYQ Factorials & Remainders 4 August 2024 · Paper I

Q34. What is the remainder when 1! + 2! + 3! + ... + 500! is divided by 8?

A. 1
B. 2
C. 3
D. 4
Correct Answer: A. 1
Explanation:
For every n ≥ 4, n! is divisible by 8.

Therefore only the first three factorials affect the remainder:

1! + 2! + 3!
= 1 + 2 + 6
= 9.

9 mod 8 = 1.

Hence the required remainder is 1.
ISRO VSSC PYQ Factorial & Unit Digit 11 February 2024 · Technical Assistant (Electronics)

Q35. What is the digit at the unit place of 0! + 1! + 2! + 3! + 4!?

A. 2
B. 3
C. 4
D. 25
Correct Answer: C. 4
Explanation:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24.

Therefore:
1+1+2+6+24 = 34.

The unit digit of 34 is 4.
CDS PYQ Number of Digits 3 February 2019 · Mathematics

Q36. Given log102 = 0.301 and log103 = 0.477, how many digits are there in 10810?

A. 19
B. 20
C. 21
D. 22
Correct Answer: C. 21
Explanation:
108 = 22 × 33.

Let N = 10810.

log N = 10 log 108
= 10[2log2 + 3log3]

= 10[2(0.301) + 3(0.477)]
= 10[0.602 + 1.431]
= 20.33.

Therefore:
Number of digits = ⌊20.33⌋ + 1
= 21.

अध्याय 1 पुनरावृत्ति : संख्या पद्धति

Chapter 1 के verified PYQs और mixed Practice MCQs हल करने से पहले पूरे Number System के महत्वपूर्ण concepts, formulas, properties और shortcuts को एक स्थान पर revise करें। यह revision sheet सभी 17 topics का सार प्रस्तुत करती है और individual study materials में दी गई पूरी theory को अनावश्यक रूप से repeat नहीं करती।

1. Number System का Classification

Number Typeअर्थ / उदाहरणमहत्वपूर्ण तथ्य
Natural Numbers1, 2, 3, 4, ...Counting numbers; smallest natural number = 1।
Whole Numbers0, 1, 2, 3, ...Natural numbers के साथ 0 भी शामिल होता है।
Integers..., −3, −2, −1, 0, 1, 2, 3, ...Negative integers, zero और positive integers सभी शामिल हैं।
Rational Numbersp/q, जहाँ q ≠ 0Decimal expansion terminating या recurring/repeating होता है।
Irrational Numbers√2, √3, π, ...Decimal expansion non-terminating और non-repeating होता है।
Real Numbersसभी rational और irrational numbersReal number line पर represent किए जा सकते हैं।
मुख्य संबंध: Natural Numbers ⊂ Whole Numbers ⊂ Integers ⊂ Rational Numbers ⊂ Real Numbers.

2. Numbers और Digits — आवश्यक तथ्य

  • Decimal system के 10 digits हैं: 0, 1, 2, 3, 4, 5, 6, 7, 8 और 9।
  • Digit एक symbol है, जबकि number एक या अधिक digits से बन सकता है।
  • Smallest positive integer = 1।
  • कोई greatest integer नहीं होता।
  • 0 में 1 digit होता है।
  • Minus sign (−) को digit नहीं गिना जाता।
  • Leading zeros digit count नहीं बढ़ाते।

3. Greatest और Smallest n-Digit Numbers

n ≥ 1 के लिए:

Smallest n-digit positive number = 10n−1

Greatest n-digit number = 10n − 1

n-digit positive integers की संख्या = 9 × 10n−1

उदाहरण: 4-digit numbers के लिए:

  • Smallest = 1000
  • Greatest = 9999
  • Total = 9000

4. Positive और Negative Numbers

  • 0 से बड़ी संख्या positive होती है।
  • 0 से छोटी संख्या negative होती है।
  • 0 न positive है और न negative।
  • Positive number a का additive inverse = −a।
  • |a|, a की zero से दूरी दर्शाता है और कभी negative नहीं होता।

5. Even और Odd Numbers

हर integer को इनमें से exactly एक form में लिखा जा सकता है:

2k — Even

2k+1 — Odd

OperationResult
Even + EvenEven
Odd + OddEven
Even + OddOdd
Even − EvenEven
Odd − OddEven
Even − OddOdd
Even × Any IntegerEven
Odd × OddOdd
  • 0 even है।
  • Even number का square even होता है।
  • Odd number का square odd होता है।
  • हर odd square को 8 से divide करने पर remainder 1 मिलता है।

6. Prime और Composite Numbers

  • Prime number के exactly 2 positive factors होते हैं — 1 और स्वयं वह संख्या।
  • Composite number के 2 से अधिक positive factors होते हैं।
  • 1 न prime है और न composite।
  • 2 smallest prime number है।
  • 2 only even prime number है।
  • 4 smallest composite number है।

7. Co-prime Numbers

दो या अधिक integers co-prime होते हैं यदि उनका HCF = 1 हो।

  • Co-prime numbers का individually prime होना आवश्यक नहीं है।
  • उदाहरण: 8 और 15 co-prime हैं।
  • किसी भी दो consecutive positive integers का HCF = 1 होता है।
  • दो distinct prime numbers हमेशा co-prime होते हैं।
Exam Trap: “Co-prime” का अर्थ यह नहीं है कि दोनों numbers prime ही हों।

8. Twin Prime Numbers

ऐसे दो prime numbers जिनका difference 2 हो, twin primes कहलाते हैं।

उदाहरण:

(3,5), (5,7), (11,13), (17,19), (29,31)

(3,5) को छोड़कर 3 से बड़े twin primes सामान्यतः इस form में होते हैं:

6n−1 और 6n+1

9. Factors और Multiples

यदि a, b को exactly divide करता है, तो:

  • a, b का factor है।
  • b, a का multiple है।

महत्वपूर्ण तथ्य:

  • 1 हर positive integer का factor है।
  • हर positive integer स्वयं अपना factor है।
  • किसी positive integer के factors finite होते हैं, जबकि multiples infinite होते हैं।

10. Prime Factorisation

1 से बड़ा हर integer, order को छोड़कर, uniquely primes के product के रूप में लिखा जा सकता है।

General form:

N = p1a1p2a2...pkak

उदाहरण:

360 = 23 × 32 × 5

11. Divisibility Rules — Essential Table

DivisorRule
2Last digit 0, 2, 4, 6 या 8 हो।
3Digits का sum 3 से divisible हो।
4Last two digits से बनी संख्या 4 से divisible हो।
5Last digit 0 या 5 हो।
6Number 2 और 3 दोनों से divisible हो।
8Last three digits से बनी संख्या 8 से divisible हो।
9Digits का sum 9 से divisible हो।
10Last digit 0 हो।
11Alternate digits के sums का difference 0 या 11 का multiple हो।
12Number 3 और 4 दोनों से divisible हो।
15Number 3 और 5 दोनों से divisible हो।
25Last two digits 00, 25, 50 या 75 हों।

12. Number of Factors

यदि:

N = paqbrc

तो positive factors की संख्या:

d(N) = (a+1)(b+1)(c+1)

उदाहरण:

72 = 23 × 32

d(72) = (3+1)(2+1) = 12

13. Odd और Even Factors

यदि:

N = 2apbqc...

तो:

Odd factors की संख्या = (b+1)(c+1)...

और:

Even factors की संख्या = Total factors − Odd factors

14. Perfect Square और Number of Factors

महत्वपूर्ण Theorem: किसी positive integer के positive factors की संख्या odd होगी यदि और केवल यदि वह perfect square हो।

सामान्यतः factors d और N/d की pairs में आते हैं। Perfect square में √N स्वयं के साथ pair बनाता है, इसलिए total factor count odd हो जाता है।

15. Sum of Factors

यदि:

N = paqb

तो:

σ(N) = (1+p+p2+...+pa)(1+q+q2+...+qb)

Geometric progression से:

1+p+...+pa = (pa+1−1)/(p−1)

16. सभी Positive Factors का Product

यदि N के d positive factors हैं, तो:

Product of all positive factors = Nd/2

यह formula perfect square के लिए भी valid है, भले ही d odd हो।

17. Division Algorithm

Positive divisor d के लिए:

N = dq+r

जहाँ:

0 ≤ r < d

  • Dividend = Divisor × Quotient + Remainder।
  • Maximum possible remainder = d−1।
  • यदि N
  • यदि remainder = 0, तो division exact है।

18. Basic Remainder Properties

यदि:

a ≡ r1 (mod m),   b ≡ r2 (mod m)

तो:

  • a+b ≡ r1+r2 (mod m)
  • a−b ≡ r1−r2 (mod m)
  • ab ≡ r1r2 (mod m)

यदि प्राप्त result m से बड़ा या negative हो, तो उसे पुनः modulo m में reduce करें।

19. Same-Remainder Property

यदि A और B को d से divide करने पर same remainder मिलता है, तो:

d divides A−B

यदि कई numbers same remainder देते हैं, तो common divisor उनके pairwise differences को divide करेगा।

20. Unit Digit Cycles

Base का Last DigitUnit-Digit CycleCycle Length
001
111
22, 4, 8, 64
33, 9, 7, 14
44, 62
551
661
77, 9, 3, 14
88, 4, 2, 64
99, 12
Exam Shortcut: Exponent को cycle length से divide करें। यदि remainder 0 हो, तो cycle का अंतिम element लें।

21. Products और Sums का Unit Digit

Product का unit digit निकालने के लिए केवल individual factors के unit digits पर्याप्त होते हैं।

उदाहरण:

23 × 47 × 16

केवल unit digits लें:

3 × 7 × 6 = 126

Required unit digit = 6।

Sum या difference वाले questions में पहले प्रत्येक term का unit digit निकालें और फिर final result को modulo 10 में reduce करें।

22. Factorial — मुख्य तथ्य

Positive integer n के लिए:

n! = n(n−1)(n−2)...3×2×1

  • 0! = 1
  • 1! = 1
  • n! = n × (n−1)!
  • n ≥ 5 होने पर n! के अंत में कम-से-कम एक zero अवश्य होता है।

23. n! में किसी Prime की Highest Power

n! में prime p की exponent होती है:

vp(n!) = ⌊n/p⌋ + ⌊n/p2⌋ + ⌊n/p3⌋ + ...

Terms तब तक लेते हैं जब तक denominator n से बड़ा न हो जाए।

उदाहरण: 10! को divide करने वाली 2 की highest power:

⌊10/2⌋ + ⌊10/4⌋ + ⌊10/8⌋ = 5+2+1 = 8

अतः:

28 divides 10!

24. Factorial में Trailing Zeros

हर trailing zero के लिए एक factor 10 = 2×5 चाहिए।

Factorials में factors 2, factors 5 से अधिक होते हैं, इसलिए factors 5 की संख्या count की जाती है।

Z(n!) = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ...

उदाहरण: 100! में trailing zeros:

⌊100/5⌋ + ⌊100/25⌋ = 20+4 = 24

25. Factorial Product और Quotient में सावधानी

Factorial products और quotients में केवल trailing-zero counts manipulate करने के बजाय prime exponents का उपयोग अधिक reliable होता है।

किसी expression X के लिए:

Trailing zeros = min(v2(X), v5(X))

Exam Trap: Factorial quotient में numerator और denominator के trailing-zero counts को सीधे subtract करना हमेशा सही नहीं होता। बची हुई powers of 2 और 5 को compare करें।

26. Number of Digits — Basic Formula

Positive integer N के लिए:

Number of digits = ⌊log10N⌋ + 1

Important boundaries:

  • 10n में n+1 digits होते हैं।
  • 10n−1 में n digits होते हैं।

27. Power में Number of Digits

an के लिए:

Digits in an = ⌊n log10a⌋ + 1

Useful common logarithms:

ValueApproximate log10
20.30103
30.47712
50.69897
70.84510

28. 2 और 5 का Useful Shortcut

चूँकि:

2n × 5n = 10n

इसलिए logarithm लगाने से पहले equal powers of 2 और 5 को pair करना चाहिए।

उदाहरण:

220 × 515 = 25 × 1015 = 32 × 1015

अतः total digits = 17।

29. Product में Digits की संख्या

यदि A में m digits और B में n digits हैं, तो AB में:

m+n−1 या m+n digits

हो सकते हैं।

इस property से कई questions exact multiplication किए बिना हल किए जा सकते हैं।

30. Factorial में Digits की संख्या

n! के लिए:

Digits in n! = ⌊log10(n!)⌋ + 1

और:

log(n!) = log 1 + log 2 + ... + log n

बहुत बड़े n के लिए Stirling's approximation उपयोगी हो सकती है, लेकिन exact digit count के लिए sufficiently accurate log-sum बेहतर है।

31. Successive Division — Two Steps

यदि:

N = d1q1 + r1

और:

q1 = d2q2 + r2

तो:

N = d1d2q2 + d1r2 + r1

Original number को directly d1d2 से divide करने पर combined remainder:

R = d1r2 + r1

32. Successive Division — Three Steps

Divisors d1, d2, d3 तथा successive remainders r1, r2, r3 के लिए:

R = d1d2r3 + d1r2 + r1

याद रखें: Successive division में दूसरा divisor, पहली division से प्राप्त quotient पर कार्य करता है; original number पर दोबारा नहीं।

33. Reverse Successive Division

यदि final quotient दिया हो, तो original number को backwards reconstruct करें।

उदाहरण: किसी number को 4 से divide करने पर remainder 3 मिलता है। Quotient को 6 से divide करने पर remainder 2 और final quotient 5 है।

First quotient = 6×5+2 = 32

Original number = 4×32+3 = 131

34. Greatest और Smallest Number Formation

  • Given digits से greatest number → digits को descending order में रखें।
  • Zero absent हो तो smallest number → digits को ascending order में रखें।
  • Zero present हो तो smallest non-zero digit पहले, फिर zero(s), फिर remaining digits ascending order में रखें।
  • Distinct-digit question में repetition तभी करें जब explicitly allowed हो।

उदाहरण: 0, 2, 5, 8 का exactly once उपयोग:

  • Greatest = 8520
  • Smallest = 2058

35. Greatest और Smallest Multiples

यदि N को d से divide करने पर remainder r मिले:

N से अधिक न होने वाला greatest multiple of d = N−r

यदि r ≠ 0:

N से बड़ा smallest multiple of d = N+(d−r)

यदि r = 0, तो N स्वयं d का multiple है।

36. Required Remainder वाली Greatest/Smallest Number

यदि condition हो:

N ≡ r (mod d)

  • Greatest-number problem में upper boundary से नीचे required residue तक आएँ।
  • Smallest-number problem में lower boundary से ऊपर required residue तक जाएँ।

उदाहरण: 17 से divide करने पर remainder 5 छोड़ने वाली greatest three-digit number:

991 = 17×58+5

37. किसी Range में Multiples की संख्या

d के positive multiples जो N से अधिक नहीं हैं:

⌊N/d⌋

A से B inclusive तक d के multiples की संख्या:

⌊B/d⌋ − ⌊(A−1)/d⌋

38. Two-Digit Number और उसका Reverse

यदि tens digit = a और units digit = b:

Original number = 10a+b

Reverse = 10b+a

Difference:

9(a−b)

Sum:

11(a+b)

अतः: Difference हमेशा 9 से और sum हमेशा 11 से divisible होता है।

39. Three-Digit Number और उसका Reverse

Original:

100a+10b+c

Reverse:

100c+10b+a

Difference:

99(a−c)

इसलिए difference हमेशा 99 से divisible होता है।

40. Digits और Zeros Append करना

यदि integer N के right side में digit d लगाया जाए:

New number = 10N+d

यदि k zeros append किए जाएँ:

New number = N×10k

यदि k-digit number N के पहले digit d लगाया जाए:

New number = d×10k+N

41. Repeated-Digit Numbers

केवल 1 से बनी n-digit number:

111...111 = (10n−1)/9

यदि digit d को n बार repeat किया जाए:

d(10n−1)/9

42. किसी Number Block को Repeat करना

यदि two-digit number N को दो बार लिखा जाए:

NN = 101N

यदि three-digit number N को दो बार लिखा जाए:

NN = 1001N

और:

1001 = 7×11×13

महत्वपूर्ण: abcabc form की हर six-digit number 7, 11 और 13 तीनों से divisible होती है।

43. Consecutive Integers

Consecutive integers को represent किया जाता है:

n, n+1, n+2, ...

  • किसी भी दो consecutive positive integers का HCF = 1 होता है।
  • दो consecutive integers में एक अवश्य even होता है।
  • तीन consecutive integers में एक 3 से divisible और कम-से-कम एक even होता है।

44. Consecutive Integers का Product

किसी भी k consecutive integers का product हमेशा:

k!

से divisible होता है।

Special cases:

  • 2 consecutive integers → divisible by 2
  • 3 consecutive integers → divisible by 6
  • 4 consecutive integers → divisible by 24
  • 5 consecutive integers → divisible by 120

45. Standard Number Sums

SeriesFormula
1+2+3+...+nn(n+1)/2
1+3+5+...+(2n−1)n2
2+4+6+...+2nn(n+1)
12+22+...+n2n(n+1)(2n+1)/6
13+23+...+n3[n(n+1)/2]2

46. Consecutive Squares

Consecutive squares का difference:

(n+1)2−n2

= 2n+1

इसलिए consecutive perfect squares का difference हमेशा odd होता है।

47. Integers के महत्वपूर्ण Forms

हर integer exactly इनमें से किसी एक form में होता है:

2k या 2k+1

हर integer modulo 3 में exactly इनमें से किसी एक form में होता है:

3k, 3k+1, 3k+2

इसी प्रकार modulo m में हर integer exactly एक residue class में होगा:

mk, mk+1, ..., mk+(m−1)

48. High-Value Number System Facts

  • 0 even है।
  • 0 न positive है और न negative।
  • 1 न prime है और न composite।
  • 2 smallest और only even prime है।
  • 4 smallest composite number है।
  • किसी भी दो consecutive positive integers का HCF = 1 होता है।
  • हर odd square ≡ 1 (mod 8)।
  • Positive integer के factors की संख्या odd होगी iff वह perfect square हो।
  • Two-digit number और उसके reverse का difference 9 से divisible होता है।
  • Two-digit number और उसके reverse का sum 11 से divisible होता है।

49. Chapter 1 — One-Minute Formula Sheet

  • Smallest n-digit number = 10n−1
  • Greatest n-digit number = 10n−1
  • Number of n-digit positive integers = 9×10n−1
  • N = dq+r, जहाँ 0≤r
  • Divisor d के लिए maximum remainder = d−1
  • यदि N=paqb..., तो factors की संख्या = (a+1)(b+1)...
  • Product of all factors = Nd(N)/2
  • Digits in N = ⌊log10N⌋+1
  • Digits in an = ⌊nlog10a⌋+1
  • vp(n!) = ⌊n/p⌋+⌊n/p2⌋+...
  • Z(n!) = ⌊n/5⌋+⌊n/25⌋+...
  • Two-step successive remainder = d1r2+r1
  • Two-digit number = 10a+b
  • Reverse difference = 9(a−b)
  • Reverse sum = 11(a+b)
  • abcabc = 1001×abc
  • 1001 = 7×11×13
  • [A,B] में d के multiples = ⌊B/d⌋−⌊(A−1)/d⌋
  • Product of k consecutive integers is divisible by k!
  • 1+2+...+n = n(n+1)/2
  • First n odd numbers का sum = n2
  • First n even numbers का sum = n(n+1)
  • First n squares का sum = n(n+1)(2n+1)/6
  • First n cubes का sum = [n(n+1)/2]2

50. Chapter 1 — सबसे महत्वपूर्ण Exam Traps

Trap 1: 1 न prime है और न composite।
Trap 2: 0 even है, लेकिन न positive है न negative।
Trap 3: Co-prime numbers का individually prime होना आवश्यक नहीं।
Trap 4: Remainder हमेशा divisor से छोटा होना चाहिए।
Trap 5: Unit-digit cycle में exponent mod cycle length = 0 हो तो cycle का अंतिम term लें।
Trap 6: n! के trailing zeros केवल n/10 से नहीं निकाले जाते; powers of 5 count करनी होती हैं।
Trap 7: 10n में n नहीं, बल्कि n+1 digits होते हैं।
Trap 8: Number-of-digits questions में logarithm को बहुत जल्दी round न करें।
Trap 9: Successive division में हर अगला divisor previous quotient पर कार्य करता है।
Trap 10: Successive remainders को सीधे add नहीं किया जाता।
Trap 11: Leading zero का उपयोग करके smaller multi-digit number नहीं बनाया जा सकता।
Trap 12: Distinct-digit questions में repetition allowed नहीं होता जब तक explicitly न कहा गया हो।
Trap 13: k consecutive integers का product k! से divisible होता है, लेकिन आवश्यक नहीं कि वह k! के बराबर हो।
Trap 14: Same remainder मिलने का अर्थ differences का divisor से divisible होना है; original numbers का divisible होना आवश्यक नहीं।
Trap 15: Factorial quotients में trailing-zero counts को हमेशा सीधे subtract नहीं किया जा सकता।

अब Chapter 1 Practice के लिए तैयार

Number System के पूरे Chapter 1 की त्वरित पुनरावृत्ति अब complete है। इसके बाद verified previous-year questions के माध्यम से concepts को test करें और फिर mixed chapter-level Practice MCQs हल करें।

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