Number Systems, Divisibility Rules, LCM, HCF & Fractions

Key Takeaways

  • The real number system classifies numerical values into Natural numbers (ℕ), Whole numbers (𝕎), Integers (ℤ), Rational numbers (ℚ), and Irrational numbers, with exactly 25 prime numbers existing below 100 and 2 standing as the sole even prime.

  • Divisibility rules allow rapid modular verification without long division, prominently featuring the alternating sum test for 11 where the absolute difference between sums of digits at odd and even positions must equal 0 or a multiple of 11.

  • The BODMAS / PEMDAS operational hierarchy strictly governs arithmetic expressions, executing operations in the immutable sequence of Brackets (vinculum/bar, parentheses, curly braces, square brackets), Orders/Powers, Division, Multiplication, Addition, and Subtraction.

  • For any two positive integers a and b, the fundamental product theorem establishes that HCF(a, b) × LCM(a, b) = a × b, while fraction operations require computing HCF as HCF(numerators)/LCM(denominators) and LCM as LCM(numerators)/HCF(denominators).

  • Recurring decimals convert into vulgar fractions through standardized algebraic expansions, where pure repeating decimals follow 0.p̄ = p/9 and mixed repeating decimals follow 0.ab̄ = (ab - a)/90.

Last updated: October 2026

Quantitative aptitude in the Odisha Sub-Ordinate Staff Selection Commission (OSSSC) Staff Nurse / Nursing Officer examination assesses high-school certificate (HSC) standard arithmetic. Clinicians routinely execute rapid mental arithmetic when calculating pediatric dosages, titrating intravenous drop rates, verifying stock inventory, and interpreting clinical laboratory indices. A solid mastery of number classifications, modular divisibility, operational precedence (BODMAS), common factors, multiples, and fractional representations forms the bedrock of numerical accuracy in healthcare practice.


1. Classification of Number Systems

All numbers utilized in standard arithmetic belong to the system of Real Numbers (ℝ), which encompasses all values that can be represented along a continuous geometric number line.

                                  Real Numbers (ℝ)
                                  /              \
                     Rational Numbers (ℚ)     Irrational Numbers
                     /                  \
             Integers (ℤ)           Fractions & Decimals
             /          \
   Negative Integers   Whole Numbers (𝕎)
                       /               \
                     Zero (0)     Natural Numbers (ℕ)
                                  /        |        \
                             Primes    Composites   Unity (1)

Core Sets of Numbers

  1. Natural Numbers (ℕ): The positive counting integers starting from unity: N={1,2,3,4,5,… }\mathbb{N} = \{1, 2, 3, 4, 5, \dots\}.
  2. Whole Numbers (𝕎): The natural numbers augmented by zero: W={0,1,2,3,4,… }\mathbb{W} = \{0, 1, 2, 3, 4, \dots\}. Zero is the only whole number that is not a natural number.
  3. Integers (ℤ): The complete set of positive integers, negative integers, and zero: Z={…,−3,−2,−1,0,1,2,3,… }\mathbb{Z} = \{\dots, -3, -2, -1, 0, 1, 2, 3, \dots\}.
    • Positive Integers: {1,2,3,… }\{1, 2, 3, \dots\}
    • Negative Integers: {−1,−2,−3,… }\{-1, -2, -3, \dots\}
    • Zero (0): Neutral integer; neither positive nor negative, but strictly even.
  4. Rational Numbers (ℚ): Any number that can be expressed in the quotient form pq\frac{p}{q}, where pp and qq are integers and q≠0q \neq 0. When converted to decimal form, rational numbers either terminate (e.g., 34=0.75\frac{3}{4} = 0.75) or repeat indefinitely in a recurring pattern (e.g., 13=0.333⋯=0.3ˉ\frac{1}{3} = 0.333\dots = 0.\bar{3}).
  5. Irrational Numbers: Numbers that cannot be expressed as a ratio of two integers. Their decimal expansions are non-terminating and non-repeating (e.g., 2≈1.4142…\sqrt{2} \approx 1.4142\dots, 3≈1.7320…\sqrt{3} \approx 1.7320\dots, π≈3.14159…\pi \approx 3.14159\dots, and Euler's constant e≈2.71828…e \approx 2.71828\dots).
  6. Real Numbers (ℝ): The unified set of all rational and irrational numbers: R=Q∪Irrationals\mathbb{R} = \mathbb{Q} \cup \text{Irrationals}.

Prime, Composite, and Related Classifications

  • Prime Numbers: Any natural number strictly greater than 1 that possesses exactly two distinct positive divisors: 1 and the number itself.
    • The number 1 is neither prime nor composite (it has only one divisor).
    • The number 2 is the smallest prime number and the only even prime number in existence. All other prime numbers are odd.
    • Prime Numbers Under 100: Exactly 25 prime numbers exist between 1 and 100: {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\}
    • Distribution: 1 to 50 contains 15 primes; 51 to 100 contains 10 primes.
  • Composite Numbers: Natural numbers greater than 1 that possess more than two distinct positive divisors (e.g., 4, 6, 8, 9, 10, 12, ...). The number 4 is the smallest composite number.
  • Twin Primes: Pairs of prime numbers that differ by exactly 2. There are 8 pairs of twin primes below 100: (3,5),(5,7),(11,13),(17,19),(29,31),(41,43),(59,61), and (71,73)(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), \text{ and } (71, 73).
  • Co-Prime (Relatively Prime) Numbers: Two positive integers aa and bb whose Highest Common Factor is 1, denoted HCF(a,b)=1\text{HCF}(a, b) = 1. The individual numbers do not need to be prime themselves. For instance, 8 (composite) and 15 (composite) are co-prime because their only common factor is 1.
  • Primality Testing Algorithm: To determine whether an integer NN is prime, test for divisibility by all prime numbers pp such that p≤Np \le \sqrt{N}. If no prime up to N\sqrt{N} divides NN, then NN is guaranteed to be prime.
    • Worked Example: Test if 173 is prime. Calculate 173≈13.15\sqrt{173} \approx 13.15. Primes to test are 2, 3, 5, 7, 11, 13. 173 ends in 3 (not div by 2 or 5); sum of digits 1+7+3=111+7+3=11 (not div by 3); 173÷7=24.71173 \div 7 = 24.71; 173÷11=15.72173 \div 11 = 15.72; 173÷13=13.30173 \div 13 = 13.30. Because none divide 173 evenly, 173 is prime.
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Hierarchical Structure of the Real Number System

2. Divisibility Rules & Systematic Verification

Divisibility tests enable rapid identification of factors and prime reductions without executing manual long division. These rules are indispensable for factoring drug dosages, scheduling shift rotations, and simplifying arithmetic fractions.

Divisibility Criteria Summary Table

DivisorMathematical Test / CriterionIllustrative Example
2The units digit is an even integer (0, 2, 4, 6, or 8).4,856 ends in 6 (even) →\rightarrow Divisible by 2.
3The sum of all constituent digits is divisible by 3.7,419: 7+4+1+9=217 + 4 + 1 + 9 = 21. Since 21÷3=721 \div 3 = 7 →\rightarrow Divisible by 3.
4The number formed by the last two digits (tens and units) is divisible by 4, or ends in double zero (00).35,728: Last two digits are 28. Since 28÷4=728 \div 4 = 7 →\rightarrow Divisible by 4.
5The units digit is either 0 or 5.12,485 ends in 5 →\rightarrow Divisible by 5.
6The number satisfies both divisibility by 2 (even units digit) AND divisibility by 3 (sum of digits divisible by 3).9,216: Ends in 6 (divisible by 2); sum 9+2+1+6=189+2+1+6 = 18 (divisible by 3) →\rightarrow Divisible by 6.
8The number formed by the last three digits is divisible by 8, or ends in triple zero (000).54,168: Last three digits are 168. Since 168÷8=21168 \div 8 = 21 →\rightarrow Divisible by 8.
9The sum of all constituent digits is divisible by 9.84,528: 8+4+5+2+8=278 + 4 + 5 + 2 + 8 = 27. Since 27÷9=327 \div 9 = 3 →\rightarrow Divisible by 9.
10The units digit is strictly 0.47,930 ends in 0 →\rightarrow Divisible by 10.
11The absolute difference between the sum of digits at odd places and the sum of digits at even places is either 0 or a multiple of 11.94,853: Odd places sum (3+8+9)=20(3 + 8 + 9) = 20; Even places sum (5+4)=9(5 + 4) = 9. Difference: 20−9=1120 - 9 = 11 →\rightarrow Divisible by 11.
12The number satisfies both divisibility by 3 AND divisibility by 4 (co-prime factors of 12).1,728: Sum 1+7+2+8=181+7+2+8=18 (divisible by 3); last two digits 28 (divisible by 4) →\rightarrow Divisible by 12.

The Alternating Sum Rule for 11 (Detailed Walkthrough)

The divisibility rule for 11 stems from the modular expansion of powers of 10: 10≡−1(mod11)10 \equiv -1 \pmod{11}, 102≡1(mod11)10^2 \equiv 1 \pmod{11}, 103≡−1(mod11)10^3 \equiv -1 \pmod{11}, and so forth. Consequently, any integer N=an10n+⋯+a110+a0N = a_n 10^n + \dots + a_1 10 + a_0 satisfies: N≡a0−a1+a2−a3+…(mod11)N \equiv a_0 - a_1 + a_2 - a_3 + \dots \pmod{11}

  • Step 1: Number digit positions from right to left (Position 1 = units digit, Position 2 = tens digit, Position 3 = hundreds digit, etc.).
  • Step 2: Calculate Sodd=sum of digits at positions 1, 3, 5, 7, …S_{\text{odd}} = \text{sum of digits at positions 1, 3, 5, 7, } \dots
  • Step 3: Calculate Seven=sum of digits at positions 2, 4, 6, 8, …S_{\text{even}} = \text{sum of digits at positions 2, 4, 6, 8, } \dots
  • Step 4: Compute the absolute difference ∣Sodd−Seven∣|S_{\text{odd}} - S_{\text{even}}|. If the result is 0, 11, 22, 33, or any multiple of 11, the entire original number is divisible by 11.

Worked Clinical Application: A batch of 85,976 surgical gloves is delivered to the hospital stores. Can this shipment be divided equally into 11 regional community health center parcels without opening individual packets?

  • Digits at odd places (from right): 6+9+8=236 + 9 + 8 = 23
  • Digits at even places (from right): 7+5=127 + 5 = 12
  • Difference: 23−12=1123 - 12 = 11
  • Conclusion: Because 11 is divisible by 11, 85,976 is exactly divisible by 11 (85,976÷11=7,81685{,}976 \div 11 = 7{,}816 gloves per center).

3. Order of Operations: BODMAS / PEMDAS Rule

When evaluating complex arithmetic expressions with multiple nested grouping symbols and operational operators, computations must proceed according to the standardized BODMAS operational hierarchy.

The Hierarchy of Precedence

  1. B — Brackets (Grouping Symbols): Evaluated strictly from the inside outward:
    • Vinculum / Bar (—): Cleared first: e.g., 8−5−2‾=8−3=58 - \overline{5 - 2} = 8 - 3 = 5.
    • Round Brackets / Parentheses ( ): Evaluated second.
    • Curly Brackets / Braces { }: Evaluated third.
    • Square Brackets / Box [ ]: Evaluated fourth.
  2. O — Orders / Of: Denotes powers, exponents, roots, and the mathematical preposition "of". Note that "of" signifies multiplication but takes operational precedence over division and multiplication (e.g., 13 of 15=5\frac{1}{3} \text{ of } 15 = 5).
  3. D — Division (÷) & M — Multiplication (×): Possess equal operational rank, evaluated strictly from left to right as they appear in the expression.
  4. A — Addition (+) & S — Subtraction (−): Possess equal operational rank, evaluated strictly from left to right.

Fundamental Algebraic Sign Rules

  • (+)×(+)=+(+) \times (+) = + and (−)×(−)=+(-) \times (-) = + (Product of like signs is positive).
  • (+)×(−)=−(+) \times (-) = - and (−)×(+)=−(-) \times (+) = - (Product of unlike signs is negative).
  • Division follows identical sign properties: (−)÷(−)=+(-) \div (-) = + and (−)÷(+)=−(-) \div (+) = -.
  • Subtracting a negative quantity is equivalent to addition: a−(−b)=a+ba - (-b) = a + b.

Step-by-Step Worked Simplification Example

Evaluate the following expression: E=48÷4 of 3+[18−{6+(14−8−3‾)}]E = 48 \div 4 \text{ of } 3 + [18 - \{6 + (14 - \overline{8 - 3})\}]

  • Step 1 (Vinculum): Evaluate the bar bracket: 8−3‾=5\overline{8 - 3} = 5. E=48÷4 of 3+[18−{6+(14−5)}]E = 48 \div 4 \text{ of } 3 + [18 - \{6 + (14 - 5)\}]
  • Step 2 (Parentheses): Evaluate round brackets: (14−5)=9(14 - 5) = 9. E=48÷4 of 3+[18−{6+9}]E = 48 \div 4 \text{ of } 3 + [18 - \{6 + 9\}]
  • Step 3 (Braces): Evaluate curly braces: {6+9}=15\{6 + 9\} = 15. E=48÷4 of 3+[18−15]E = 48 \div 4 \text{ of } 3 + [18 - 15]
  • Step 4 (Square Brackets): Evaluate square brackets: [18−15]=3[18 - 15] = 3. E=48÷4 of 3+3E = 48 \div 4 \text{ of } 3 + 3
  • Step 5 ("Of" Precedence): Evaluate "of" before division: 4 of 3=4×3=124 \text{ of } 3 = 4 \times 3 = 12. E=48÷12+3E = 48 \div 12 + 3
  • Step 6 (Division): Perform division: 48÷12=448 \div 12 = 4. E=4+3E = 4 + 3
  • Step 7 (Addition): Compute final sum: 4+3=74 + 3 = 7.
  • Final Result: 7.

4. Highest Common Factor (HCF) & Least Common Multiple (LCM)

Finding common measures and synchronized intervals forms a routine component of resource scheduling and packaging in healthcare administration.

Mathematical Definitions

  • Highest Common Factor (HCF / GCD): The greatest positive integer that divides each of two or more given integers without leaving a remainder. HCF represents the largest possible common grouping size.
  • Least Common Multiple (LCM): The smallest positive integer that is divisible by each of two or more given integers. LCM represents the earliest point of synchronization or alignment.

Analytical Methods for Computation

1. Prime Factorization Method

Express each integer as a product of prime powers:

  • HCF\text{HCF} is obtained by taking the lowest power of each common prime factor.
  • LCM\text{LCM} is obtained by taking the highest power of every prime factor present across all numbers.

Worked Example: Find HCF and LCM of 72, 108, and 180.

  • 72=23×3272 = 2^3 \times 3^2
  • 108=22×33108 = 2^2 \times 3^3
  • 180=22×32×51180 = 2^2 \times 3^2 \times 5^1
  • HCF=2min⁡(3,2,2)×3min⁡(2,3,2)×5min⁡(0,0,1)=22×32×1=4×9=36\text{HCF} = 2^{\min(3,2,2)} \times 3^{\min(2,3,2)} \times 5^{\min(0,0,1)} = 2^2 \times 3^2 \times 1 = 4 \times 9 = \mathbf{36}.
  • LCM=2max⁡(3,2,2)×3max⁡(2,3,2)×5max⁡(0,0,1)=23×33×51=8×27×5=1,080\text{LCM} = 2^{\max(3,2,2)} \times 3^{\max(2,3,2)} \times 5^{\max(0,0,1)} = 2^3 \times 3^3 \times 5^1 = 8 \times 27 \times 5 = \mathbf{1{,}080}.

2. The Fundamental Product Theorem for Two Numbers

For any two positive integers aa and bb, the product of their HCF and LCM is identically equal to the product of the two numbers: HCF(a,b)×LCM(a,b)=a×b\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b   ⟹  a=HCF×LCMbandLCM=a×bHCF\implies a = \frac{\text{HCF} \times \text{LCM}}{b} \quad \text{and} \quad \text{LCM} = \frac{a \times b}{\text{HCF}} (Note: This direct product identity holds strictly for two numbers; it does not generalize to sets of three or more numbers).

HCF and LCM of Fractions

When dealing with rational quantities, formulas operate across numerators and denominators: HCF of Fractions=HCF of NumeratorsLCM of Denominators\text{HCF of Fractions} = \frac{\text{HCF of Numerators}}{\text{LCM of Denominators}} LCM of Fractions=LCM of NumeratorsHCF of Denominators\text{LCM of Fractions} = \frac{\text{LCM of Numerators}}{\text{HCF of Denominators}} (Prerequisite: All fractions must first be reduced to their simplest irreducible form before applying these formulas).

Worked Example: Find HCF and LCM of 23,89, and 1681\frac{2}{3}, \frac{8}{9}, \text{ and } \frac{16}{81}.

  • All fractions are already irreducible.
  • Numerators: {2,8,16}\{2, 8, 16\}. Denominators: {3,9,81}\{3, 9, 81\}.
  • HCF of numerators(2,8,16)=2\text{HCF of numerators}(2, 8, 16) = 2.
  • LCM of denominators(3,9,81)=81\text{LCM of denominators}(3, 9, 81) = 81.
  • LCM of numerators(2,8,16)=16\text{LCM of numerators}(2, 8, 16) = 16.
  • HCF of denominators(3,9,81)=3\text{HCF of denominators}(3, 9, 81) = 3.
  • Therefore: HCF=281andLCM=163\text{HCF} = \frac{2}{81} \quad \text{and} \quad \text{LCM} = \frac{16}{3}

Standard Application Word Problems

  1. Synchronized Periodic Events (Bells / Alarms / Medication Rounds):
    • Problem: In an intensive care unit, three infusion pump alarms chime at intervals of 12 seconds, 15 seconds, and 20 seconds respectively. If they all chime simultaneously at 08:00 AM, at what time will they next chime together?
    • Method: The synchronized interval is given by LCM(12,15,20)\text{LCM}(12, 15, 20).
      • 12=22×312 = 2^2 \times 3
      • 15=3×515 = 3 \times 5
      • 20=22×520 = 2^2 \times 5
      • LCM=22×3×5=60 seconds=1 minute\text{LCM} = 2^2 \times 3 \times 5 = 60\text{ seconds} = 1\text{ minute}.
    • Solution: The alarms will chime together next at 08:01 AM.
  2. Partitioning and Equal Distribution (Maximum Measuring Capacity):
    • Problem: A nursing supervisor has 120 ampoules of Paracetamol, 144 ampoules of Tramadol, and 192 ampoules of Ondansetron. She wishes to package them into emergency crash carts such that each cart receives identical contents with no ampoules left over. What is the maximum number of crash carts that can be equipped?
    • Method: The maximum uniform quantity is the HCF(120,144,192)\text{HCF}(120, 144, 192).
      • 120=23×3×5120 = 2^3 \times 3 \times 5
      • 144=24×32144 = 2^4 \times 3^2
      • 192=26×3192 = 2^6 \times 3
      • HCF=23×3=8×3=24\text{HCF} = 2^3 \times 3 = 8 \times 3 = 24.
    • Solution: At most 24 crash carts can be equipped, each holding 5 Paracetamol, 6 Tramadol and 8 Ondansetron ampoules (120÷24=5120 \div 24 = 5, 144÷24=6144 \div 24 = 6, 192÷24=8192 \div 24 = 8).
  3. Modular Remainder Problems:
    • Finding the greatest number dividing x,y,zx, y, z leaving remainders a,b,ca, b, c: Required Divisor=HCF(x−a,y−b,z−c)\text{Required Divisor} = \text{HCF}(x - a, y - b, z - c)
    • Finding the least number divided by x,y,zx, y, z leaving constant remainder rr: Required Number=LCM(x,y,z)+r\text{Required Number} = \text{LCM}(x, y, z) + r

5. Fractions, Decimals & Recurring Decimal Expansions

Fractions quantify rational portions of a whole, operating throughout pharmacokinetics, laboratory dilutions, and demographic calculations.

Classification of Fractions

  • Proper Fraction: Numerator is strictly less than denominator (pq\frac{p}{q} with p<qp < q), such as 37\frac{3}{7}. The value is always strictly less than 1.
  • Improper Fraction: Numerator is greater than or equal to denominator (p≥qp \ge q), such as 114\frac{11}{4}. The value is ≥1\ge 1.
  • Mixed Fraction: Combination of an integer whole number and a proper fraction: 234=2+34=2×4+34=1142 \frac{3}{4} = 2 + \frac{3}{4} = \frac{2 \times 4 + 3}{4} = \frac{11}{4}.
  • Equivalent Fractions: Fractions that represent the identical value when reduced to lowest terms: 23=46=69=2030\frac{2}{3} = \frac{4}{6} = \frac{6}{9} = \frac{20}{30}.

Comparing Fractions: Rapid Techniques

  1. Cross-Multiplication Method (Best for Two Fractions):
    • To compare ab\frac{a}{b} and cd\frac{c}{d}:
      • Compute product a×da \times d and product b×cb \times c.
      • If a×d>b×ca \times d > b \times c, then ab>cd\frac{a}{b} > \frac{c}{d}.
      • If a×d<b×ca \times d < b \times c, then ab<cd\frac{a}{b} < \frac{c}{d}.
    • Example: Compare 58\frac{5}{8} and 711\frac{7}{11}. Compute 5×11=555 \times 11 = 55 and 8×7=568 \times 7 = 56. Since 55<5655 < 56, it follows that 58<711\frac{5}{8} < \frac{7}{11}.
  2. Equating Denominators via LCM (Best for Three or More Fractions):
    • Find LCM of all denominators, scale numerators proportionately, and compare resulting numerators.

Decimals & Recurring Decimal Conversion

Decimals are either terminating (denominator in irreducible form contains only prime factors of 2 and/or 5) or non-terminating recurring.

1. Pure Recurring Decimals

A decimal in which all digits immediately following the decimal point repeat indefinitely.

  • Conversion Rule: Place the repeating digits in the numerator, and place as many 9s in the denominator as there are repeating digits: 0.aˉ=a9,0.ab‾=ab99,0.abc‾=abc9990.\bar{a} = \frac{a}{9}, \quad 0.\overline{ab} = \frac{ab}{99}, \quad 0.\overline{abc} = \frac{abc}{999}
  • Examples:
    • 0.6ˉ=69=230.\bar{6} = \frac{6}{9} = \frac{2}{3}
    • 0.36‾=3699=4110.\overline{36} = \frac{36}{99} = \frac{4}{11}
    • 0.142857‾=142857999999=170.\overline{142857} = \frac{142857}{999999} = \frac{1}{7}

2. Mixed Recurring Decimals

A decimal in which some digits after the decimal point do not repeat, followed by digits that repeat indefinitely.

  • Conversion Rule: In the numerator, subtract the non-repeating digits from the entire sequence of digits up to the end of the first repeating block. In the denominator, write as many 9s as there are repeating digits, followed by as many 0s as there are non-repeating digits: 0.abˉ=ab−a900.a\bar{b} = \frac{ab - a}{90} 0.abcˉ=abc−ab9000.ab\bar{c} = \frac{abc - ab}{900} 0.abc‾=abc−a9900.a\overline{bc} = \frac{abc - a}{990}
  • Worked Examples:
    • Convert 0.47ˉ0.4\bar{7} to a vulgar fraction: 0.47ˉ=47−490=43900.4\bar{7} = \frac{47 - 4}{90} = \frac{43}{90}
    • Convert 0.16ˉ0.1\bar{6} to a vulgar fraction: 0.16ˉ=16−190=1590=160.1\bar{6} = \frac{16 - 1}{90} = \frac{15}{90} = \frac{1}{6}
    • Convert 0.123ˉ0.12\bar{3} to a vulgar fraction: 0.123ˉ=123−12900=111900=373000.12\bar{3} = \frac{123 - 12}{900} = \frac{111}{900} = \frac{37}{300}
Test Your Knowledge

Which of the following numbers is divisible by 11 according to the alternating sum divisibility test?

A

62,941

B

75,418

C

83,742

D

94,853

Test Your Knowledge

The Highest Common Factor (HCF) and Least Common Multiple (LCM) of two numbers are 12 and 240 respectively. If one of the numbers is 48, what is the value of the other number?

A

45

B

60

C

72

D

80

Test Your Knowledge

What is the vulgar fraction equivalent of the mixed recurring decimal 0.47̄ (where only the digit 7 repeats indefinitely) in its simplest form?

A

43/99

B

47/99

C

47/90

D

43/90

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