Number System, BODMAS, HCF and LCM

Common to all departments Mathematics (class 10 level) Chapter 1/13 15 min read Source: Semi-Official
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This chapter is the base for every other arithmetic topic: kinds of numbers, divisibility tests, unit digits of large powers, counting factors, the order of operations (BODMAS), fractions and recurring decimals, and HCF and LCM with their word problems.

Kinds of Numbers

TypeMeaningExamples
Natural numbers (N)Counting numbers starting from 11, 2, 3, …
Whole numbers (W)Natural numbers and zero0, 1, 2, …
Integers (Z)Positive and negative whole numbers and zero…, −2, −1, 0, 1, 2, …
Rational numbers (Q)Can be written as p/q with q ≠ 0; decimals terminate or repeat½, −3, 0.75
Irrational numbersCannot be written as p/q; decimals neither end nor repeat√2, √3, π
Prime numbersExactly two factors: 1 and the number itself2, 3, 5, 7, 11
Composite numbersMore than two factors4, 6, 8, 9
Co-prime numbersTwo numbers whose HCF is 1(8, 9), (14, 15)
Facts to Remember
  • 1 is neither prime nor composite; 2 is the only even prime.
  • There are 25 primes from 1 to 100, and 4 primes from 1 to 10 (2, 3, 5, 7).
  • Odd × odd = odd; even × any integer = even; odd + odd = even.
  • Sum of the first n natural numbers = n(n + 1)/2; sum of their cubes = [n(n + 1)/2]².

Divisibility Rules

DivisorTestExample
2Last digit even3748
3Sum of digits divisible by 35316 → 15
4Last two digits divisible by 47312 → 12
5Last digit 0 or 54835
6Divisible by both 2 and 34632
8Last three digits divisible by 854136 → 136
9Sum of digits divisible by 98127 → 18
11Difference between sums of alternate digits is 0 or a multiple of 117469 → 13 − 13 = 0
12Divisible by both 3 and 41452

Unit Digit of a Power

Last digit of baseCycle of unit digitsCycle length
0, 1, 5, 6Always the same digit1
22, 4, 8, 64
33, 9, 7, 14
77, 9, 3, 14
88, 4, 2, 64
44 (odd power), 6 (even power)2
99 (odd power), 1 (even power)2
Method
Divide the power by the cycle length; the remainder picks the digit (remainder 0 → last digit of the cycle)
Example: 7⁹⁵ → 95 ÷ 4 leaves 3 → third digit of (7, 9, 3, 1) = 3.

Factors and Remainders

Number and sum of factors
If N = pᵃ × qᵇ × rᶜ, number of factors = (a + 1)(b + 1)(c + 1)
Sum of factors = (1 + p + … + pᵃ)(1 + q + … + qᵇ)(1 + r + … + rᶜ). Example: 360 = 2³ × 3² × 5 has 4 × 3 × 2 = 24 factors, and their sum is 15 × 13 × 6 = 1170.

BODMAS: Order of Operations

Order
Brackets → Of / Orders (powers, roots) → Division and Multiplication → Addition and Subtraction
Division and multiplication have equal priority and are done from left to right; so are addition and subtraction. Brackets are removed in the order: bar (vinculum), ( ), { }, [ ].
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Common Trap
20 ÷ 5 × 2 = 4 × 2 = 8, not 20 ÷ 10 = 2. Work left to right when ÷ and × appear together.

Fractions and Recurring Decimals

HCF and LCM

HCF (highest common factor)LCM (least common multiple)
MeaningLargest number dividing all the given numbersSmallest number divisible by all the given numbers
By prime factorsCommon primes with the lowest powersAll primes with the highest powers
Of fractionsHCF of numerators ÷ LCM of denominatorsLCM of numerators ÷ HCF of denominators
Typical useLargest tile, biggest container, equal rowsBells or lights together again, smallest number divisible by several
Key relation (two numbers only)
HCF × LCM = product of the two numbers
Example: HCF 23, LCM 1449, one number 161 → other = 23 × 1449 ÷ 161 = 207. This does not hold for three numbers.
Problem typeAnswer
Largest number dividing a, b, c leaving the same remainderHCF of the differences (b − a), (c − b), (c − a)
Largest number dividing a, b, c leaving remainders p, q, rHCF of (a − p), (b − q), (c − r)
Smallest number leaving the same remainder r when divided by several numbersLCM + r
Smallest number where (divisor − remainder) is the same constant k for every divisorLCM − k

Worked Examples

  1. Simplify 18 ÷ 3 × (4 + 2) − 5². Bracket: 18 ÷ 3 × 6 − 5²; power: 18 ÷ 3 × 6 − 25; left to right: 6 × 6 − 25 = 36 − 25 = 11.
  2. Three lights turn green every 36, 48 and 72 seconds and are together at 9:00 AM. LCM = 2⁴ × 3² = 144 s = 2 min 24 s, so next together at 9:02:24 AM.
  3. A courtyard 48 m × 36 m is paved with the largest square tiles. Side = HCF(48, 36) = 12 m; tiles = 4 × 3 = 12.
  4. Largest number dividing 445, 572 and 699 leaving remainders 4, 5 and 6: HCF(441, 567, 693) = 3² × 7 = 63.
  5. Smallest 4-digit number divisible by 12, 15, 18 and 27: LCM = 2² × 3³ × 5 = 540; 1000 is not a multiple, the next multiple is 1080.
  6. Least number which leaves 3 when divided by 5, 6, 7 and 8 but is divisible by 9: LCM = 840, numbers are 840k + 3; k = 2 gives 1683 = 9 × 187.
  7. Unit digit of 2¹⁰⁰: 100 ÷ 4 leaves 0, so take the last digit of the cycle (2, 4, 8, 6) = 6.
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Common Mistakes
Treating 1 as prime; doing × before ÷ instead of left to right; using HCF × LCM = product for three numbers; swapping the fraction rules for HCF and LCM; forgetting that remainder 0 in a unit-digit question means the last digit of the cycle.
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Exam Focus
Expect one or two questions on BODMAS simplification, one on HCF/LCM word problems (bells, tiles, remainders) and one on divisibility or unit digit. Memorise squares up to 30 and cubes up to 15 to save time.

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