Number System, HCF–LCM & Simplification

This is the foundation floor of the whole maths section. Percentages, ratios and data questions all sit on top of it — but the number system also gives marks directly: a divisibility rule, a unit digit, an HCF word problem, a BODMAS simplification. None of it needs cleverness; it needs a small set of rules applied fast and without panic. Every technique below is shown with the full arithmetic, so work each example out on paper once and the method sticks.

Types of Numbers, Place Value & Primality

Before anything else, get the family tree of numbers straight. Each family sits inside the next one: natural → whole → integers → rational → real. Irrational numbers join the rationals to complete the real numbers. Most "which of the following is true" questions are answered just by knowing which family contains which.

The families of numbers
TypeDefinitionExamples
Natural (N)Counting numbers, starting at 11, 2, 3, 4, …
Whole (W)Natural numbers plus 00, 1, 2, 3, …
Integers (Z)Whole numbers plus their negatives… −3, −2, −1, 0, 1, 2, 3 …
Rational (Q)Can be written p/q with q ≠ 0; decimal terminates or recurs3/4 = 0.75, −5, 0.333…
IrrationalCannot be written as p/q; decimal is non-terminating, non-recurring√2, √3, √5, π
Real (R)All rationals and all irrationals together−7, 0, 2/3, √2, π
PrimeExactly two factors — 1 and itself2, 3, 5, 7, 11, 13, 17, 19
CompositeMore than two factors4, 6, 8, 9, 10, 12
Co-primeA pair whose HCF is 1 (need not be prime themselves)(8, 15), (9, 16), (7, 12)
Perfect numberSum of all its factors other than itself equals the number6 = 1+2+3; 28 = 1+2+4+7+14; then 496, 8128
⚠️ Four traps set every single year
(1) 1 is neither prime nor composite — it has only one factor.
(2) 2 is the only even prime, and it is also the smallest prime.
(3) 0 is a whole number but not a natural number, and 0 is an integer, so "every integer is a natural number" is false.
(4) π is irrational; 22/7 is only a convenient approximation, not its value.

Place value vs face value. The face value of a digit is the digit itself, wherever it stands. The place value is the digit multiplied by the value of its position. Take 78,432: the digit 4 sits in the hundreds place, so its face value is 4 and its place value is 4 × 100 = 400. The difference asked for in exams is therefore 400 − 4 = 396. Note that the face value of 0 is 0 and its place value is also 0 — the one digit for which the two are always equal.

🧠 Is it prime? The √n test
To test whether n is prime you only need to divide by primes up to √n — no further. Example: is 197 prime? √197 is a little over 14, so test 2, 3, 5, 7, 11, 13 only. 197 is odd (not ÷2); digit sum 1+9+7 = 17, not a multiple of 3; it does not end in 0 or 5; 7 × 28 = 196, remainder 1; for 11 the alternating sum 1 − 9 + 7 = −1; 13 × 15 = 195, remainder 2. All fail, so 197 is prime. Compare 391: √391 is under 20, and testing up to 19 finds 391 = 17 × 23, so it is composite.
ℹ️ Note
Useful counts to have ready: there are 25 primes below 100 and 15 primes below 50. Twin primes are prime pairs differing by 2 — (3, 5), (5, 7), (11, 13), (17, 19). And any composite number has exactly one prime factorisation, e.g. 360 = 2³ × 3² × 5 — this single fact powers the whole HCF–LCM section.

Divisibility Rules — the Ten-Second Tests

A divisibility rule saves you an entire long division. In the exam they appear directly ("which of these is divisible by 11?") and indirectly — while cancelling a big fraction or factorising for HCF. Learn the table, then check every rule on the example given beside it.

Divisibility rules with a worked example each
ByRuleWorked example
2Last digit is even (0, 2, 4, 6, 8)1,234 ends in 4 → yes (1234 ÷ 2 = 617)
3Sum of digits divisible by 34,152 → 4+1+5+2 = 12 → yes (4152 ÷ 3 = 1384)
4Last two digits form a multiple of 43,116 → 16 ÷ 4 = 4 → yes (3116 ÷ 4 = 779)
5Last digit is 0 or 52,345 ends in 5 → yes (2345 ÷ 5 = 469)
6Divisible by 2 and 3 both1,332 is even and 1+3+3+2 = 9 → yes (1332 ÷ 6 = 222)
7Double the last digit, subtract from the rest; repeat1,673 → 167 − 6 = 161 → 16 − 2 = 14 → yes (1673 ÷ 7 = 239)
8Last three digits form a multiple of 87,120 → 120 ÷ 8 = 15 → yes (7120 ÷ 8 = 890)
9Sum of digits divisible by 95,832 → 5+8+3+2 = 18 → yes (5832 ÷ 9 = 648)
10Last digit is 04,570 ends in 0 → yes (4570 ÷ 10 = 457)
11(Sum of digits in odd places) − (sum in even places) = 0 or a multiple of 111,331 → (1+3) − (3+1) = 0 → yes (1331 ÷ 11 = 121)
13Multiply the last digit by 4 and add to the rest; repeat2,197 → 219 + 28 = 247 → 24 + 28 = 52 = 13 × 4 → yes (2197 ÷ 13 = 169)

Worked example (the 11-rule on a bigger number). Is 61,809 divisible by 11? Count places from the right: 9 is 1st, 0 is 2nd, 8 is 3rd, 1 is 4th, 6 is 5th. Odd places give 9 + 8 + 6 = 23; even places give 0 + 1 = 1. Difference = 23 − 1 = 22, which is 11 × 2. So yes — and indeed 61,809 ÷ 11 = 5,619.

  • For a composite divisor, split it into co-prime factors and test both: 12 = 3 and 4; 15 = 3 and 5; 18 = 2 and 9; 45 = 5 and 9; 72 = 8 and 9.
  • Check: is 1,332 divisible by 12? It is divisible by 4 (last two digits 32 ÷ 4 = 8) and by 3 (digit sum 9). So yes: 1332 ÷ 12 = 111.
  • If a number is divisible by 9, it is automatically divisible by 3 — but not the reverse. 4,152 passes the 3-test (sum 12) and fails the 9-test.
⚠️ Exam trap
Two errors cost easy marks. First: for 6 you must check both 2 and 3 — 4,152 has digit sum 12 (divisible by 3) and is even, so it passes; but 4,153 has the same digit sum and fails because it is odd. Second: to split a composite divisor the factors must be co-prime. For 12, use 3 × 4 (co-prime), never 2 × 6 — 18 is divisible by both 2 and 6 yet 18 is not divisible by 12.

HCF & LCM and Their Classic Word Problems

The HCF (highest common factor, म.स.) is the biggest number that divides all the given numbers. The LCM (lowest common multiple, ल.स.) is the smallest number that all of them divide into. Think of a tiled floor: the HCF is the largest tile that fits the room exactly with no cutting; the LCM is the smallest floor size on which several tile designs all end together.

Method 1 — prime factorisation (worked). Take 72 and 120. Factorise: 72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3², and 120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5. For the HCF take every common prime to its lowest power: 2³ × 3¹ = 8 × 3 = 24. For the LCM take every prime appearing anywhere, to its highest power: 2³ × 3² × 5 = 8 × 9 × 5 = 360. Verify with the relation HCF × LCM = product of the two numbers: 24 × 360 = 8,640 and 72 × 120 = 8,640. ✓

Method 2 — division method (worked). For the HCF of the same pair, divide the larger by the smaller and keep dividing by the remainder: 120 ÷ 72 → remainder 48; 72 ÷ 48 → remainder 24; 48 ÷ 24 → remainder 0. The last non-zero remainder, 24, is the HCF — matching Method 1. For the LCM of three or more numbers, use common division. Take 12, 15, 20: divide by 2 → 6, 15, 10; by 2 again → 3, 15, 5; by 3 → 1, 5, 5; by 5 → 1, 1, 1. Multiply all the divisors: 2 × 2 × 3 × 5 = 60.

⚠️ Exam trap
HCF × LCM = product of the numbers holds for exactly TWO numbers. It fails for three. Check 12, 15, 20: HCF = 1, LCM = 60, so HCF × LCM = 60 — but 12 × 15 × 20 = 3,600. A favourite exam question uses the valid two-number form in reverse: HCF = 24, LCM = 360, one number = 72, find the other. Product = 24 × 360 = 8,640, so the other number = 8,640 ÷ 72 = 120.

HCF and LCM of fractions. The rules cross over, which is exactly why students forget them: HCF of fractions = HCF of numerators ÷ LCM of denominators, while LCM of fractions = LCM of numerators ÷ HCF of denominators. Worked example with 2/3, 8/9, 16/81: HCF of numerators (2, 8, 16) = 2 and LCM of denominators (3, 9, 81) = 81, so HCF = 2/81. LCM of numerators = 16 and HCF of denominators = 3, so LCM = 16/3. Sanity check: the HCF must be the smaller of the two answers, and 2/81 is indeed tiny while 16/3 is large.

Which tool? Read the signal words
Signal in the questionUse
Largest / greatest size that measures or divides exactlyHCF of the quantities
Least number divisible by all; bells / lights / runners meeting againLCM of the intervals
Least number leaving the same remainder r with each divisorLCM + r
Least number whose remainders are each short of the divisor by the same kLCM − k
Greatest number dividing a, b, c leaving the same (unknown) remainderHCF of the differences (b−a), (c−b), (c−a)
Greatest number dividing a, b leaving given remainders r₁, r₂HCF of (a − r₁) and (b − r₂)
  1. Longest tape (HCF). Find the longest tape that can measure exactly 7 m, 3 m 85 cm and 12 m 95 cm. Convert to one unit: 700, 385, 1295 cm. Factorise: 700 = 2² × 5² × 7; 385 = 5 × 7 × 11; 1295 = 5 × 7 × 37. Common part = 5 × 7 = 35 cm. Check: 700 ÷ 35 = 20, 385 ÷ 35 = 11, 1295 ÷ 35 = 37 — all whole. Answer: 35 cm.
  2. Bells tolling together (LCM). Four bells toll at intervals of 6, 8, 12 and 18 seconds and toll together at 9:00:00 am. When do they next toll together? LCM: 6 = 2 × 3, 8 = 2³, 12 = 2² × 3, 18 = 2 × 3², so LCM = 2³ × 3² = 8 × 9 = 72 seconds. Next together at 9:01:12 am. In one full hour after the start they toll together 3600 ÷ 72 = 50 more times.
  3. Least number leaving the same remainder (LCM + r). Find the least number which, divided by 12, 15, 20 and 54, leaves remainder 4 each time. LCM: 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5, 54 = 2 × 3³ → LCM = 2² × 3³ × 5 = 4 × 27 × 5 = 540. Add the remainder: 540 + 4 = 544. Check: 544 = 12 × 45 + 4 = 15 × 36 + 4 = 20 × 27 + 4 = 54 × 10 + 4. ✓
  4. Greatest number leaving the same remainder (HCF of differences). Find the greatest number that divides 43, 91 and 183 leaving the same remainder. Differences: 91 − 43 = 48, 183 − 91 = 92, 183 − 43 = 140. Their HCF: 48 = 2⁴ × 3, 92 = 2² × 23, 140 = 2² × 5 × 7 → HCF = 2² = 4. Check: 43 = 4 × 10 + 3, 91 = 4 × 22 + 3, 183 = 4 × 45 + 3 — the same remainder 3 each time. ✓
  5. Greatest number leaving given remainders (HCF after subtracting). Find the greatest number that divides 285 and 1249 leaving remainders 9 and 7 respectively. Subtract first: 285 − 9 = 276, 1249 − 7 = 1242. Factorise: 276 = 2² × 3 × 23 and 1242 = 2 × 3³ × 23 → HCF = 2 × 3 × 23 = 138. Check: 285 = 138 × 2 + 9 and 1249 = 138 × 9 + 7. ✓
🧠 When the remainders look different but are not
Find the least number which, divided by 6, 7 and 8, leaves remainders 4, 5 and 6. The remainders look random — but notice 6 − 4 = 2, 7 − 5 = 2, 8 − 6 = 2. Each remainder is short of its divisor by the same k = 2. So the answer is LCM − 2. LCM(6, 7, 8) = 2³ × 3 × 7 = 168, giving 168 − 2 = 166. Check: 166 = 6 × 27 + 4 = 7 × 23 + 5 = 8 × 20 + 6. ✓ Always test the differences before you start guessing.

Remainders, Unit Digits & Trailing Zeros

Remainder of a product. You never have to multiply the big numbers. Take the remainder of each factor first, multiply those small remainders, then take the remainder again. Worked example: find the remainder when 17 × 23 × 29 is divided by 7. Remainders: 17 ÷ 7 → 3; 23 ÷ 7 → 2; 29 ÷ 7 → 1. Multiply: 3 × 2 × 1 = 6, which is already less than 7, so the remainder is 6. (The long way confirms it: 17 × 23 × 29 = 11,339 and 7 × 1,619 = 11,333, leaving 6.)

Remainders of powers — use the cycle. Find the remainder when 3¹⁰⁰ is divided by 5. List the remainders of successive powers: 3¹ → 3, 3² = 9 → 4, 3³ = 27 → 2, 3⁴ = 81 → 1. Once you hit 1 the pattern restarts, so the cycle length is 4. Now 100 ÷ 4 leaves remainder 0, meaning 3¹⁰⁰ lands on the last member of the cycle → remainder 1. Same idea for 2¹⁰⁰ ÷ 3: powers of 2 give 2, 1, 2, 1 … a cycle of 2, and 100 is even → remainder 1.

Unit-digit cycles — the heart of the "cycle of 4" method
Unit digit of baseCycle of unit digitsCycle length
0, 1, 5, 6Never changes — 0, 1, 5, 6 respectively1
44, 6, 4, 6 … (odd power → 4, even power → 6)2
99, 1, 9, 1 … (odd power → 9, even power → 1)2
22, 4, 8, 64
33, 9, 7, 14
77, 9, 3, 14
88, 4, 2, 64

Worked example — the unit digit of 7¹⁰⁵. Step 1: the unit digit of the base is 7, whose cycle is 7, 9, 3, 1 (length 4). Step 2: divide the power by 4 — 105 ÷ 4 = 26 remainder 1. Step 3: remainder 1 → take the 1st member of the cycle → 7. So 7¹⁰⁵ ends in 7. Two more, same three steps: 2⁶³ → cycle 2, 4, 8, 6; 63 ÷ 4 leaves 3 → 3rd member → 8. 13⁵⁷ → base unit digit 3, cycle 3, 9, 7, 1; 57 ÷ 4 leaves 1 → 3.

🧠 The cycle-of-4 recipe
Only the last digit of the base and the power ÷ 4 remainder matter. Remainder 1, 2 or 3 → take the 1st, 2nd or 3rd member of the cycle. Remainder 0 → take the 4th (last) member. Never write "remainder 0 means the answer is 0" — that is the single commonest slip. For example 4⁸⁸: 88 ÷ 4 leaves 0, so take the last of 4, 6 → 6.

Trailing zeros in a factorial. A zero at the end comes from a factor of 10 = 2 × 5. In a factorial the 2s are plentiful and the 5s are scarce, so simply count the 5s: zeros in n! = [n/5] + [n/25] + [n/125] + …, taking only whole parts. Worked example — 100!: [100/5] = 20, [100/25] = 4, [100/125] = 0. Total = 20 + 4 = 24 trailing zeros. Two quick checks of the same method: 25! gives 5 + 1 = 6 zeros, and 50! gives 10 + 2 = 12 zeros.

⚠️ Exam trap
Do not stop at [n/5]. For 100! the answer 20 is wrong — 25, 50, 75 and 100 each contribute an extra 5, adding 4 more zeros for a total of 24. Also, "number of zeros" means zeros at the end; 100! contains many other zeros scattered inside it, which the question is not asking about.

Simplification: BODMAS, Fractions & Decimals

A simplification question has no trick at all — it only tests whether you follow the agreed order of operations. BODMAS = Brackets → Of / Order (powers and roots) → Division → Multiplication → Addition → Subtraction. Within brackets, work from the innermost outwards: ( ) then { } then [ ]. Division and multiplication rank equally, and so do addition and subtraction — when two equals meet, go strictly left to right.

Worked example 1 (nested brackets). Simplify 15 − [8 − {6 − (5 − 3 + 2)}]. Innermost round bracket first, left to right: 5 − 3 = 2, then 2 + 2 = 4. Next the curly bracket: 6 − 4 = 2. Next the square bracket: 8 − 2 = 6. Finally 15 − 6 = 9.

⚠️ The 5 − 3 + 2 trap
In the example above, 5 − 3 + 2 = 4, not 0. Many students see the addition sign and do 3 + 2 = 5 first, getting 5 − 5 = 0 — and then the whole answer becomes 15 − [8 − 6] = 13 instead of 9. Addition does not outrank subtraction; they are equals, so you move left to right. The same warning applies to 36 ÷ 6 × 3: it is (36 ÷ 6) × 3 = 18, not 36 ÷ 18 = 2.

Worked example 2 ("of" and powers). Simplify 48 ÷ 4 of 3 + 5 × 2 − 7. "Of" is handled with the O step, before division: 4 of 3 = 4 × 3 = 12. Now 48 ÷ 12 = 4; then 5 × 2 = 10; then addition and subtraction left to right: 4 + 10 = 14, 14 − 7 = 7. Worked example 3 (with a power). Simplify [72 ÷ (8 + 4)] × 3 − 2² + 5: bracket 8 + 4 = 12; 72 ÷ 12 = 6; power 2² = 4; so 6 × 3 = 18, then 18 − 4 = 14, then 14 + 5 = 19.

  • Fraction → decimal, and when it stops. A fraction in lowest terms terminates if its denominator has only 2s and 5s as prime factors. So 7/8 = 7/2³ = 0.875 terminates and 3/16 = 0.1875 terminates, but 5/12 = 5/(2² × 3) = 0.41666… recurs because of the 3.
  • Pure recurring decimal → fraction. Put the repeating block over as many 9s as there are repeating digits. 0.6̄ = 6/9 = 2/3; 0.36̄ (i.e. 0.363636…) = 36/99 = 4/11; 0.7̄ = 7/9.
  • Mixed recurring decimal → fraction. Numerator = (all digits after the point) − (the non-repeating digits); denominator = one 9 per repeating digit followed by one 0 per non-repeating digit. So 0.27̄ (0.2777…) = (27 − 2)/90 = 25/90 = 5/18, and 0.16̄ (0.1666…) = (16 − 1)/90 = 15/90 = 1/6.
  • Ordering fractions. Arrange 3/5, 5/8, 7/12 in ascending order. LCM of 5, 8, 12 is 120: 3/5 = 72/120, 5/8 = 75/120, 7/12 = 70/120. So 7/12 < 3/5 < 5/8 (as decimals 0.583 < 0.6 < 0.625).
🧠 Compare two fractions by cross-multiplying
To compare a/b and c/d (all positive), cross-multiply: compare a × d with c × b, and the bigger product belongs to the bigger fraction. Compare 5/8 and 7/11: 5 × 11 = 55 against 7 × 8 = 56. Since 55 < 56, 5/8 < 7/11. One more: 13/17 versus 15/19 → 13 × 19 = 247 against 15 × 17 = 255, so 13/17 < 15/19. No LCM, no decimals, one line.

Surds, Indices, Squares & Roots

Laws of exponents, each verified on a small number
LawWorked check
aᵐ × aⁿ = a^(m+n)2³ × 2⁴ = 2⁷ = 128 (8 × 16 = 128 ✓)
aᵐ ÷ aⁿ = a^(m−n)3⁵ ÷ 3² = 3³ = 27 (243 ÷ 9 = 27 ✓)
(aᵐ)ⁿ = a^(m×n)(2²)³ = 2⁶ = 64 (4³ = 64 ✓)
(a × b)ⁿ = aⁿ × bⁿ(2 × 5)² = 10² = 100 (4 × 25 = 100 ✓)
a⁰ = 1 for any a ≠ 07⁰ = 1, and (−5)⁰ = 1
a^(−n) = 1 / aⁿ2^(−3) = 1/2³ = 1/8
a^(1/n) = ⁿ√a8^(1/3) = ∛8 = 2
a^(m/n) = (ⁿ√a)ᵐ27^(2/3) = (∛27)² = 3² = 9; 16^(3/4) = 2³ = 8; 32^(2/5) = 2² = 4

Worked example (solving with indices). If 2^(x−1) + 2^(x+1) = 1280, find x. Take out the smaller power: 2^(x−1) × (1 + 2²) = 2^(x−1) × 5 = 1280. So 2^(x−1) = 1280 ÷ 5 = 256 = 2⁸, giving x − 1 = 8 and x = 9. Check: 2⁸ + 2¹⁰ = 256 + 1024 = 1280. ✓ The whole method is: make both sides powers of the same base, then equate the exponents. Similarly 3^(x+2) = 81 = 3⁴ → x + 2 = 4 → x = 2.

Surds — simplify, then add. A surd is an irrational root such as √2 or ∛5. You can only add surds after making the part under the root identical: √8 + √18 = √(4 × 2) + √(9 × 2) = 2√2 + 3√2 = 5√2. Also √50 = √(25 × 2) = 5√2, so √50 − √18 = 5√2 − 3√2 = 2√2.

Rationalising a denominator (worked). Multiply top and bottom by the conjugate — the same expression with the middle sign flipped — because (a − b)(a + b) = a² − b² kills the roots. Simplify (√5 + √3) / (√5 − √3): multiply above and below by (√5 + √3). Denominator = (√5)² − (√3)² = 5 − 3 = 2. Numerator = (√5 + √3)² = 5 + 2√15 + 3 = 8 + 2√15. So the value = (8 + 2√15)/2 = 4 + √15 ≈ 7.87. Simpler case: 1/(√7 − √3) = (√7 + √3)/(7 − 3) = (√7 + √3)/4.

Comparing surds (worked). Which is greater, ∛3 or √2? Make the root indices equal by using the LCM of 3 and 2, which is 6: ∛3 = 3^(1/3) = 3^(2/6) = ⁶√(3²) = ⁶√9, and √2 = 2^(1/2) = 2^(3/6) = ⁶√(2³) = ⁶√8. Now compare the insides: 9 > 8, so ∛3 > √2 (1.442 versus 1.414). Add ⁶√5 to the race and it is smallest, since 5 < 8 < 9.

🧠 Squaring any number ending in 5
Drop the 5, multiply what is left by the next whole number, and write 25 after it. 35²: 3 × 4 = 12 → 1225. 65²: 6 × 7 = 42 → 4225. 85²: 8 × 9 = 72 → 7225. 95²: 9 × 10 = 90 → 9025. It works past two digits too — 115²: 11 × 12 = 132 → 13225.

Roots by prime factorisation (worked). For a square root, factorise and pair the primes: 1764 = 2 × 2 × 3 × 3 × 7 × 7 = 2² × 3² × 7², so √1764 = 2 × 3 × 7 = 42. For a cube root, group the primes in threes: 3375 = 3 × 3 × 3 × 5 × 5 × 5 = 3³ × 5³, so ∛3375 = 3 × 5 = 15. And difference of squares saves whole minutes: 68² − 32² = (68 + 32)(68 − 32) = 100 × 36 = 3600 — no squaring needed at all.

⚠️ Exam trap
Roots and squares do not distribute over addition. (a + b)² ≠ a² + b² — check (3 + 4)² = 49 while 9 + 16 = 25. Likewise √(a + b) ≠ √a + √b — check √(9 + 16) = √25 = 5 while 3 + 4 = 7. And √(a × b) = √a × √b is valid, which is why √50 = √25 × √2 works. Know which one distributes.

Progressions & Exam Speed Tricks

An arithmetic progression (AP) is a list where each term rises by the same fixed amount d. Two formulas do everything: the nth term aₙ = a + (n − 1)d and the sum Sₙ = n/2 × (first term + last term), which is the same as n/2 × [2a + (n − 1)d]. Worked example: for 7, 11, 15, … find the 20th term and the sum of 20 terms. Here a = 7 and d = 4. So a₂₀ = 7 + (20 − 1) × 4 = 7 + 76 = 83, and S₂₀ = 20/2 × (7 + 83) = 10 × 90 = 900.

Worked example (how many terms?). How many terms are there in 3, 7, 11, …, 99, and what is their sum? Rearranging the nth-term formula gives n = (last − first)/d + 1 = (99 − 3)/4 + 1 = 96/4 + 1 = 24 + 1 = 25 terms. Then the sum = 25/2 × (3 + 99) = 25/2 × 102 = 25 × 51 = 1275. Notice the trick in the last step: pair the first and last term, halve the count, multiply — never add 25 numbers one by one.

The standard sums worth memorising
SumFormulaWorked value
First n natural numbers 1 + 2 + … + nn(n + 1)/2n = 100 → (100 × 101)/2 = 5050
First n odd numbers 1 + 3 + 5 + …n²1 + 3 + … + 19 (10 terms) = 10² = 100
First n even numbers 2 + 4 + 6 + …n(n + 1)2 + 4 + … + 20 (10 terms) = 10 × 11 = 110
Sum of squares 1² + 2² + … + n²n(n + 1)(2n + 1)/6n = 10 → (10 × 11 × 21)/6 = 2310/6 = 385
Sum of cubes 1³ + 2³ + … + n³[n(n + 1)/2]²n = 5 → (15)² = 225 (1+8+27+64+125 = 225 ✓)
🧠 Odd numbers build perfect squares
The sum of the first n odd numbers is exactly n² — 1 = 1², 1+3 = 2², 1+3+5 = 3², 1+3+5+7 = 4². So if a question asks for 1 + 3 + 5 + … + 29, just count the terms: the odd numbers up to 29 are (29 + 1)/2 = 15 of them, so the sum is 15² = 225. No formula sheet required.
  • Multiply by 11 (two digits). Write the two digits apart and put their sum in the middle. 43 × 11 → 4 _ 3 with 4+3 = 7 → 473. If the middle sum exceeds 9, carry over: 87 × 11 → 8 _ 7 with 8+7 = 15 → carry the 1 → 957. Check 78 × 11 → 7+8 = 15 → 858.
  • Multiply by 25, 5 and 125. Since 25 = 100/4: 48 × 25 = 4800 ÷ 4 = 1200. Since 5 = 10/2: 86 × 5 = 860 ÷ 2 = 430. Since 125 = 1000/8: 64 × 125 = 64000 ÷ 8 = 8000.
  • Multiply by 9, 99. Use "one more, then subtract": 47 × 9 = 470 − 47 = 423; 36 × 99 = 3600 − 36 = 3564.
  • Difference of squares, and squares near 100. 51² − 49² = (51+49)(51−49) = 100 × 2 = 200. And 98² = (100 − 2)² = 10000 − 400 + 4 = 9604; 103² = 10000 + 600 + 9 = 10609.
Percentage–fraction equivalents beyond the basics
PercentageFractionUse it like this
6¼% (6.25%)1/166.25% of 480 = 480 ÷ 16 = 30
8⅓%1/128⅓% of 600 = 600 ÷ 12 = 50
11⅑%1/911⅑% of 720 = 720 ÷ 9 = 80
14²⁄₇%1/714²⁄₇% of 490 = 490 ÷ 7 = 70
16⅔%1/616⅔% of 540 = 540 ÷ 6 = 90
37.5%3/837.5% of 640 = 640 × 3 ÷ 8 = 240
62.5%5/862.5% of 320 = 320 × 5 ÷ 8 = 200
87.5%7/887.5% of 240 = 240 × 7 ÷ 8 = 210
🎯 Why it matters
Number-system questions are the cheapest marks in the paper. There is nothing to update, nothing to forget from the news, and the same dozen patterns return year after year — a divisibility test, a unit digit, an HCF word problem, a BODMAS line. Fifteen minutes of daily drill turns this section into marks you can bank before you even open the reasoning set, leaving more clock for the questions that actually need thinking.

Key takeaways

  • 1 is neither prime nor composite; 2 is the only even prime; 0 is whole but not natural. π is irrational — 22/7 is only an approximation.
  • Divisibility: 3 & 9 by digit sum, 4 by last two digits, 8 by last three, 11 by the alternating sum. For 6 check 2 and 3; split composite divisors into co-prime factors.
  • HCF × LCM = product — for two numbers only. HCF of fractions = HCF of numerators / LCM of denominators; LCM of fractions is the reverse. Example: 72 & 120 → HCF 24, LCM 360.
  • Word problems: largest measuring size → HCF; bells together → LCM (6,8,12,18 → 72 s); same remainder r → LCM + r (→ 544); same shortfall k → LCM − k (→ 166); same unknown remainder → HCF of differences (→ 4).
  • Unit digit: divide the power by 4 — remainder 1/2/3 picks that member of the cycle, remainder 0 picks the last. 7¹⁰⁵ → cycle 7,9,3,1 and 105 ÷ 4 leaves 1 → 7. Zeros in n! = [n/5] + [n/25] + … ; 100! → 20 + 4 = 24.
  • BODMAS: brackets inside-out, then Of/powers, then ÷ and × left to right, then + and − left to right. 15 − [8 − {6 − (5 − 3 + 2)}] = 9, because 5 − 3 + 2 = 4, not 0.
  • Shortcuts to own: squaring numbers ending in 5 (65² = 6×7 then 25 = 4225), difference of squares (68² − 32² = 100 × 36 = 3600), ×11 by adding the digits (43 × 11 = 473), and AP sum = n/2 × (first + last) (7, 11, …, 83 → 900).

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. What is the unit digit of 7¹⁰⁵?

    1. 7
    2. 9
    3. 3
    4. 1
    Show answer

    Answer: A — 7

    The unit digits of powers of 7 cycle as 7, 9, 3, 1 (length 4). Divide the power by 4: 105 ÷ 4 = 26 remainder 1, so take the 1st member of the cycle → 7.

  2. Using the shortcut for squaring numbers ending in 5, the value of 65² is:

    1. 4025
    2. 4125
    3. 4225
    4. 4325
    Show answer

    Answer: C — 4225

    Drop the 5, multiply 6 by the next number 7 to get 42, then write 25 after it: 4225. (Check: 65 × 65 = 4225.)

  3. Simplify: 15 − [8 − {6 − (5 − 3 + 2)}]

    1. 7
    2. 9
    3. 11
    4. 13
    Show answer

    Answer: B — 9

    Innermost bracket left to right: 5 − 3 + 2 = 4 (not 0). Then 6 − 4 = 2; then 8 − 2 = 6; then 15 − 6 = 9.

  4. What is the length of the longest tape that can measure exactly 7 m, 3 m 85 cm and 12 m 95 cm?

    1. 45 cm
    2. 35 cm
    3. 25 cm
    4. 15 cm
    Show answer

    Answer: B — 35 cm

    Convert to centimetres: 700, 385, 1295. Then 700 = 2² × 5² × 7, 385 = 5 × 7 × 11 and 1295 = 5 × 7 × 37, so the HCF is 5 × 7 = 35 cm. Check: 700 ÷ 35 = 20, 385 ÷ 35 = 11, 1295 ÷ 35 = 37.

  5. The HCF of two numbers is 24 and their LCM is 360. If one number is 72, the other number is:

    1. 90
    2. 100
    3. 108
    4. 120
    Show answer

    Answer: D — 120

    For two numbers, HCF × LCM = product of the numbers. So the product = 24 × 360 = 8640, and the other number = 8640 ÷ 72 = 120.

  6. Consider the following two statements:
    1. 1331 is divisible by 11.
    2. 2197 is divisible by 9.
    Which is/are correct?

    1. Only 1
    2. Only 2
    3. Both 1 and 2
    4. Neither 1 nor 2
    Show answer

    Answer: A — Only 1

    For 1331 the alternating sum is (1 + 3) − (3 + 1) = 0, so it is divisible by 11 (1331 ÷ 11 = 121) — statement 1 is correct. For 2197 the digit sum is 2 + 1 + 9 + 7 = 19, which is not a multiple of 9, so statement 2 is wrong. (2197 is in fact divisible by 13: 2197 ÷ 13 = 169.)

  7. The number of zeros at the end of 100! (100 factorial) is:

    1. 20
    2. 22
    3. 24
    4. 25
    Show answer

    Answer: C — 24

    Count the factors of 5: [100/5] = 20, [100/25] = 4, [100/125] = 0. Total = 20 + 4 = 24. Stopping at 20 is the usual mistake — 25, 50, 75 and 100 each supply an extra 5.

  8. The sum of the first 20 terms of the arithmetic progression 7, 11, 15, … is:

    1. 880
    2. 900
    3. 920
    4. 940
    Show answer

    Answer: B — 900

    Here a = 7 and d = 4. The 20th term = 7 + 19 × 4 = 83, so the sum = (20/2) × (7 + 83) = 10 × 90 = 900.