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compulsory · CSS study notes

4. Integers

CSS GSA — Paper 7, Part-II, Section VI, Sub-topic 4 (Integers)

Past Paper Weightage

  • Foundational — rarely asked standalone, but errors here cascade into BODMAS/Algebra questions. Quick review, not heavy practice, is enough.

Key Rules

  • Integers = {..., -3, -2, -1, 0, 1, 2, 3, ...} — whole numbers and their negatives, no fractions/decimals
  • Addition/Subtraction: same sign → add magnitudes, keep sign; different signs → subtract smaller magnitude from larger, keep sign of larger
  • Multiplication/Division sign rules: (+)×(+) = +, (−)×(−) = +, (+)×(−) = −, (−)×(+) = − (same signs → positive, different signs → negative)
  • Order of magnitude: −5 < −2 < 0 < 3 (more negative = smaller)
  • Absolute value |x|: distance from zero, always non-negative — |−7| = 7

Prime Numbers and Number Properties

  • Prime number: a whole number greater than 1 with exactly two factors — 1 and itself (e.g., 2, 3, 5, 7, 11, 13...). 2 is the only even prime number; every other even number is divisible by 2, so it has more than two factors.
  • Composite number: a whole number greater than 1 with more than two factors (e.g., 4, 6, 8, 9)
  • 1 is neither prime nor composite (only one factor: itself) — a very common exam trap
  • Quick primality test (shortcut): to check if a number N is prime, test divisibility only by prime numbers up to √N — if none divide evenly, N is prime
    • Example: Is 97 prime?
      • Step 1: √97 ≈ 9.8, so test primes up to 9: 2, 3, 5, 7.
      • Step 2: 97 is odd (not divisible by 2); digit sum 9+7=16, not divisible by 3; doesn't end in 0/5 (not divisible by 5); 97÷7 ≈ 13.86 (not divisible by 7).
      • Step 3: No prime up to √97 divides it evenly, so 97 is prime.
  • Even/Odd properties (useful shortcuts): even±even=even, odd±odd=even, even±odd=odd; even×anything=even; odd×odd=odd
  • Divisibility rules (shortcuts): by 2 (last digit even), by 3 (digit sum divisible by 3), by 5 (ends in 0 or 5), by 9 (digit sum divisible by 9), by 10 (ends in 0)

Worked Example

  • (−8) + 5 − (−3) = −8 + 5 + 3 = 0
  • (−4) × (−6) ÷ 3 = 24 ÷ 3 = 8

Past Paper Style Numericals (Solved)

  1. Evaluate: (−15) × (−4) + (−20) ÷ 5
    • Step 1: Apply BODMAS — do multiplication and division before addition. (−15)×(−4): two negatives multiply to a positive → +60.
    • Step 2: (−20)÷5: negative ÷ positive = negative → −4.
    • Step 3: Add the results: 60 + (−4) = 56.
  2. The temperature was −7°C at night, rose by 15°C by noon, then dropped 6°C by evening. Find the evening temperature.
    • Step 1: Start at −7°C.
    • Step 2: Rise of 15°C: −7 + 15 = 8°C (noon temperature).
    • Step 3: Drop of 6°C: 8 − 6 = 2°C (evening temperature).
  3. Evaluate: (−6)² − (−4)³
    • Step 1: Compute (−6)² = (−6)×(−6) = 36 (negative × negative = positive).
    • Step 2: Compute (−4)³ = (−4)×(−4)×(−4) = 16×(−4) = −64.
    • Step 3: Subtract: 36 − (−64) = 36 + 64 = 100.
  4. A submarine at −120 m dives another 45 m, then rises 60 m. Find its final depth.
    • Step 1: Start at −120 m.
    • Step 2: Dives (goes further negative) 45 m: −120 − 45 = −165 m.
    • Step 3: Rises (becomes less negative) 60 m: −165 + 60 = −105 m.

Remainders

  • Division rule: Dividend = Divisor × Quotient + Remainder, and the remainder is always smaller than the divisor (0 ≤ remainder < divisor)
    • Example: 47 ÷ 6 → 47 = 6 × 7 + 5, so the remainder is 5
  • Work with remainders only
    • The remainder of a sum or product equals the remainder of the sum or product of the remainders
    • Example: remainder of 23 × 47 ÷ 5: 23 leaves 3 and 47 leaves 2; 3 × 2 = 6, which leaves 1
  • Remainders of powers: find the repeating cycle
    • Example: 2¹⁰ ÷ 7: powers of 2 leave 2, 4, 1, 2, 4, 1 … (cycle of 3); 10 is one more than a multiple of 3, so the remainder is 2
  • Last digits repeat in cycles: 2 → 2, 4, 8, 6; 3 → 3, 9, 7, 1; 7 → 7, 9, 3, 1
  • Number from remainders
    • Smallest number leaving remainder 3 when divided by 4, 5 and 6 = LCM(4, 5, 6) + 3 = 60 + 3 = 63
    • If every remainder is one less than its divisor, the number is LCM − 1
  • Shortcuts with digits: the remainder on division by 9 equals the remainder of the digit sum (1234: digit sum 10 → remainder 1); by 5 or 10 look at the last digit; by 4 look at the last two digits
  • Link: the 'coincide again' question (CSS 2019 Q6(a)) uses LCM, the same idea as 'number with a given remainder'

MCQ Traps

  • Subtracting a negative = adding a positive: a − (−b) = a + b
  • Even number of negative signs in a product → positive result; odd number → negative

Self-Test Questions

  1. Evaluate: (−12) + 7 − (−5) × 2
  2. Is −15 or −3 the smaller integer, and why?

Past-Paper MCQs (FPSC)

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The FPSC MCQ papers you shared (2024 and 2025) had no direct integer question.

Past-Paper Questions with Answers (FPSC)

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Every question below appeared in the FPSC papers you shared; the year and question number are in each title. Answers are worked out by me.

  • CSS 2022 Q7(a): The sum of three consecutive odd numbers is 273. Find them. (5)
    1. Let the numbers be n − 2, n, n + 2; the sum is 3n = 273, so n = 91
    2. The numbers are 89, 91, 93
  • CSS 2019 Q6(a): A clock ticks every 5 seconds, a tap drips every 7th second and a dog snores every 12th second; all three happen together at midnight. After how many seconds do they next coincide? (5)
    1. Find the LCM of 5, 7 and 12 (all share no common factor)
    2. LCM = 5 × 7 × 12 = 420 seconds = 7 minutes
  • CSS 2019 Q8(d): Each packet of washing powder carries a token, and 4 tokens can be exchanged for a free packet. How many free packets will I get if I buy 64 packets? (5)
    1. 64 tokens ÷ 4 = 16 free packets; these 16 carry 16 tokens
    2. 16 ÷ 4 = 4 more free packets; these carry 4 tokens
    3. 4 ÷ 4 = 1 more free packet
    4. Total free packets = 16 + 4 + 1 = 21 (1 token is left over)

CSS Exam Perspective

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  • Most repeated areas: consecutive-number problems, LCM/HCF word problems, token/exchange puzzles
  • Tip: for 'coincide again' use the LCM; for consecutive odd or even numbers use n − 2, n, n + 2