📑 Topics on this page (Jump to Topic):
📌 1. Types of Numbers
| Type | Definition | Example |
|---|---|---|
| Natural (N) | Counting numbers starting from 1 | 1, 2, 3, … |
| Whole (W) | Natural numbers including 0 | 0, 1, 2, … |
| Integers (Z) | Positive, negative and zero | …, −2, −1, 0, 1, … |
| Rational (Q) | p/q form, q ≠ 0; decimal terminates or repeats | 3/4, 0.333…, −5 |
| Irrational | Not p/q; non-terminating non-repeating decimal | √2, π, √3 |
| Prime | Exactly two factors: 1 and itself | 2, 3, 5, 7, 11 |
| Composite | More than two factors | 4, 6, 8, 9 |
| Co-prime | Pair whose HCF = 1 | (8, 15), (4, 9) |
- 1 is neither prime nor composite. 2 is the only even prime.
- There are 25 primes from 1 to 100 and 15 from 1 to 50.
- Prime test: divide by primes up to √n. E.g. 221 = 13 × 17 (not prime).
⚠️ Exam Traps:
- 0 is a whole number but not a natural number.
- π = 22/7 is only an approximation; π is irrational.
- Sum/product of two irrationals can be rational: √2 × √2 = 2.
📌 2. Divisibility Rules
| Number | Rule |
|---|---|
| 2 | Unit digit even (0,2,4,6,8) |
| 3 / 9 | Sum of digits divisible by 3 / 9 |
| 4 | Last two digits divisible by 4 |
| 8 | Last three digits divisible by 8 |
| 5 | Unit digit 0 or 5 |
| 6 | Divisible by both 2 and 3 |
| 11 | (Sum of digits at odd places − sum at even places) = 0 or multiple of 11 |
| 7 / 13 | Subtract 2 × unit digit (7) / add 4 × unit digit (13) and check |
| 12, 15, 18, 24… | Check co-prime factors: 12 = 3×4, 24 = 3×8, 18 = 2×9 |
⚠️ Exam Traps:
- For 24, testing with 4 × 6 is wrong (not co-prime); use 3 × 8.
✏️ Example: For 5k4 to be divisible by 9, k = ? 5+k+4 = 9+k → k = 0 or 9.
📌 3. Unit Digit & Cyclicity
| Last digit | Cycle | Cycle length |
|---|---|---|
| 0, 1, 5, 6 | Always the same digit | 1 |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 4 | 4, 6 (odd power 4, even power 6) | 2 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
| 9 | 9, 1 (odd 9, even 1) | 2 |
💡 Shortcut Trick: Divide the power by 4, take remainder r (use 4 if r = 0) and find the unit digit of last-digit^r.
✏️ Example: 783: 83 ÷ 4 leaves 3 → 7³ = 343 → unit digit 3.
📌 4. Remainders, Factors & Zeros
📘 Formulas:
- Dividend = Divisor × Quotient + Remainder
- If N = pa qb rc, number of factors = (a+1)(b+1)(c+1)
- Sum of factors = (p0+…+pa)(q0+…+qb)…
- Trailing zeros in n! = [n/5] + [n/25] + [n/125] + …
- (an + bn) is divisible by (a + b) for odd n; (an − bn) always by (a − b) and also by (a + b) for even n
- Sum of first n natural numbers = n(n+1)/2; squares = n(n+1)(2n+1)/6; cubes = [n(n+1)/2]²
✏️ Example: A number divided by 899 leaves 63; what is the remainder when divided by 29? 899 = 29 × 31, so remainder = 63 mod 29 = 5.
⚠️ Exam Traps:
- A remainder is never negative; −3 mod 7 = 4.
- Trailing zeros of 100! = 20 + 4 = 24 (not just 20).
📌 5. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. What is the unit digit of 27173? Practice
Correct answer: (B)
Solution: Cycle of 7: 7, 9, 3, 1. 173 ÷ 4 leaves 1 → unit digit of 71 = 7.
Solution: Cycle of 7: 7, 9, 3, 1. 173 ÷ 4 leaves 1 → unit digit of 71 = 7.
Q2. How many trailing zeros are there in 256!? Practice
Correct answer: (C)
Solution: Zeros = [256/5] + [256/25] + … = 51 + 10 + 2 = 63.
Solution: Zeros = [256/5] + [256/25] + … = 51 + 10 + 2 = 63.
Q3. The total number of factors of 14 is: Practice
Correct answer: (D)
Solution: 14 = 21 × 71 → factors = (1+1) × (1+1) = 4.
Solution: 14 = 21 × 71 → factors = (1+1) × (1+1) = 4.
Q4. If the number 18*69 is divisible by 11, the value of * is: Practice
Correct answer: (A)
Solution: With * = 4, (sum of odd-place digits − sum of even-place digits) is 0 or a multiple of 11, so 18469 is divisible by 11.
Solution: With * = 4, (sum of odd-place digits − sum of even-place digits) is 0 or a multiple of 11, so 18469 is divisible by 11.
Q5. A number when divided by 323 leaves remainder 115. What will be the remainder when the same number is divided by 19? Practice
Correct answer: (B)
Solution: 323 = 19 × 17, so N = 323k + 115. 323k is divisible by 19 → remainder = 115 mod 19 = 1.
Solution: 323 = 19 × 17, so N = 323k + 115. 323k is divisible by 19 → remainder = 115 mod 19 = 1.
Q6. Sum of all natural numbers from 1 to 20 is: Practice
Correct answer: (C)
Solution: n(n+1)/2 = 20×21/2 = 210.
Solution: n(n+1)/2 = 20×21/2 = 210.
Q7. Which of the following is a prime number? Practice
Correct answer: (D)
Solution: 269 is not divisible by any prime up to √269, so it is prime. Others: 187 = 11 × 17, 437 = 19 × 23, 119 = 7 × 17.
Solution: 269 is not divisible by any prime up to √269, so it is prime. Others: 187 = 11 × 17, 437 = 19 × 23, 119 = 7 × 17.
Q8. (299 + 1) is definitely divisible by: Practice
Correct answer: (A)
Solution: For odd n, aⁿ + bⁿ is divisible by (a + b) → 29 + 1 = 30.
Solution: For odd n, aⁿ + bⁿ is divisible by (a + b) → 29 + 1 = 30.
Q9. The unit digit of 4955 × 5422 × 9552 is: Practice
Correct answer: (B)
Solution: Individual unit digits: 9, 6, 5. Unit digit of their product 270 = 0.
Solution: Individual unit digits: 9, 6, 5. Unit digit of their product 270 = 0.
Q10. The smallest value of * that makes 48377* divisible by 9 is: Practice
Correct answer: (C)
Solution: Sum of known digits = 29. To reach the next multiple of 9 we need 7.
Solution: Sum of known digits = 29. To reach the next multiple of 9 we need 7.
Q11. Express 0.2777… in p/q form: Practice
Correct answer: (D)
Solution: Converting the recurring/terminating decimal gives 0.2777… = 5/18.
Solution: Converting the recurring/terminating decimal gives 0.2777… = 5/18.
Q12. In a division sum, the divisor is 36, the quotient is 16 and the remainder is 1. Find the dividend. Practice
Correct answer: (A)
Solution: Dividend = divisor × quotient + remainder = 36 × 16 + 1 = 577.
Solution: Dividend = divisor × quotient + remainder = 36 × 16 + 1 = 577.
Q13. The largest 5-digit number exactly divisible by 37 is: Practice
Correct answer: (B)
Solution: 99999 ÷ 37 leaves 25. 99999 − 25 = 99974.
Solution: 99999 ÷ 37 leaves 25. 99999 − 25 = 99974.
Q14. What is the unit digit of 1362? Practice
Correct answer: (C)
Solution: Cycle of 3: 3, 9, 7, 1. 62 ÷ 4 leaves 2 → unit digit of 32 = 9.
Solution: Cycle of 3: 3, 9, 7, 1. 62 ÷ 4 leaves 2 → unit digit of 32 = 9.
Q15. How many trailing zeros are there in 125!? Practice
Correct answer: (D)
Solution: Zeros = [125/5] + [125/25] + … = 25 + 5 + 1 = 31.
Solution: Zeros = [125/5] + [125/25] + … = 25 + 5 + 1 = 31.
Q16. The total number of factors of 75 is: Practice
Correct answer: (A)
Solution: 75 = 31 × 52 → factors = (1+1) × (2+1) = 6.
Solution: 75 = 31 × 52 → factors = (1+1) × (2+1) = 6.
Q17. If the number 551*7 is divisible by 11, the value of * is: Practice
Correct answer: (B)
Solution: With * = 8, (sum of odd-place digits − sum of even-place digits) is 0 or a multiple of 11, so 55187 is divisible by 11.
Solution: With * = 8, (sum of odd-place digits − sum of even-place digits) is 0 or a multiple of 11, so 55187 is divisible by 11.
Q18. A number when divided by 391 leaves remainder 220. What will be the remainder when the same number is divided by 23? Practice
Correct answer: (C)
Solution: 391 = 23 × 17, so N = 391k + 220. 391k is divisible by 23 → remainder = 220 mod 23 = 13.
Solution: 391 = 23 × 17, so N = 391k + 220. 391k is divisible by 23 → remainder = 220 mod 23 = 13.
Q19. The sum of the first 40 odd natural numbers is: Practice
Correct answer: (D)
Solution: n² = 40² = 1600.
Solution: n² = 40² = 1600.
Q20. (145 − 1) is definitely divisible by: Practice
Correct answer: (A)
Solution: aⁿ − bⁿ is always divisible by (a − b) → 14 − 1 = 13. For odd n, not by (a + 1).
Solution: aⁿ − bⁿ is always divisible by (a − b) → 14 − 1 = 13. For odd n, not by (a + 1).