Prelims · 8 Mental Ability · 8.1 Number system and simplification
8.1.1 Number types and divisibility rules
Know which family a number belongs to and test divisibility from its digits, so that "find the missing digit", "nearest multiple" and "how many numbers between" questions take seconds.
- APPSC Group 2 · Prelims
- Standard level
- 8 min read
- Free
Core facts
- Natural numbers: 1, 2, 3, … Whole numbers: 0, 1, 2, … Integers: …, −2, −1, 0, 1, 2, …
- Rational: can be written p/q with q ≠ 0; its decimal terminates or recurs. Irrational: non-terminating, non-recurring (√2, √3, π).
- Prime: exactly two factors (1 and itself). 1 is neither prime nor composite.
- 2 is the only even prime; 3 is the first odd prime; 4 is the smallest composite; 9 is the smallest odd composite.
- There are 25 primes below 100 and 15 primes below 50.
- Co-prime: two numbers whose HCF is 1 (8 and 15). They need not be prime. Twin primes differ by 2 (11, 13).
- Prime test: check divisibility only by primes up to √n.
- Parity: odd ± odd = even; even ± odd = odd; odd × odd = odd; any product with an even factor is even.
- Face value of a digit is the digit itself. Place value = digit × its place.
- Divisibility tests:
| By | Rule |
|---|---|
| 2, 5, 10 | last digit even / 0 or 5 / 0 |
| 3, 9 | digit sum divisible by 3 / 9 |
| 4, 25 | last two digits divisible by 4 / 25 |
| 8, 16 | last three / last four digits divisible by 8 / 16 |
| 11 | (sum of alternate digits) − (sum of the others) is 0 or a multiple of 11 |
| 6, 12, 15, 18, 24 | 2 and 3 / 3 and 4 / 3 and 5 / 2 and 9 / 3 and 8 |
- Combine two tests only when the two divisors are co-prime (24 = 3 × 8, not 4 × 6).
Example 1. Is 247 a prime number? Solution. √247 is about 15.7, so test the primes 2, 3, 5, 7, 11, 13. 247 = 13 × 19. Answer: No, it is composite.
Example 2. In 46,218, what is the place value of 6 minus its face value? Solution. 6 is in the thousands place: place value 6,000, face value 6. Answer: 5,994
Example 3. Is 73,416 divisible by 8? Solution. Last three digits 416 = 8 × 52. Answer: Yes.
Explained
Type 1: one missing digit, one divisor. Write the test as an equation in the digit.
Example 4. 58x3 is divisible by 11. Find x. (a) 2 (b) 4 (c) 6 (d) 8 Solution. Alternate sums: (5 + x) − (8 + 3) = x − 6. It must be 0 or a multiple of 11, so x = 6. Check: 5,863 = 11 × 533. Answer: (c) 6
Type 2: one digit, two divisors. Apply the stricter test first, then confirm with the other.
Example 5. Replace * in 74381 so the number is divisible by both 3 and 11. (a) 1 (b) 4 (c) 7 (d) 9 Solution.* For 11: (7 + 4 + 8) − (* + 3 + 1) = 15 − , so = 4. For 3: digit sum 23 + 4 = 27, divisible by 3. Answer: (b) 4
Type 3: two digits, two divisors. 72 = 8 × 9. Fix the last digit with the 8-test, then the other with the 9-test.
Example 6. The number 6x31y is divisible by 72. Find x + y. (a) 6 (b) 8 (c) 10 (d) 12 Solution. For 8: 31y must be a multiple of 8; only 312 works, so y = 2. For 9: 6 + x + 3 + 1 + 2 = 12 + x, so x = 6. Answer: (b) 8
Type 4: multiples in a range. Multiples of n from a to b (both included) = ⌊b/n⌋ − ⌊(a − 1)/n⌋.
Example 7. How many multiples of 9 lie strictly between 200 and 500? Solution. ⌊499/9⌋ − ⌊200/9⌋ = 55 − 22. Answer: 33
Example 8. How many numbers from 1 to 300 are divisible by 4 but not by 6? Solution. Multiples of 4: 75. Divisible by both 4 and 6 means divisible by LCM 12: 25. 75 − 25 = 50. Answer: 50
Type 5: sum of multiples. Terms = (last − first)/n + 1; sum = terms × (first + last)/2.
Example 9. Find the sum of all three-digit multiples of 11. Solution. First 110, last 990; terms = 880/11 + 1 = 81. Sum = 81 × 1,100/2 = 81 × 550. Answer: 44,550
Type 6: nearest, smallest and largest multiples. Divide, then move to the multiple below or above.
Example 10. (i) Which number nearest to 4,000 is divisible by 23? (ii) What is the least number to add to 2,500 to make it divisible by 47? Solution. (i) 4,000 = 23 × 173 + 21. The multiple below is 3,979 (21 away); above is 4,002 (2 away). (ii) 2,500 = 47 × 53 + 9, so add 47 − 9 = 38. Answer: (i) 4,002 (ii) 38
- Smallest five-digit multiple of 37: 10,000 = 37 × 270 + 10, so take 37 × 271 = 10,027. Largest four-digit multiple of 29: 9,999 = 29 × 344 + 23, so 9,999 − 23 = 9,976.
Speed tips.
- For a missing digit, use the strictest test first (8, 11), then the digit-sum test.
- For "how many between", write the floor formula once; never list the numbers.
- "Strictly between" excludes both ends; "from … to" includes both.
Common traps
- ✗ "1 is a prime number." ✓ 1 is neither prime nor composite; 2 is the smallest prime.
- ✗ "Co-prime numbers must both be prime." ✓ Co-prime means HCF = 1 (8 and 9 are co-prime).
- ✗ "Divisible by 4 and 6, so divisible by 24." ✓ Only by LCM 12. Combine tests only for co-prime divisors.
- ✗ "Divisible by 6 and 9: count multiples of 54." ✓ Count multiples of the LCM, 18.
- ✗ "Multiples of 9 between 200 and 500 = 500/9 − 200/9 ≈ 33.3." ✓ Use whole parts: ⌊499/9⌋ − ⌊200/9⌋ = 33.
- ✗ "Place value of 6 in 46,218 is 6." ✓ That is the face value; the place value is 6,000.
Quick revision
- 1 is neither prime nor composite; 2 is the only even prime.
- 25 primes below 100; 15 below 50.
- Rational = terminating or recurring decimal.
- 4: last two digits; 8: last three; 16: last four.
- 3 and 9: digit sum. 11: alternate-digit difference 0 or 11k.
- Combine tests only for co-prime divisors.
- Multiples of n in [a, b] = ⌊b/n⌋ − ⌊(a − 1)/n⌋.
- "Divisible by both" means divisible by the LCM.
- abcabc = abc × 1,001 = abc × 7 × 11 × 13.
- Check primes only up to √n.
Go deeper: Deeper (beyond what APPSC Group 2 usually asks)
Deeper Advanced
Inclusion–exclusion. |A ∪ B ∪ C| = |A| + |B| + |C| − |A∩B| − |B∩C| − |C∩A| + |A∩B∩C|. Intersections use the LCM of the divisors.
Example 11. How many integers from 1 to 200 are divisible by none of 2, 3 and 5? (a) 46 (b) 54 (c) 60 (d) 66 Solution. Single counts: 100 + 66 + 40 = 206. Pairs (6, 10, 15): 33 + 20 + 13 = 66. All three (30): 6. At least one = 206 − 66 + 6 = 146. None = 200 − 146. Answer: (b) 54
Example 12. Caselet. In a school game, a number from 1 to 100 is "lucky" if it is a multiple of 4 or its digit sum is 8 (or both). (i) How many lucky numbers are there? (ii) How many satisfy both rules? Solution. Multiples of 4: 25. Digit sum 8: 8, 17, 26, 35, 44, 53, 62, 71, 80 (9 numbers). Of these, 8, 44 and 80 are multiples of 4. Lucky = 25 + 9 − 3. Answer: (i) 31 (ii) 3
- List the shorter set in full; the overlap is then easy to see.
Example 13. Which of these divides 286,286 exactly: 7, 11, 13 or all three? Solution. A six-digit number of the form abcabc equals abc × 1,001, and 1,001 = 7 × 11 × 13. 286,286 = 286 × 1,001. Answer: all three
Example 14. What is the smallest number made only of the digit 1 that is divisible by 7? Solution. Test 1, 11, 111, … : 111,111 = 7 × 15,873 is the first. (111,111 = 1,001 × 111.) Answer: 111,111 (six ones)
Example 15. For how many positive integers x is x² + 15 exactly divisible by x? Solution. x divides x², so x must divide 15: x = 1, 3, 5, 15. Answer: 4
- Divisibility by 7 or 13 for long numbers: split into groups of three digits from the right and take the alternating sum (1,001 again).
- Product of n consecutive integers is divisible by n!; so the product of any three consecutive integers is divisible by 6.
- Disguised composites used as "which is not prime": 91 (7 × 13), 119 (7 × 17), 133 (7 × 19), 143 (11 × 13), 161 (7 × 23), 187 (11 × 17), 209 (11 × 19), 221 (13 × 17), 247 (13 × 19), 323 (17 × 19).
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