Number System
CSAT · Quantitative Aptitude
Introduction
CSAT numeracy is pitched at Class X level, but the questions reward speed: knowing divisibility rules, remainders and unit-digit cycles lets you answer in seconds what would otherwise need long multiplication. Number-system ideas also sit underneath almost every other quant topic.
Key concepts
Types of numbers
| Type | Meaning | Examples |
|---|---|---|
| Natural | Counting numbers | 1, 2, 3 … |
| Whole | Natural numbers and 0 | 0, 1, 2 … |
| Integers | Whole numbers and negatives | … −2, −1, 0, 1 … |
| Prime | Exactly two factors (1 and itself) | 2, 3, 5, 7, 11 … (2 is the only even prime) |
| Composite | More than two factors | 4, 6, 8, 9 … (1 is neither prime nor composite) |
| Co-prime | HCF of the pair is 1 | 8 and 15 |
Divisibility rules
| Divisor | Rule |
|---|---|
| 2, 4, 8 | Last 1, 2, 3 digits divisible by 2, 4, 8 |
| 3, 9 | Sum of digits divisible by 3, 9 |
| 5, 10 | Ends in 0/5; ends in 0 |
| 6 | Divisible by both 2 and 3 |
| 11 | (Sum of odd-place digits) − (sum of even-place digits) is 0 or a multiple of 11 |
| 7, 13 | Test directly or use remainders (1001 = 7 × 11 × 13) |
Remainders
The remainder of a sum or product equals the remainder of the sum or product of the individual remainders. So 1,234 × 5,678 ÷ 9 only needs 1 × 8 = 8 (digit sums 10 → 1 and 26 → 8).
Unit digits
Unit digits of powers repeat in cycles: 2 → 2, 4, 8, 6; 3 → 3, 9, 7, 1; 7 → 7, 9, 3, 1; 8 → 8, 4, 2, 6 (length 4); 4 and 9 alternate (length 2); 0, 1, 5, 6 never change. Divide the power by the cycle length and use the remainder (remainder 0 = last element).
Counting factors
If N = a^p × b^q × c^r (prime factorisation), the number of factors is (p + 1)(q + 1)(r + 1). Only perfect squares have an odd number of factors.
- Dividend = Divisor × Quotient + Remainder
- Sum of first n natural numbers = n(n + 1)/2
- Sum of first n odd numbers = n²; first n even numbers = n(n + 1)
- Sum of squares 1² + … + n² = n(n + 1)(2n + 1)/6
- Number of factors of a^p·b^q = (p + 1)(q + 1)
Solved examples
Example 1
How many factors (divisors) does 308 have?
(a) 12 (b) 4 (c) 10 (d) 14
- Prime factorisation: 308 = 2^2 × 7 × 11.
- Number of factors = product of (each power + 1).
- (2+1) × (1+1) × (1+1) = 12.
Answer: (a) 12
Example 2
What is the least number that must be added to 9577 to make it exactly divisible by 37?
(a) 31 (b) 7 (c) 36 (d) 6
- 9577 ÷ 37 = 258 remainder 31.
- To reach the next multiple, add 37 − 31 = 6.
- Check: 9577 + 6 = 9583 = 37 × 259.
Answer: (d) 6
Example 3
What is the sum of the first 36 odd numbers?
(a) 1296 (b) 1260 (c) 1332 (d) 1426
- Use the formula: sum = n² = 36² = 1296.
Answer: (a) 1296
Example 4
What is the unit digit of 87^140?
(a) 2 (b) 3 (c) 1 (d) 7
- Only the unit digit 7 matters.
- Powers of 7 end in the cycle 7, 9, 3, 1 (length 4).
- 140 ÷ 4 leaves remainder 4 (a remainder of 0 means the last term of the cycle).
- So the unit digit is 1.
Answer: (c) 1
Example 5
What is the sum of the first 22 natural numbers?
(a) 253 (b) 231 (c) 275 (d) 485
- Use the formula: sum = n(n+1)/2 = 22×23/2 = 253.
Answer: (a) 253
Example 6
What is the remainder when 684 × 251 × 250 is divided by 13?
(a) 2 (b) 6 (c) 8 (d) 5
- Replace each number by its remainder on division by 13: 684 → 8, 251 → 4, 250 → 3.
- 8 × 4 × 3 = 96.
- 96 ÷ 13 leaves remainder 5.
Answer: (d) 5
Example 7
What is the unit digit of 4^145?
(a) 6 (b) 4 (c) 7 (d) 5
- Only the unit digit 4 matters.
- Powers of 4 end in the cycle 4, 6 (length 2).
- 145 ÷ 2 leaves remainder 1 (a remainder of 0 means the last term of the cycle).
- So the unit digit is 4.
Answer: (b) 4
Example 8
What is the sum of the first 27 even numbers?
(a) 730 (b) 756 (c) 729 (d) 783
- Use the formula: sum = n(n+1) = 27×28 = 756.
Answer: (b) 756
- To find a remainder, reduce each number first – never multiply out large numbers.
- For unit digits, keep only the last digit of the base and the power mod 4.
- Least number to add to make N divisible by d = d − (N mod d); least to subtract = N mod d.
- Check options against divisibility rules to eliminate quickly.
- 1 is not a prime; 2 is the only even prime.
- A remainder of 0 in the unit-digit cycle means the last element of the cycle, not the first.
- Divisibility by 6 needs both 2 and 3 – a number divisible by 3 alone is not enough.
- Reading 'factors' as 'prime factors' – they are different counts.
Practice set
Attempt all 12 questions (about 90 seconds each), then check the worked solutions below.
Q1. What is the unit digit of 99^191?
(a) 9 (b) 11 (c) 1 (d) 10
Q2. What is the remainder when 567 × 852 × 204 is divided by 7?
(a) 1 (b) 0 (c) 6 (d) 3
Q3. What is the least number that must be added to 1866 to make it exactly divisible by 17?
(a) 13 (b) 5 (c) 4 (d) 16
Q4. What is the least number that must be added to 7367 to make it exactly divisible by 7?
(a) 6 (b) 4 (c) 3 (d) 5
Q5. How many factors (divisors) does 7546 have?
(a) 5 (b) 18 (c) 14 (d) 16
Q6. What is the unit digit of 4^163?
(a) 4 (b) 7 (c) 5 (d) 6
Q7. How many factors (divisors) does 7203 have?
(a) 12 (b) 8 (c) 5 (d) 10
Q8. How many factors (divisors) does 594 have?
(a) 18 (b) 14 (c) 16 (d) 5
Q9. What is the least number that must be added to 5135 to make it exactly divisible by 23?
(a) 6 (b) 17 (c) 18 (d) 22
Q10. What is the sum of the first 35 odd numbers?
(a) 1190 (b) 1347 (c) 1260 (d) 1225
Q11. What is the least number that must be added to 8100 to make it exactly divisible by 7?
(a) 7 (b) 8 (c) 6 (d) 1
Q12. How many factors (divisors) does 4312 have?
(a) 24 (b) 22 (c) 26 (d) 6
Q1 – (a) 9. Only the unit digit 9 matters. Powers of 9 end in the cycle 9, 1 (length 2). 191 ÷ 2 leaves remainder 1 (a remainder of 0 means the last term of the cycle). So the unit digit is 9.
Q2 – (b) 0. Replace each number by its remainder on division by 7: 567 → 0, 852 → 5, 204 → 1. 0 × 5 × 1 = 0. 0 ÷ 7 leaves remainder 0.
Q3 – (c) 4. 1866 ÷ 17 = 109 remainder 13. To reach the next multiple, add 17 − 13 = 4. Check: 1866 + 4 = 1870 = 17 × 110.
Q4 – (b) 4. 7367 ÷ 7 = 1052 remainder 3. To reach the next multiple, add 7 − 3 = 4. Check: 7367 + 4 = 7371 = 7 × 1053.
Q5 – (d) 16. Prime factorisation: 7546 = 2 × 7^3 × 11. Number of factors = product of (each power + 1). (1+1) × (3+1) × (1+1) = 16.
Q6 – (a) 4. Only the unit digit 4 matters. Powers of 4 end in the cycle 4, 6 (length 2). 163 ÷ 2 leaves remainder 1 (a remainder of 0 means the last term of the cycle). So the unit digit is 4.
Q7 – (d) 10. Prime factorisation: 7203 = 3 × 7^4. Number of factors = product of (each power + 1). (1+1) × (4+1) = 10.
Q8 – (c) 16. Prime factorisation: 594 = 2 × 3^3 × 11. Number of factors = product of (each power + 1). (1+1) × (3+1) × (1+1) = 16.
Q9 – (b) 17. 5135 ÷ 23 = 223 remainder 6. To reach the next multiple, add 23 − 6 = 17. Check: 5135 + 17 = 5152 = 23 × 224.
Q10 – (d) 1225. Use the formula: sum = n² = 35² = 1225.
Q11 – (c) 6. 8100 ÷ 7 = 1157 remainder 1. To reach the next multiple, add 7 − 1 = 6. Check: 8100 + 6 = 8106 = 7 × 1158.
Q12 – (a) 24. Prime factorisation: 4312 = 2^3 × 7^2 × 11. Number of factors = product of (each power + 1). (3+1) × (2+1) × (1+1) = 24.
- Remainder of product = product of remainders (then reduce)
- Unit digit cycles: 2,3,7,8 → 4; 4,9 → 2; 0,1,5,6 → 1
- Factors of a^p b^q = (p+1)(q+1)
- Σn = n(n+1)/2; Σodd = n²
