CSAT· Prelims
Number System for CSAT
The number system is the backbone of CSAT quant: divisibility rules, primes, factors, HCF and LCM, remainders and unit digits explained with worked examples, common traps and practice questions.
Number system is the study of how numbers are classified, broken into factors, and divided with remainders. It covers divisibility rules, primes, factors, HCF and LCM, remainders and unit digits. It matters for CSAT because Paper II is a qualifying paper that needs a 33 percent score, and number-system questions are among the most frequent and most scoring questions on it: they test pure logic, not memory, and each one yields to a standard trick.
Divisibility rules
A divisibility rule is a short test that tells you whether a number divides another number without actually doing the division. The rules most asked in CSAT are: a number is divisible by 2 if its last digit is even; by 3 if the sum of its digits is divisible by 3; by 4 if the last two digits form a number divisible by 4; by 5 if it ends in 0 or 5; by 8 if the last three digits form a number divisible by 8; by 9 if the sum of its digits is divisible by 9; by 10 if it ends in 0; and by 11 if the difference between the sum of digits in odd positions and the sum of digits in even positions is 0 or divisible by 11.
Example 1. Check whether 5,148 is divisible by 3, 4 and 9. The digit sum is 5+1+4+8 = 18, which is divisible by both 3 and 9, so the number is divisible by 3 and by 9. The last two digits are 48, and 48 is divisible by 4, so the number is divisible by 4 as well. Hence 5,148 is divisible by all three.
Prime numbers and co-primes
A prime number is a whole number greater than 1 that has exactly two divisors, 1 and itself: 2, 3, 5, 7, 11, 13 and so on. A composite number is a whole number greater than 1 that is not prime, meaning it has at least one more divisor: 4, 6, 8, 9, 10 and so on. Note that 1 is neither prime nor composite. Two numbers are co-prime if they share no common factor other than 1, even when neither number is itself prime; for instance 8 and 15 are co-prime because the only factor common to both is 1.
Example 2. Which of the following are co-prime pairs: (14, 21), (8, 15), (16, 25)? The pair 14 and 21 share the factor 7, so they are not co-prime. The pairs 8 and 15 share only 1, and 16 and 25 share only 1, so both are co-prime pairs. Hence the answer is (8, 15) and (16, 25).
Number of factors and factor rules
The number of factors of a number is found from its prime factorisation. If a number is written as p^a times q^b times r^c where p, q, r are distinct primes, then the count of its positive divisors is (a+1)(b+1)(c+1). This works because for each prime you may choose its exponent as anything from 0 up to its full power. For example, 72 = 2^3 x 3^2, so it has (3+1)(2+1) = 12 factors.
Example 3. How many factors does 180 have? Write 180 = 18 x 10 = 2 x 3^2 x 2 x 5 = 2^2 x 3^2 x 5^1. The factor count is (2+1)(2+1)(1+1) = 3 x 3 x 2 = 18. So 180 has 18 positive factors.
HCF and LCM
The HCF (highest common factor, also called GCD) of two numbers is the largest number that divides both of them. The LCM (lowest common multiple) of two numbers is the smallest number that is a multiple of both. The key link is: HCF x LCM = product of the two numbers. So if you know one of them, the other follows at once.
Example 4. The HCF of two numbers is 12 and their product is 3,600. What is their LCM? Using HCF x LCM = product, we get 12 x LCM = 3,600, so LCM = 3,600 / 12 = 300. Hence the LCM is 300.
Three speed rules for HCF and LCM: (1) the HCF of two co-prime numbers is 1 and their LCM is their product; (2) if one number divides the other, the smaller is the HCF and the larger is the LCM; (3) to find the greatest number that leaves the same remainder r when dividing two numbers a and b, take the HCF of (a-r) and (b-r).
Remainders and unit digits
A remainder is what is left over after dividing one number by another. Two shortcut ideas dominate CSAT questions. First, cyclicity of unit digits: the last digits of powers of a number repeat in a cycle, so the unit digit of x^y depends only on y modulo the cycle length. Second, the remainder of a product is the remainder of the product of remainders: when you divide a x b by n, you may first replace a and b by their remainders on division by n.
Example 5. Find the unit digit of 7^2025. Powers of 7 end in 7, 9, 3, 1 and then repeat, a cycle of 4. Divide 2025 by 4: 2025 = 4 x 506 + 1, remainder 1. The first position in the cycle is 7, so the unit digit is 7.
Example 6. Find the remainder when 43 x 67 is divided by 10. 43 leaves remainder 3 and 67 leaves remainder 7 when divided by 10. Their product 3 x 7 = 21 leaves remainder 1 on division by 10. So the remainder is 1.
BODMAS and surds
BODMAS is the order of operations: Brackets, Orders (powers and roots), Division, Multiplication, Addition, Subtraction, evaluated strictly in that order. A surd is an irrational root of a rational number, such as the square root of 2 or the cube root of 3, written with a radical sign. A recurring decimal is a decimal whose digits repeat forever, like 0.333..., and every recurring decimal equals a rational fraction: for example, 0.666... = 2/3 and 0.1666... = 1/6.
Example 7. Simplify 24 / 4(2 + 4). First the bracket: 2 + 4 = 6. Then division: 24 / 4 = 6. Then multiplication: 6 x 6 = 36. So the value is 36.
Common traps and speed tips
- Trap 1: 1 is not prime. When asked for the number of primes below 20, do not count 1. The primes below 20 are 2, 3, 5, 7, 11, 13, 17 and 19, which is 8, not 9.
- Trap 2: even numbers divisible by 4. Every number divisible by 4 is even, but not every even number is divisible by 4. Divisibility by 2 does not imply divisibility by 4.
- Trap 3: co-prime does not mean prime. 8 and 15 are co-prime though neither is prime. Read the option words exactly.
- Tip 1: memorise key fraction decimals. Knowing 1/2, 1/3, 1/4, 1/6, 1/7, 1/8, 1/9, 1/11 as decimals saves minutes in remainder and percentage questions.
- Tip 2: write the prime factorisation first. Factor-count, HCF, LCM and divisibility questions all become one-line questions once the number is written as powers of primes.
The digit sum of a number is 42. Which of the following statements is definitely true?
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Answer: (C) The digit sum 42 is divisible by 3 but not by 9 (4+2=6), so the number is divisible by 3 but not necessarily by 9. Divisibility by 6 or 18 would need an even number, which is not guaranteed.
How many of the following are prime numbers: 51, 53, 57, 59, 61?
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Answer: (B) 51 = 3 x 17 and 57 = 3 x 19 are composites; 53, 59 and 61 have no divisors other than 1 and themselves, so the primes among them are 53, 59 and 61, a total of 3.
The number of factors of 96 is:
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Answer: (B) 96 = 2^5 x 3^1, so the factor count is (5+1)(1+1) = 6 x 2 = 12.
The HCF of two numbers is 15 and their LCM is 450. If one number is 75, the other number is:
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Answer: (A) Using HCF x LCM = product of the two numbers: 15 x 450 = 75 x other number, so the other number = (15 x 450) / 75 = 90.
The unit digit of 9^203 is:
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Answer: (A) Powers of 9 alternate between unit digits 9 and 1: 9^1 = 9, 9^2 = 81 (ends 1). Odd powers end in 9, and 203 is odd, so the unit digit is 9.
The remainder when 17^4 is divided by 16 is:
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Answer: (A) 17 leaves remainder 1 when divided by 16, so 17^4 leaves remainder 1^4 = 1. Hence the remainder is 1.
Find the value of (36 / 6) x (2 + 5) - 4^2.
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Answer: (A) BODMAS: bracket first, (2+5) = 7; then 36/6 = 6; then 6 x 7 = 42; then 4^2 = 16; finally 42 - 16 = 26.
Answer key
- Q1 - (c). The digit sum 42 is divisible by 3 but not by 9 (4+2=6), so the number is divisible by 3 but not necessarily by 9. Divisibility by 6 or 18 would need an even number, which is not guaranteed.
- Q2 - (b). 51 = 3 x 17 and 57 = 3 x 19 are composites; 53, 59 and 61 have no divisors other than 1 and themselves, so the primes among them are 53, 59 and 61, a total of 3.
- Q3 - (b). 96 = 2^5 x 3^1, so the factor count is (5+1)(1+1) = 6 x 2 = 12.
- Q4 - (a). Using HCF x LCM = product of the two numbers: 15 x 450 = 75 x other number, so the other number = (15 x 450) / 75 = 90.
- Q5 - (a). Powers of 9 alternate between unit digits 9 and 1: 9^1 = 9, 9^2 = 81 (ends 1). Odd powers end in 9, and 203 is odd, so the unit digit is 9.
- Q6 - (a). 17 leaves remainder 1 when divided by 16, so 17^4 leaves remainder 1^4 = 1. Hence the remainder is 1.
- Q7 - (a). BODMAS: bracket first, (2+5) = 7; then 36/6 = 6; then 6 x 7 = 42; then 4^2 = 16; finally 42 - 16 = 26.
Can a number be divisible by both 4 and 9 but not by 36?
No. If a number is divisible by both 4 and 9, and 4 and 9 are co-prime (they share no common factor other than 1), then it is divisible by their product 36. This is a general rule: when two divisors are co-prime, divisibility by both implies divisibility by the product.
Why is 2 the only even prime number?
A prime number has exactly two divisors. Any even number greater than 2 is divisible by 1, 2 and itself, which is at least three divisors, so it cannot be prime. The number 2 itself has exactly the divisors 1 and 2, so it qualifies.
How do you find the greatest number that divides 245 and 315 leaving the same remainder?
First find the difference: 315 - 245 = 70. The required number must divide this difference, because both numbers leave the same remainder on it. Take the factors of 70 (1, 2, 5, 7, 10, 14, 35, 70) and check which one actually leaves the same remainder; 70 gives 245 = 70 x 3 + 35 and 315 = 70 x 4 + 35, so the answer is 70.
Is 0 an even number?
Yes. An even number is one that is exactly divisible by 2, and 0 divided by 2 leaves remainder 0. However, 0 is not a prime number, not a composite number, and it is not positive.