Number System

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SSC CGL Number System Notes and Free Practice

Free SSC CGL Number System notes: divisibility, HCF and LCM, remainders, unit digits, factors and decimals. Learn with 12 worked examples and a 15-question concept check.

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Number System notes: learn, check, revise

Start with divisibility and prime factorization, then move to HCF, LCM, remainders and powers. Each section below contains a rule, a worked example and a mistake to avoid. These are independently written learning materials, not an official SSC paper.

Content updated 30 October 2026.

Try the 15-question concept check

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Integers, prime numbers and co-primes

Integers include negative whole numbers, zero and positive whole numbers. A prime is a positive integer greater than 1 with exactly two positive divisors. The number 1 is neither prime nor composite; 2 is the only even prime. Two positive integers are co-prime when their HCF is 1; neither has to be prime.

Rule: To test whether n > 1 is prime, try prime divisors no larger than the square root of n.

Worked example

Is 143 prime? Are 14 and 25 co-prime?

Show solution

The square root of 143 lies between 11 and 12. Check 2, 3, 5, 7 and 11: 143 = 11 × 13, so it is composite. Meanwhile 14 = 2 × 7 and 25 = 5² share no prime factor, so their HCF is 1.

Common mistake: Odd does not mean prime. Co-prime does not mean both numbers are prime.

Divisibility tests and missing digits

For 2, check the last digit; for 3 or 9, check the digit sum; for 4, check the last two digits; for 8, check the last three digits. A number divisible by both 2 and 3 is divisible by 6. For 11, subtract the sums of digits in alternating positions: the difference must be a multiple of 11, including zero.

Rule: For divisibility by 12, test 3 and 4. Testing 2 and 6 is insufficient because their LCM is only 6.

Worked example

Which digit d makes 52d4 divisible by 12?

Show solution

For divisibility by 3, 5 + 2 + d + 4 = 11 + d must be a multiple of 3, so d is 1, 4 or 7. For divisibility by 4, the ending d4 must be a multiple of 4. Of 14, 44 and 74, only 44 works. Thus d = 4 and 5244 = 12 × 437.

Common mistake: A digit is an integer from 0 to 9. A leading digit cannot be zero.

HCF: largest equal groups and Euclid's algorithm

The highest common factor is the largest positive integer that divides every given number exactly. Use it when dividing quantities into the largest possible equal pieces without leftovers. For large numbers, repeatedly replace the pair by the divisor and remainder until the remainder is zero.

Rule: For prime factorizations, keep only common primes, using their smallest exponents.

Worked example

Find the HCF of 391 and 221.

Show solution

391 = 221 × 1 + 170; 221 = 170 × 1 + 51; 170 = 51 × 3 + 17; 51 = 17 × 3. The last non-zero remainder is 17, so HCF = 17. Check: 391 = 17 × 23 and 221 = 17 × 13.

Common mistake: The HCF cannot exceed the smallest positive input. A common factor is not necessarily the highest one.

LCM: repeating events and common multiples

The least common multiple is the smallest positive number divisible by all the inputs. It gives the first common repeat when events start together and repeat at fixed integer intervals. In prime factorization, retain every prime that occurs and use its largest exponent.

Rule: For two positive integers a and b: HCF(a,b) × LCM(a,b) = a × b.

Worked example

Three signals flash together at noon and then every 14, 21 and 30 seconds. When do they next flash together?

Show solution

14 = 2 × 7, 21 = 3 × 7 and 30 = 2 × 3 × 5. Their LCM is 2 × 3 × 5 × 7 = 210 seconds, or 3 minutes 30 seconds. The next common flash is 12:03:30.

Common mistake: The product identity is for two numbers. It cannot be applied unchanged to three numbers.

Finding a divisor from a common remainder

If division of N by d leaves remainder r, then N − r is divisible by d. Subtract the common remainder from each number and take the HCF to find the greatest possible divisor. Finally check that the proposed remainder is smaller than the divisor.

Rule: For a positive divisor d, the remainder satisfies 0 ≤ r < d.

Worked example

Find the greatest divisor that leaves remainder 7 when dividing both 127 and 187.

Show solution

The divisor must divide 120 and 180. HCF(120, 180) = 60. Since 7 < 60, it is valid: 127 = 60 × 2 + 7 and 187 = 60 × 3 + 7.

Common mistake: Taking the HCF of the original numbers ignores the remainder. A candidate divisor no larger than the remainder is invalid.

Remainders of sums, products and differences

Reduce each term modulo the same positive divisor before adding or multiplying. You can also subtract remainders, but a negative result must be adjusted by a multiple of the divisor to obtain the standard non-negative remainder.

Rule: N = dq + r, where q is an integer and 0 ≤ r < d. Reduce the final result again if necessary.

Worked example

A leaves remainder 2 and B leaves remainder 8 on division by 13. What remainder does A − B leave?

Show solution

Write A = 13m + 2 and B = 13n + 8. Then A − B = 13(m − n) − 6 = 13(m − n − 1) + 7. The remainder is 7, not −6.

Common mistake: You cannot generally divide remainders as ordinary fractions. Addition and multiplication rules do not justify that shortcut.

Unit digits and cycles of powers

For a unit digit, work modulo 10 and list successive powers until the pattern repeats. For another divisor, find the cycle using that divisor. The relevant cycle length may be 1, 2, 4 or something else; do not assume every cycle has four terms.

Rule: For a cycle of length k and a positive exponent, remainder zero on division by k selects the last term of the cycle.

Worked example

Find the unit digit of 7^38.

Show solution

The unit digits of positive powers of 7 repeat as 7, 9, 3, 1. Since 38 leaves remainder 2 on division by 4, choose the second term, 9. Computing the full power is unnecessary.

Common mistake: An exponent divisible by the cycle length does not give a unit digit of zero. Also, a non-zero base raised to exponent zero equals 1.

Counting positive factors

If N = p^a × q^b for distinct primes p and q, each divisor chooses an exponent from 0 to a for p and from 0 to b for q. These choices are independent. Multiply the counts, including the zero-exponent choices, to count all positive divisors.

Rule: If N = p₁^a₁ × ... × pₖ^aₖ, its number of positive factors is (a₁ + 1) × ... × (aₖ + 1).

Worked example

How many positive factors of 600 are odd?

Show solution

600 = 2³ × 3 × 5². An odd factor must use exponent zero for 2. There are 2 choices for the exponent of 3 and 3 for the exponent of 5, giving 2 × 3 = 6 odd factors: 1, 3, 5, 15, 25 and 75.

Common mistake: The formula counts positive factors, including 1 and N. Counting prime factors is a different task.

Smallest multiplier for a perfect square or cube

A positive integer is a perfect square exactly when every prime exponent in its factorization is even. For a perfect cube, every exponent must be a multiple of 3. To find the smallest multiplier, supply only the missing prime powers.

Rule: For a square, an odd exponent needs one more copy of that prime. For a cube, raise each exponent to the next multiple of 3.

Worked example

What is the smallest positive integer by which 648 must be multiplied to make a perfect cube?

Show solution

648 = 2³ × 3⁴. The exponent of 2 already meets the cube condition. The exponent of 3 needs to increase from 4 to 6, so multiply by 3² = 9. Check: 648 × 9 = 5832 = 18³.

Common mistake: Multiplying and dividing are different questions. Do not add a prime that is already at a valid exponent.

Trailing zeros in a factorial

A trailing zero comes from a factor of 10 = 2 × 5. In n!, there are at least as many factors of 2 as factors of 5, so count the fives. Multiples of 25 contribute an extra five, multiples of 125 contribute another, and so on.

Rule: Zeros in n! = floor(n/5) + floor(n/25) + floor(n/125) + ...; stop when a term is zero.

Worked example

How many trailing zeros does 82! have?

Show solution

There are floor(82/5) = 16 contributions from multiples of 5 and floor(82/25) = 3 extra contributions from 25, 50 and 75. The next term is zero. Total = 16 + 3 = 19.

Common mistake: This counts trailing zeros, not every zero anywhere in the decimal expansion. For a general product, compare both the number of twos and fives.

Terminating and recurring decimals

First reduce a rational number to lowest terms. Its decimal terminates exactly when the positive denominator contains no prime factors other than 2 and 5. If the denominator is 2^a × 5^b, the shortest terminating form needs max(a,b) decimal places. A repeating block can be converted to a fraction by multiplying and subtracting.

Rule: If x = 0.454545... then 100x − x = 45, so x = 45/99 = 5/11.

Worked example

Does 39/156 terminate? If it does, how many decimal places are needed?

Show solution

Divide numerator and denominator by 39 to obtain 1/4. The denominator is 2², so the decimal terminates in two places: 0.25. Testing the unreduced denominator would incorrectly retain a factor of 13.

Common mistake: A finite approximation on a calculator does not prove that a decimal terminates. Reduce the fraction first.

Reconstructing numbers from HCF and LCM

If two positive integers have HCF h, write them as hu and hv, where u and v are co-prime. Their LCM is huv. Therefore uv = LCM/HCF. Enumerate factor pairs of that quotient and reject pairs with a common factor greater than 1.

Rule: For unordered pairs, count (u,v) and (v,u) once. LCM must be a multiple of HCF.

Worked example

Two positive integers have HCF 9 and LCM 108. Which unordered pairs are possible?

Show solution

uv = 108/9 = 12. The factor pairs are (1,12), (2,6) and (3,4). Reject (2,6), since its HCF is 2. Multiplying the valid pairs by 9 gives (9,108) and (27,36). Both have HCF 9 and LCM 108.

Common mistake: Using every factor pair gives extra answers whose HCF is larger than the one specified.

How to use your practice results

First solve without a stopwatch and write down the rule used. After submission, identify whether each error came from the concept, arithmetic or wording. Use the revision links on the test page to revisit that concept, then explain the solution without looking. Practise a different set before repeating the same questions.

Number System concept check More HCF and LCM practice Next: simplification practice

Practice Test 1