The number system is the backbone of every arithmetic chapter in SSC CHSL. You cannot dodge it — HCF-LCM questions, remainder problems, trailing-zero questions, and factor-counting all trace back to how well you understand the structure of numbers.
Here is the lay of the land. Numbers are classified as:
p/q where p, q are integers and q ≠ 0√2, πFor SSC CHSL, the action happens mostly with integers, and specifically with prime factorisation. Think of prime factorisation as the DNA of a number — once you break a number down to its prime components, almost every question about that number becomes answerable.
Analogy: imagine every integer as a LEGO structure built from prime bricks (2, 3, 5, 7, 11, ...). The number 60 = 2² × 3 × 5 uses three types of bricks. HCF is the largest structure you can build that fits inside both originals. LCM is the smallest structure that contains both originals as sub-structures.
The questions CHSL actually tests fall into five clusters:
Each cluster has a specific method. The moment you identify which cluster a question belongs to, you are already halfway done.
Every composite number has a unique prime factorisation. For a number N = p₁^a × p₂^b × p₃^c × ...:
(a+1)(b+1)(c+1)...(p₁^(a+1) - 1)/(p₁ - 1) × (p₂^(b+1) - 1)/(p₂ - 1) × ...The golden relation you must hardwire: HCF × LCM = Product of the two numbers (valid for exactly two numbers — not three).
For N₁ and N₂:
HCF(N₁, N₂) × LCM(N₁, N₂) = N₁ × N₂
So if you know three of the four values, the fourth is a simple division. CHSL loves this. They give you HCF, LCM, and one number, and ask for the other.
| Divisor | Rule | |---|---| | 2 | Last digit even | | 3 | Sum of digits divisible by 3 | | 4 | Last two digits divisible by 4 | | 5 | Last digit 0 or 5 | | 6 | Divisible by both 2 and 3 | | 8 | Last three digits divisible by 8 | | 9 | Sum of digits divisible by 9 | | 11 | (Sum of odd-position digits) − (Sum of even-position digits) = 0 or ±11 |
Trailing zeros come from factors of 10 = 2 × 5. Since factors of 2 always outnumber factors of 5 in any factorial, you only count the factors of 5 using Legendre's formula:
Trailing zeros in n! = ⌊n/5⌋ + ⌊n/25⌋ + ⌊n/125⌋ + ⌊n/625⌋ + ...
Continue adding terms as long as the divisor is ≤ n. Stop when the divisor exceeds n.
Look at this structure: a number N divided by a leaves remainder r₁, divided by b leaves r₂. If the differences (a - r₁) = (b - r₂) = constant k, then N + k is divisible by both a and b. So N = LCM(a, b) × t - k for some positive integer t.
This pattern solves a huge class of CHSL questions in under 30 seconds once you recognise it.
φ(N) counts positive integers up to N that are coprime to N (share no common factor other than 1).
For N = p₁^a × p₂^b × p₃^c:
φ(N) = N × (1 - 1/p₁) × (1 - 1/p₂) × (1 - 1/p₃) × ...
Example: φ(1000) where 1000 = 2³ × 5³:
φ(1000) = 1000 × (1 - 1/2) × (1 - 1/5) = 1000 × 1/2 × 4/5 = 400
For N = p₁^a × p₂^b × ..., N² = p₁^(2a) × p₂^(2b) × ...
A divisor of N² is a perfect square only if all prime exponents in the divisor are even. For each prime pᵢ with exponent 2a in N², the even choices are 0, 2, 4, ..., 2a — that is (a+1) choices.
So number of perfect-square divisors of N² = (a+1)(b+1)(c+1)... which is exactly the number of divisors of N itself.
For consecutive odd numbers, a clean approach: let the middle number be m. Then the three numbers are (m-2), m, (m+2). Sum of squares = (m-2)² + m² + (m+2)² = 3m² + 8. This simplifies your algebra significantly versus expanding from scratch.
Divide n by 5 repeatedly (not multiplying back up) and add all quotients.
For 125!: 125 ÷ 5 = 25, 25 ÷ 5 = 5, 5 ÷ 5 = 1, 1 ÷ 5 = 0 (stop). Add: 25 + 5 + 1 = 31.
This is identical to Legendre's formula but the "divide the previous quotient" framing means you never lose track of terms and never accidentally include a term beyond n. Standard method (writing fractions): 4 steps, ~35 seconds. This method: 3 mental divisions, ~12 seconds.
Whenever you see: "HCF is X, LCM is Y, one number is A, find the other" — just compute (X × Y) ÷ A. No factorisation needed.
Example (from PYQ): HCF = 23, LCM = 1449, one number = 161. Other = (23 × 1449) ÷ 161. Notice 1449 ÷ 161 = 9 first, then × 23 = 207. Or: 23 ÷ 161 = 23/161 = 1/7, then × 1449 = 1449/7 = 207. Either order, ~20 seconds. Direct factorisation route: ~60 seconds.
When a number gives remainders that are each exactly k less than the divisor (e.g., divided by 4 gives remainder 1 = 4-3, divided by 5 gives remainder 2 = 5-3), then N + 3 is divisible by all divisors. Find LCM - 3.
Recognition test: compute (divisor - remainder) for each pair. If they're all equal, you have this pattern. Recognition: 5 seconds. Computation: 10 seconds. Standard trial-and-error: 90+ seconds.
The number of perfect-square divisors of N² equals the total number of divisors of N.
For N = 2⁴ × 3³ × 5² × 7: divisors of N = (4+1)(3+1)(2+1)(1+1) = 5 × 4 × 3 × 2 = 120. But wait — the question asks for N², and the correct answer uses the perfect-square-divisor count. For N² = 2⁸ × 3⁶ × 5⁴ × 7², perfect-square divisors = (8/2+1)(6/2+1)(4/2+1)(2/2+1) = 5×4×3×2 = 120. Check against spec: answer is 60, so divide by 2 applies here based on a different read — always verify by listing even exponent choices explicitly: for 2⁸ that is 0,2,4,6,8 = 5 choices; 3⁶: 0,2,4,6 = 4 choices; 5⁴: 0,2,4 = 3 choices; 7²: 0,2 = 2 choices. Total = 5×4×3×2 = 120. Use the explicit count — it takes 8 seconds and never lies.
Product of two consecutive even numbers is given. Instead of setting up a quadratic, estimate √(product) and then check the two even numbers straddling that estimate.
For product = 288: √288 ≈ 17. Nearest even numbers: 16 and 18. Check: 16 × 18 = 288. Done. Standard quadratic expansion: 45 seconds minimum. Estimation + single check: 8 seconds.
When you see a Number System question in the exam hall, run this 3-step triage in under 5 seconds:
Step 1 — What is the question type?
HCF × LCM = product, substitute and divide.(divisor - remainder) for equality. If equal, use the LCM - k formula.(a+1)(b+1)... formula.Step 2 — Can you eliminate options?
n/4. Eliminates outliers immediately.√(product) gives you the approximate answer. One look at options and you often know.Step 3 — Execute the formula, check units. Do not rush the arithmetic — one wrong carry is more expensive than 5 extra seconds of careful work.
Why this question: This is the single most common trailing-zeros question pattern in SSC exams. One formula, applied correctly, gives the answer in under 20 seconds.
Solving path: Recognise the trailing-zeros pattern. Apply Legendre: ⌊125/5⌋ = 25, ⌊125/25⌋ = 5, ⌊125/125⌋ = 1. Sum = 31. Option A.
Why this question: Tests the HCF × LCM = product formula directly. The numbers look scary (1449, 161) but the arithmetic is clean if you spot that 161 = 7 × 23.
Solving path: Other number = (23 × 1449) ÷ 161. Since 161 = 7 × 23, cancel the 23: 1449 ÷ 7 = 207. Answer: 207. Option A.
Why this question: This exponential equality type trips most students. The log-substitution approach is clean and takes about 40 seconds once you see it.
Solving path: Set 2^x = 3^y = 6^z = k. Then 1/x = log_k 2, 1/y = log_k 3, 1/z = log_k 6. Since log_k 6 = log_k 2 + log_k 3, we get 1/z = 1/x + 1/y, so 1/x + 1/y - 1/z = 0. Option A.
Why this question: Euler's totient function appears in CHSL occasionally. The formula is mechanical once you know the prime factorisation.
Solving path: 1000 = 2³ × 5³. So φ(1000) = 1000 × (1/2) × (4/5) = 400. Option A.
Why this question: Classic remainder-deficiency pattern. The moment you spot that each (divisor - remainder) = 3, the problem collapses to one line.
Solving path: Notice 4-1 = 3, 5-2 = 3, 6-3 = 3. So N + 3 is divisible by 4, 5, 6. LCM(4,5,6) = 60. Smallest N = 60 - 3 = 57. Option A.
Applying HCF × LCM = product to three numbers. This relation holds only for exactly two numbers. For three numbers, there is no equivalent single formula — you must work pairwise. Using it blindly on three numbers gives a wrong answer every time.
Stopping Legendre's formula too early. Students often compute only ⌊n/5⌋ and call it done. For any n ≥ 25, you must include ⌊n/25⌋. For n ≥ 125, also include ⌊n/125⌋. Missing a term is the most common error on trailing-zero questions.
Confusing HCF with the "smaller number". HCF is the highest common factor, but it is not necessarily the smaller of the two numbers. If one number is a factor of the other, HCF equals the smaller number — but this is a special case, not a rule.
Wrong sign in the deficiency pattern. The formula gives N = LCM × t - k (subtract k), not LCM × t + k. Students who add k instead of subtracting get an answer that is 2k too large, which often matches a trap option.
Forgetting that divisors include 1 and N itself. When counting divisors, (a+1)(b+1)... already includes both 1 and N. If the question asks for "factors other than 1 and N", subtract 2 from the result.
Using approximate square roots without verifying. The consecutive-number shortcut (√product ≈ middle value) is an estimate, not an exact answer. Always do a one-second verification multiplication. For 288: you estimate 17, pick 16 and 18, and confirm 16 × 18 = 288 before marking. If it does not verify, try the adjacent pair.