Numbers are the bedrock of CSAT arithmetic — and the UPSC question-setters know it. Every year, two to four questions in the Prelims Paper-II touch some corner of the number system: a divisibility condition, a remainder trap, an LCM-HCF identity. Get comfortable here and you've secured easy marks.
Here's the mental model that unifies everything: every integer lives on a number line, and it leaves a "fingerprint" when divided by any fixed number. That fingerprint is the remainder — and most number-system questions are really about tracking remainders through arithmetic operations.
Think of it like a clock. The clock face has 12 positions (0 through 11). Once you pass 12, you "wrap around." If it's 10 o'clock now and you add 5 hours, you don't get 15 — you get 3. That wrapping-around behavior is modular arithmetic, and divisibility rules are just shortcuts to check whether the remainder is zero.
The number system you need for CSAT spans:
1, 2, 3, ...0, 1, 2, 3, ......, −2, −1, 0, 1, 2, ...p/q where q ≠ 0For CSAT, the action happens almost entirely in positive integers. The three engines driving most questions are:
N = Divisor × Quotient + Remainder — the single most-tested identity in this chapterA useful analogy: LCM is the smallest "meeting point" of two recurring events (say, two buses running on different cycles — when do they depart together?), while HCF is the largest common "unit" you can use to measure both quantities without a remainder.
You do not have time to divide in the exam hall. These rules let you filter in seconds:
| Divisor | Rule |
|---------|------|
| 2 | Last digit is even |
| 3 | Sum of digits divisible by 3 |
| 4 | Last two digits divisible by 4 |
| 5 | Last digit is 0 or 5 |
| 6 | Divisible by both 2 and 3 |
| 7 | No clean one-liner; use repeated subtraction of 2 × last digit from the rest |
| 8 | Last three digits divisible by 8 |
| 9 | Sum of digits divisible by 9 |
| 10 | Last digit is 0 |
| 11 | Alternating digit sum (odd positions minus even positions) divisible by 11 |
| 12 | Divisible by both 3 and 4 |
Key composite rule: divisibility by a composite number n works via co-prime factors. For n = 12: check divisibility by 4 and by 3 separately (4 and 3 are co-prime). Do not check by 2 and 6 — they are not co-prime and will give false positives.
This is the skeleton of almost every "remainder" problem:
N = dividend (the number being divided)D = divisorQ = quotientR = remainderIf a question tells you the quotient and remainder, reconstruct N immediately. Do not try to guess — just compute.
For any two positive integers a and b:
This is the single most-tested identity in LCM-HCF questions. Know it cold. If three of the four values are given, the fourth drops out in one step.
How to find HCF: Use the Euclidean algorithm — repeatedly divide the larger by the smaller and take remainders until you hit zero. The last non-zero remainder is the HCF.
Example: HCF(360, 60). Divide 360 by 60: remainder 0. So HCF = 60.
Example: HCF(72, 60). Divide 72 by 60: remainder 12. Divide 60 by 12: remainder 0. So HCF = 12.
How to find LCM: Prime factorize both numbers, take the highest power of each prime that appears.
LCM(8, 9): 8 = 2³, 9 = 3². LCM = 2³ × 3² = 72.
When a question says "divisible by both X and Y," it means divisible by LCM(X, Y). When it says "not divisible by Z," you count using inclusion-exclusion.
Count of multiples of k in range [a, b]:
Or equivalently, find the smallest multiple of k that is ≥ a, find the largest that is ≤ b, then use:
When you have conditions like "divisible by 6 AND leaves remainder 4 when divided by 5," you're solving a system of congruences. CSAT doesn't expect you to name CRT, but you need the method:
N = 6k6k ≡ 4 (mod 5) → k ≡ 4 (mod 5) → k = 5m + 4N = 6(5m + 4) = 30m + 24The combined cycle is lcm(6, 5) = 30. General form: N = 30m + 24. Now plug in values of m until you satisfy the range condition.
The number of factors of N = p₁^a × p₂^b × p₃^c × ... is (a+1)(b+1)(c+1)....
This sometimes appears in CSAT as "how many divisors does N have?" or embedded in a counting problem. Quick check: 72 = 2³ × 3² has (3+1)(2+1) = 12 factors.
When checking divisibility by a composite number, always split into co-prime factors. For 12: check 4 AND 3. For 18: check 9 AND 2. For 24: check 8 AND 3. The shortcut is that checking two co-prime factors is always valid, while non-co-prime pairs (like 2 and 6 for 12) are not.
Worked example: Is 1836 divisible by 12? Sum of digits = 1+8+3+6 = 18 (divisible by 3, check). Last two digits = 36 = 9×4 (divisible by 4, check). So yes. Standard approach would require long division: ~45 seconds. This check: ~8 seconds.
When a question gives LCM, HCF, and one number, you do not need to factorize anything. Directly apply: other number = (LCM × HCF) / given number. One multiplication, one division — done.
Worked example: LCM = 360, HCF = 12, one number = 60. Other = (360 × 12)/60 = 4320/60 = 72. Standard method (factorize both numbers and rebuild): 4–5 steps. This method: 2 steps, under 10 seconds.
To count multiples of k from a to b (inclusive): find the first multiple ≥ a (call it F), find the last multiple ≤ b (call it L), then count = (L − F)/k + 1.
Worked example: Multiples of 72 from 1000 to 9999. First: ceil(1000/72)×72 = 14×72 = 1008. Last: floor(9999/72)×72 = 138×72 = 9936. Count = (9936−1008)/72 + 1 = 124 + 1 = 125. Attempting to list them: impossible in exam time. This formula: ~30 seconds.
If a question tells you "divided by D gives quotient Q and remainder R," your very first move is always N = D×Q + R. Write the number down. Every subsequent step becomes trivial.
Worked example: "Divided by 13, quotient 15, remainder 8." N = 13×15 + 8 = 195 + 8 = 203. Now any follow-up question (divide by 12, by 7, by 11) is just one more division. Without this, students try to reverse-engineer from options: 4 attempts × 30 seconds each = 2 minutes wasted.
When you have two remainder conditions with moduli m₁ and m₂, the combined solution repeats every lcm(m₁, m₂). Find ONE solution that satisfies both conditions, then all solutions are of the form: N = lcm(m₁, m₂)×t + N₀.
Worked example: N ≡ 0 (mod 6) and N ≡ 4 (mod 5). lcm(6,5) = 30. First solution: N = 24 (check: 24/6 = 4 rem 0, 24/5 = 4 rem 4). General form: N = 30t + 24. For N > 100: t = 3 gives N = 114. Trial-and-error through multiples of 6: testing 102, 108, 114... — 3 checks. This method: 1 formula, answer in 20 seconds.
In the exam hall, read the question and immediately classify it:
Step 1 — What type is it?
Step 2 — Is there an exclusion condition?
Step 3 — Range check
Step 4 — Verify with options
One rule that holds across all variants: never start computing before you've identified which of these four types the question is. Twenty seconds of classification saves two minutes of wrong-track computation.
Why this question: This is the canonical "count integers satisfying multiple divisibility conditions with an exclusion" — the most structurally complete question type in this chapter.
Solving path: Identify that "divisible by both 8 and 9" means divisible by LCM(8, 9) = 72. Count four-digit multiples of 72: smallest is 1008 (14×72), largest is 9936 (138×72). Count = (9936 − 1008)/72 + 1 = 125. Now exclude those also divisible by 5: LCM(72, 5) = 360. Smallest four-digit multiple of 360 is 1080, largest is 9720. Count = (9720 − 1080)/360 + 1 = 25. Final answer = 125 − 25 = 100. The key decision point: recognizing that "divisible by both 8 and 9" requires LCM, not just multiplication.
Why this question: Tests whether you can apply the division algorithm in forward and then backward direction. Students who don't reconstruct N get stuck.
Solving path: The moment you see "quotient" and "remainder," write N = D×Q + R. Here: N = 13×15 + 8 = 203. Now the question becomes simply: what is 203 mod 12? 203 = 12×16 + 11. Remainder = 11. The trap is overthinking this — it's two multiplications and one division, nothing more.
Why this question: The LCM × HCF = product identity in its purest exam form. One step if you know the identity; five steps if you don't.
Solving path: Apply LCM × HCF = a × b directly. 360 × 12 = 60 × b. 4320 = 60b. b = 72. Verify: HCF(60, 72) — use Euclidean algorithm: 72 = 1×60 + 12, 60 = 5×12 + 0. HCF = 12. LCM = (60 × 72)/12 = 360. Both checks pass. Verification takes 15 extra seconds and eliminates any arithmetic error.
Why this question: Simultaneous congruence conditions — the CRT approach without needing to name it. Tests whether you can merge two modular conditions into a single general form.
Solving path: Two conditions: N ≡ 0 (mod 6) and N ≡ 4 (mod 5). Write N = 6k. Substitute: 6k ≡ 4 (mod 5). Since 6 ≡ 1 (mod 5), this gives k ≡ 4 (mod 5), so k = 5m + 4. Back-substitute: N = 6(5m + 4) = 30m + 24. For N > 100: 30m + 24 > 100, so m > 2.53, meaning m = 3 gives N = 114. Quick verify: 114/6 = 19 (no remainder), 114/5 = 22 remainder 4. Both pass. The elimination trap: option D (104) — check 104/6 = 17 remainder 2. Fails immediately.
Splitting composite divisors into non-co-prime factors: Checking "divisible by 12" using divisors 2 and 6 is wrong — they share the factor 2. Always split into co-prime pairs (for 12: use 3 and 4).
Off-by-one in range counting: Forgetting the "+1" in the formula (Last − First)/k + 1 is extremely common. If you get an answer that's one less than an option, this is your error.
Using LCM instead of product in HCF-LCM identity: The identity is LCM × HCF = a × b (the product of the two numbers), not LCM × HCF = LCM or any other form. Write it out explicitly before substituting.
Not reconstructing N before proceeding in division algorithm questions: Students sometimes try to work backwards from options when the quotient and remainder are given. Reconstruct N first — it takes five seconds and eliminates all ambiguity.
Assuming "divisible by both" means divisible by the product: LCM(8, 9) = 72, not 8×9 = 72 (they happen to be equal here because 8 and 9 are co-prime). But LCM(6, 9) = 18, not 54. Always compute LCM properly.
Stopping after finding the general form in CRT problems without applying the range condition: The general form N = 30m + 24 is not the answer — you must then find which value of m satisfies the stated range constraint. This step is skipped under time pressure, leading to answers that are valid solutions to the congruence but outside the required range.