Think of numbers as a family tree. At the base are the counting numbers you learned as a child — 1, 2, 3, 4... These are called natural numbers (प्राकृतिक संख्याएं). Add zero to this family and you get whole numbers (पूर्ण संख्याएं): 0, 1, 2, 3...
Now extend the tree in the negative direction — include -1, -2, -3 and so on alongside the positives and zero — and you have integers (पूर्णांक): ...-3, -2, -1, 0, 1, 2, 3...
Within natural numbers, there's an important sub-family split: numbers that are "gatekeepers" — divisible only by 1 and themselves — are called prime numbers (अभाज्य संख्याएं). Numbers with more than two factors are composite numbers (भाज्य संख्याएं). The number 1 belongs to neither group — that fact trips up a surprising number of candidates.
Here's a clean analogy. Imagine a town with houses numbered from 1 onward. The prime-number houses have only two keys: the master key (1) and their own key. Composite-number houses have extra keys — meaning more people (factors) can open them. House 1 is a special case: the administration decided it doesn't fit either category.
Why does the UP Police paper care about this? Because questions built around number types — "which of these is prime?", "find the smallest n-digit number divisible by x", "what multiplies 1200 into a perfect square?" — all test whether you've understood this family tree clearly, not just memorized it. A blurry understanding gets you misled by attractive wrong options every single time.
Quick reference table:
| Set | Includes | Example |
|---|---|---|
| Natural Numbers (N) | 1, 2, 3, ... | 7, 45, 100 |
| Whole Numbers (W) | 0, 1, 2, 3, ... | 0, 7, 100 |
| Integers (Z) | ...-2, -1, 0, 1, 2... | -5, 0, 12 |
| Primes | p > 1, factors only 1 and p | 2, 3, 5, 7, 11 |
| Composite | n > 1, more than 2 factors | 4, 6, 9, 15 |
Natural vs. Whole: The only difference is zero. Whole numbers include 0; natural numbers do not. In most UP Police questions, "which set does 0 belong to?" is a trap — 0 is whole but not natural.
Integers: Every natural number is an integer. Every whole number is an integer. But -7 is an integer without being natural or whole. Keep the subset relationship clear: N ⊂ W ⊂ Z.
A prime is any integer greater than 1 that has exactly two distinct positive factors: 1 and itself.
Common primes to memorize up to 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. That's 15 primes.
2 is the only even prime. Every other even number is divisible by 2, giving it at least three factors (1, 2, itself), making it composite.
Quick primality check for any number N: Divide N by all primes up to √N. If none divide evenly, N is prime.
Example — is 97 prime? √97 ≈ 9.8. Check primes up to 9: 2, 3, 5, 7. 97 ÷ 2 = no, 97 ÷ 3 = no, 97 ÷ 5 = no, 97 ÷ 7 = no. So 97 is prime.
Any 3-digit number ABC repeated to form ABCABC satisfies:
ABCABC = ABC × 1001
And 1001 = 7 × 11 × 13, so LCM(7, 11, 13) = 1001.
Therefore, ABCABC ÷ LCM(7,11,13) = ABC × 1001 ÷ 1001 = ABC.
This is not a coincidence — it's a direct property of place value. ABCABC = ABC × 1000 + ABC = ABC × (1000 + 1) = ABC × 1001. This exact pattern has appeared in UP Police exams. Recognize it instantly.
A perfect square has every prime factor appearing an even number of times in its prime factorization.
1200 = 2⁴ × 3¹ × 5²
Powers: 4 (even), 1 (odd), 2 (even). The odd-powered factor is 3¹. To make it even, multiply by 3 once. So the multiplier is 3.
1200 × 3 = 3600 = 60². Done.
The method: do the prime factorization, identify all factors with odd exponents, multiply those together — the result is your answer.
| Digits | Smallest | Largest | |---|---|---| | 1 | 1 | 9 | | 2 | 10 | 99 | | 3 | 100 | 999 | | 4 | 1,000 | 9,999 | | 5 | 10,000 | 99,999 |
Largest 5-digit = 99,999. Smallest 4-digit = 1,000. Difference = 98,999. Lock these boundaries in — "largest n-digit" and "smallest n-digit" questions are mechanical if you know the table.
Divide the smallest n-digit number by X. If it divides evenly, that number is your answer. If not, take the remainder r, and the answer is (smallest n-digit number) + (X - r).
Example: Smallest 4-digit number divisible by 6.
1000 ÷ 6 = 166 remainder 4.
Next = 1000 + (6 - 4) = 1000 + 2 = 1002. Check: 1002 ÷ 6 = 167. Correct.
For questions involving successive additions and subtractions, always process left to right, column by column. Carry and borrow precisely. The fastest error source on decimal questions is sloppy column alignment.
90.53 + 67.14 = 157.67
157.67 - 70.16 = 87.51
No shortcut here — just careful column arithmetic.
When you see ABCABC ÷ LCM(7,11,13), write the answer as ABC immediately. The factorization ABCABC = ABC × 1001 and LCM(7,11,13) = 1001 collapse the problem to one step. Standard method: factorize 6-digit number + compute LCM separately = ~60 seconds. Pattern recognition: 5 seconds. One fact saves 55 seconds.
Prime-factorize the number. Highlight every prime whose exponent is odd. Multiply those highlighted primes together (once each). That product is the multiplier needed for a perfect square. For 1200 = 2⁴×3¹×5², only 3 has an odd exponent, so multiply by 3. Standard method: trial and error multiplying by each option = 45 seconds. Odd-exponent hunt: 15 seconds.
To test if N is prime, calculate √N roughly, then divide N only by primes up to that root. For N = 143: √143 ≈ 12. Primes up to 12: 2, 3, 5, 7, 11. 143 ÷ 11 = 13. So 143 = 11×13 — composite. Without this trick, you might test all numbers up to 143. With it, you stop at 11. Step count: from ~140 checks down to 5 checks.
Formula: If (smallest n-digit number) mod X = r, then answer = (smallest n-digit number) + (X - r). If r = 0, the smallest number itself qualifies. Applied to smallest 4-digit divisible by 6: 1000 mod 6 = 4, so answer = 1000 + (6-4) = 1002. This is 3 arithmetic steps versus checking 1000, 1001, 1002, 1003... one by one, which averages 3-5 guesses but is error-prone under exam pressure.
When asked for "sum of digits of number N", never compute digit by digit separately — read the digits visually and add mentally. For 10010: glance left to right, 1+0+0+1+0 = 2. Two non-zero digits, both are 1. Train this visual scan: one pass through the number = done. Standard approach (write and add each): 4 steps. Visual scan: 1 step. Saves ~10 seconds per such sub-problem, which matters when two sub-problems are multiplied together.
When you see a Number System question in the exam hall, run this 4-step decision:
Step 1 — Identify the question type.
Is it asking about: (a) type/classification, (b) a property like ABCABC, (c) perfect square multiplier, (d) n-digit boundary, or (e) arithmetic operations?
Step 2 — Classification questions.
Recall the definition precisely. Prime = greater than 1, exactly 2 factors. Composite = more than 2 factors. 1 = neither. Eliminate options that describe even numbers or non-divisibility — these are partial truths used as traps.
Step 3 — Pattern questions (ABCABC, perfect square).
Apply the stored pattern directly. Do not re-derive — if you know 1001 = 7×11×13, you're done in one line.
Step 4 — Arithmetic questions (balances, word-count products).
Write down the operation, align decimal points, compute once. Do not do mental arithmetic on multi-step decimal problems under exam pressure — a 10-second written computation beats a 30-second mental error.
If two options look close (like 87.51 and 78.51), you made a carry/borrow error — go back and recheck the column where they differ.
Why this question: The most fundamental definition question — gets the definition of "prime" exactly right versus common misconceptions. Tests conceptual clarity, not computation.
Solving path: Read all four options. Option A gives the textbook definition: greater than 1, divisible only by 1 and itself. Option B (even number) fails because 4, 6, 8 are even and not prime. Option C (more than two factors) is the definition of composite, not prime. Option D (not divisible by 2) misses the point — 9 is not divisible by 2 but is not prime (9 = 3×3). Select A instantly.
Why this question: Tests multi-step word problems requiring careful subtraction from total. A common reasoning-wrapped-as-arithmetic type.
Solving path: 8 insects × 1 head = 8 heads. Total heads = 10. Heads belonging to 2 creatures = 10 - 8 = 2. Select option A. Do not overthink — the arithmetic is one subtraction.
Why this question: Three-factor multiplication product. Foundational for word-count, area, and inventory problems.
Solving path: Total words = 300 × 20 × 10. Do it in two steps: 300 × 20 = 6,000. Then 6,000 × 10 = 60,000. Match to option A. Trap: option B is 6,000 (forgot to multiply by 10) and option D is 600,000 (multiplied by 100 instead of 10).
Why this question: The ABCABC pattern is a recurring favorite. This is the exact question type that looks hard but becomes trivial with one stored fact.
Solving path: ABCABC = ABC × 1001. LCM(7, 11, 13) = 1001. Division gives ABC. No calculation needed beyond knowing this identity.
Why this question: Perfect square multiplier — prime factorization applied. This question type appears across multiple UP Police years.
Solving path: Factorize 1200. 1200 = 12 × 100 = 4 × 3 × 4 × 25 = 2⁴ × 3 × 5². Exponents: 4 (even), 1 (odd), 2 (even). Only 3 has an odd exponent. Multiply by 3. Verify: 1200 × 3 = 3600 = 60². Select option B.
Calling 1 a prime number. The definition explicitly requires a prime to be greater than 1 and have exactly two distinct factors. The number 1 has only one factor (itself), so it has exactly one factor, not two. It is neither prime nor composite. This trap appears in classification questions with option choices designed to catch the hasty reader.
Confusing "not divisible by 2" with "prime." Many candidates think all odd numbers are prime. Counter-examples: 9 = 3×3, 15 = 3×5, 25 = 5×5, 35 = 5×7. These are all odd and composite.
Forgetting that 2 is prime. Because it's even, candidates often mentally exclude it. 2 is the only even prime — it has exactly two factors: 1 and 2.
ABCABC division: trying to compute LCM from scratch. LCM(7, 11, 13) = 7 × 11 × 13 = 1001 because 7, 11, 13 are all prime and share no common factors. If you don't know this, you waste 90 seconds computing. Memorize: 7 × 11 × 13 = 1001.
Perfect square factorization: forgetting to check all prime factors. A common error is finding one odd-exponent prime and stopping. You must check every prime in the factorization — if 1200 had both 3¹ and 7¹, the multiplier would be 3×7 = 21, not just 3.
Decimal column misalignment. In balance problems like 90.53 + 67.14 - 70.16, candidates rush the decimal addition and mis-carry. Write it out vertically, align decimal points, and compute one operation at a time. The wrong options are designed to match specific carry errors — seeing your computed answer among the options does not confirm you are correct.