Number System
In short: Number types, primes, divisibility, factors, unit digit and basic remainders.
Arithmetic — Constable › Foundations & Geometry · 32 min read · Free to read · English & Telugu
Bilingual study notes for the Andhra Pradesh Police SI & Constable exams (SLPRB AP) by AP Police Hub, a product of Anvil Academy.
Classification, primes, divisibility, factors, unit digits and basic remainders — the complete Constable ladder, kept simple and exam-ready.
1. Introduction & why it matters
Number System is the very first step of the Arithmetic paper, and the friendliest. It is built on simple rules — how numbers are grouped, which numbers are prime, when one number divides another — and once you know the rules the questions take only a few seconds. In the APSLPRB Constable Prelims this section reliably gives you a few easy marks, so it is worth mastering fully before anything harder.
2. Classification of numbers
Every number belongs to a family. Counting numbers grow into whole numbers, then integers, then rational and irrational numbers. Knowing exactly which box a number sits in answers a whole set of true/false questions on sight.
The number families| Family | Meaning | Examples |
|---|
| Natural (N) | Counting numbers from 1 | 1, 2, 3, … |
| Whole (W) | Naturals together with 0 | 0, 1, 2, 3, … |
| Integers (Z) | Positive, negative and zero — no fractions | …, −2, −1, 0, 1, 2, … |
| Rational (Q) | Can be written as \frac{p}{q}, q\ne 0; decimal terminates or repeats | \frac{3}{4}, −6, 0.25, \sqrt{9} |
| Irrational | Not expressible as \frac{p}{q}; decimal never ends and never repeats | \sqrt{2}, \sqrt{5}, \pi |
Special integer families
| Term | Definition | Note |
|---|
| Even / Odd | Divisible by 2 / not divisible by 2 | 0 is even |
| Prime | Exactly two factors: 1 and itself | 2 = smallest & only even prime |
| Composite | More than two factors | 4 = smallest; 9 = smallest odd |
| Co-prime | Two numbers with HCF = 1 | (8, 9), (4, 15) — need not be prime |
| Twin primes | Two primes differing by 2 | (3, 5), (5, 7), (11, 13) |
3. Handy facts & formulae
A short toolkit of facts covers almost every Constable-level Number-System question. Learn each one with the situation it is used in — that is what makes it stick.
Formula → where it is used| Fact / Formula | Statement | Used for |
|---|
| Sum of first n naturals | 1+2+\cdots+n=\dfrac{n(n+1)}{2} | quick series sums |
| Sum of first n odd numbers | 1+3+5+\cdots=n^{2} | square patterns |
| Difference of squares | a^2-b^2=(a-b)(a+b) | quick factorising |
| Consecutive product | product of 2 consecutive integers is always even | even/odd questions |
| a^n-b^n divisibility | a^n-b^n is always divisible by (a-b) | power divisibility |
4. Models — primes, parity & decimals
Model 1 — True/false about number types
Test each option against a definition or a single counter-example. One counter-example is enough to break a claim that says 'always'.
Model 1
Question: Which one is NOT correct? (a) 0 is neither positive nor negative (b) 1 is neither prime nor composite (c) the sum of two odd numbers is always odd (d) 2 is the only even prime number.
Approach: A statement that says something is 'always' true is broken the moment we find one case where it fails. Three of the four options are standard, well-known facts; the fourth makes a parity claim we can test in a single line by adding two odd numbers.
- Options (a), (b) and (d) each restate a definition every student knows, so all three are true:
3 + 5 = 8 \;(\text{even}) — Option (c) claims two odds always give an odd sum; testing it with the smallest odds shows otherwise:- Odd plus odd is always even, so the claim in (c) is the false one:
Answer: Statement (c) is the one that is NOT correct.
Shortcut: Two odds always add to an even number — the '±' parity table settles it in one look.
Model 2 — Prime test & the 6k±1 rule
To check if n is prime, trial-divide by primes only up to \sqrt{n}. Also: every prime greater than 3 has the form 6k\pm1 — but the reverse is not true (a 6k\pm1 number can still be composite, e.g. 25=6\times4+1).
Model 2
Question: Is 161 a prime number?
Approach: To decide whether a number is prime we only need to try prime divisors up to its square root — any factor larger than the root must pair with one smaller. For 161 the root is about 12.7, so a short list of primes settles it.
- The primes up to 12.7 are just 2, 3, 5, 7 and 11:
- It is odd, its digit-sum 8 is not a multiple of 3, and it does not end in 0 or 5, so 2, 3 and 5 are ruled out:
161 = 7 \times 23 — Trying 7 succeeds — it divides 161 exactly:- Because 161 has a factor other than 1 and itself, it cannot be prime:
Answer: 161 is not a prime number; it equals 7\times23.
Shortcut: Handy to remember: the smallest 3-digit prime is 101, the largest is 997.
Model 3 — Prime puzzles using parity
Every prime except 2 is odd. So when a sum of primes turns out even or odd, whether the number 2 is included is the entire trick.
Model 3
Question: The sum of two prime numbers is 39. Find the two primes.
Approach: The sum of two odd numbers is always even, yet 39 is odd. The only way a pair of primes can add to an odd total is if one of them is the single even prime, 2 — so we fix that and find its partner.
39 - 2 = 37 — Fixing one prime as 2, the other must be 39-2=37:2 + 37 = 39 — Since 37 is itself prime, the pair 2+37=39 is valid:
Answer: The two primes are 2 and 37.
Shortcut: If the sum of two primes is ODD, one of them is always 2.
Model 4 — Recurring decimal → fraction
For a pure repeating block, write the block over as many 9's as its length, then reduce. So 0.\overline{ab}=\dfrac{ab}{99}, 0.\overline{a}=\dfrac{a}{9}.
Model 4
Question: Express 0.\overline{27} (that is 0.2727\ldots) as a fraction in lowest terms.
Approach: A pure recurring decimal can be turned into a fraction by a standard rule: write the repeating block over as many 9's as it has digits, then reduce. The block '27' has two digits, so it sits over 99.
0.\overline{27} = \frac{27}{99} — The two-digit block 27 goes over two nines:= \frac{3}{11} — Both 27 and 99 share a factor of 9, so dividing top and bottom by 9 gives the lowest terms:
Answer: In lowest terms, 0.\overline{27}=\dfrac{3}{11}.
Shortcut: Check by dividing back: 3\div11=0.2727\ldots ✓
5. Models — divisibility & digits
Divisibility rules you must know cold| Divisor | Rule |
|---|
| 2 | last digit is even (0,2,4,6,8) |
| 3, 9 | digit-sum divisible by 3 / by 9 |
| 4 | last 2 digits divisible by 4 |
| 5 | last digit is 0 or 5 |
| 8 | last 3 digits divisible by 8 |
| 11 | (sum of odd-place digits) − (sum of even-place digits) = 0 or a multiple of 11 |
| Composite (6, 12, 15…) | split into co-prime factors and test each (6 = 2×3, 12 = 3×4, 15 = 3×5) |
Model 5 — Find the missing digit (3 / 9 rule)
If a number is divisible by 9, its digit-sum is a multiple of 9. Add the known digits, then choose the missing digit so the total reaches the next multiple of 9.
Model 5
Question: If the 3-digit number 34x is divisible by 9, find the digit x.
Approach: A number is divisible by 9 exactly when its digits add to a multiple of 9. We total the known digits, then pick the single missing digit that pushes the sum up to the next multiple of 9.
3 + 4 + x = 7 + x — The known digits 3 and 4 add to 7, so the digit-sum is 7+x:7 + x = 9 \Rightarrow x = 2 — The next multiple of 9 at or above 7 is 9, so 7+x=9 forces x=2:- As a check, 342\div9=38 divides exactly, confirming the digit:
Answer: The missing digit is x=2.
Model 6 — Composite divisor: split into co-prime factors
To test divisibility by a composite number, split it into co-prime factors and check each separately. For 6, check 2 AND 3.
Model 6
Question: Is 34578 divisible by 6?
Approach: Six is not prime, so we split it into its co-prime factors 2 and 3 and test each rule separately; a number divisible by both is divisible by 6. The factors must share no common factor, which 2 and 3 do not.
- First the 2-rule: the last digit 8 is even, so 34578 passes it:
3+4+5+7+8 = 27 — Now the 3-rule: the digits add to 27:- Since 27 is a multiple of 3, that rule passes too, and both being satisfied means 6 divides the number:
Answer: Yes — since it passes both the 2 and 3 tests, 34578 is divisible by 6.
Shortcut: Never split a composite into factors that share a factor — 12 = 3×4 works (co-prime), but 12 = 2×6 does NOT (both even).
Model 7 — The 11 rule
For 11: add the digits in the odd places and the digits in the even places separately; their difference must be 0 or a multiple of 11.
Model 7
Question: If the number 2143x is divisible by 11, find the digit x.
Approach: The rule for 11 works on alternating digit places: sum the digits in the odd positions, sum those in the even positions, and their difference must be 0 or a multiple of 11. We count positions from the right and solve for the missing digit.
x + 4 + 2 = x + 6 — Counting from the right, the odd-place digits are x, 4 and 2, adding to x+6:3 + 1 = 4 — The even-place digits 3 and 1 add to 4:(x+6) - 4 = x + 2 = 11 \Rightarrow x = 9 — The difference x+2 must be a multiple of 11; the only single digit that works is x=9, giving 11:- The check 21439\div11=1949 divides exactly, confirming x=9:
Answer: The missing digit is x=9.
6. Models — unit digit & cyclicity
The unit digit of a power repeats in a cycle. Digits 0, 1, 5, 6 keep the same unit digit at every power. Digits 4 and 9 flip in a cycle of length 2. Digits 2, 3, 7, 8 repeat in a cycle of length 4 — divide the power by 4 and use the remainder.
Cyclicity table| Unit digit of base | Cycle | Cycle length |
|---|
| 0, 1, 5, 6 | same digit always | 1 |
| 4 | 4, 6, 4, 6… | 2 (odd power→4, even→6) |
| 9 | 9, 1, 9, 1… | 2 (odd power→9, even→1) |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
Model 8 — Unit digit of a single power
Model 8
Question: What is the unit digit of 3^{24}?
Approach: The unit digit of a power of 3 repeats in the cycle 3, 9, 7, 1, which has length 4. So the position within the cycle is decided by the remainder when the exponent is divided by 4.
24 \div 4 = 6 \text{ remainder } 0 — Dividing the exponent 24 by the cycle length 4 leaves remainder 0:\text{last of cycle} = 1 — A remainder of 0 points to the last digit of the cycle 3, 9, 7, 1, which is 1:
Answer: The unit digit of 3^{24} is 1.
Shortcut: When the remainder is 0, always use the LAST digit of the cycle, not the first.
Model 9 — Unit digit of a product
Model 9
Question: What is the unit digit of 17^{12}\times24^{13}?
Approach: Only the unit digit of each base matters, so we reduce 17 to 7 and 24 to 4, work out each power's unit digit from its cycle, and finally multiply those two unit digits — taking the unit digit of that product.
12 \div 4 = 3 \text{ r } 0 \Rightarrow \text{last of } 7,9,3,1 = 1 — The first base ends in 7 (cycle 7, 9, 3, 1); 12\div4 leaves remainder 0, so its unit digit is the last of the cycle, 1:\text{odd power of 4} \Rightarrow 4 — The second base ends in 4, whose odd powers all end in 4, so 24^{13} has unit digit 4:1 \times 4 = 4 — Multiplying the two unit digits, 1\times4=4, gives the unit digit of the whole product:
Answer: The unit digit of 17^{12}\times24^{13} is 4.
Shortcut: Any product with a factor ending in 5 and any even factor ends in 0 (each 5×2 makes a trailing zero).
7. Models — factors of a number
Everything about factors starts from the prime factorisation N=a^{x}\times b^{y}\times c^{z} (with a,b,c prime). From it, factor questions become one-line formulas.
Factor formulae (for N=a^{x}b^{y}c^{z})| Quantity | Formula |
|---|
| Number of factors (NOF) | (x+1)(y+1)(z+1) |
| Sum of factors (SOF) | \dfrac{a^{x+1}-1}{a-1}\cdot\dfrac{b^{y+1}-1}{b-1}\cdot\dfrac{c^{z+1}-1}{c-1} |
| Product of factors | N^{\,\text{NOF}/2} |
Model 10 — Number & sum of factors
Model 10
Question: Find the number of factors and the sum of all factors of 360.
Approach: Every factor question flows from the prime factorisation. Once 360 is written as 2^3\times3^2\times5^1, the count of factors comes from adding one to each exponent and multiplying, and the sum of factors from the standard geometric-series product.
360 = 2^3 \times 3^2 \times 5^1 — Break 360 into its prime powers:\text{NOF} = (3+1)(2+1)(1+1) = 4\times3\times2 = 24 — Adding 1 to each exponent and multiplying counts every factor: 4\times3\times2=24:\text{SOF} = \frac{2^4-1}{2-1}\cdot\frac{3^3-1}{3-1}\cdot\frac{5^2-1}{5-1} — The sum of factors multiplies one geometric-series term for each prime:= 15 \times 13 \times 6 — Each term evaluates to 15, 13 and 6:= 1170 — and their product is 1170:
Answer: 360 has 24 factors, and they add up to 1170.
Model 11 — Product of all factors
Factors pair up so that each pair multiplies to N. With NOF factors there are \tfrac{\text{NOF}}{2} such pairs, so the product of all factors is N^{\text{NOF}/2}.
Model 11
Question: What is the product of all the factors of 24?
Approach: Factors pair off so that each pair multiplies to the number itself. With n factors there are n/2 such pairs, so the product of all factors is N^{n/2}. We only need the count of factors from the prime factorisation.
24 = 2^3 \times 3^1 — Write 24 in prime powers:\text{NOF} = (3+1)(1+1) = 8 — Adding 1 to each exponent gives 8 factors in all:24^{\,8/2} = 24^{4} — With 8 factors forming 4 pairs, the product of all of them is 24^{8/2}=24^4:
Answer: The product of all the factors of 24 is 24^{4}.
Shortcut: Quick check with factors of 24: (1,24),(2,12),(3,8),(4,6) — four pairs, each product 24, so 24^{4}.
8. Models — basic remainders
Model 12 — Dividend = DQ + R puzzles
Model 12
Question: In a division, the divisor is 5 times the quotient and 3 times the remainder. If the remainder is 40, find the dividend.
Approach: The whole thing rests on the division law, dividend = divisor × quotient + remainder. We are given the remainder and two links tying the divisor to the quotient and to the remainder, so we unpick them one at a time and assemble the dividend at the end.
\text{Divisor} = 3 \times 40 = 120 — The divisor is three times the remainder, so it is 3\times40=120:\text{Quotient} = 120 \div 5 = 24 — The same divisor is five times the quotient, so the quotient is 120\div5=24:\text{Dividend} = 120 \times 24 + 40 — Putting the pieces into the division law:= 2880 + 40 = 2920 — which works out to 2920:
Answer: The dividend is 2920.
Model 13 — Remainder of a product
Replace each factor by its own remainder, multiply the small remainders, then take the remainder once more. Never multiply the big numbers first.
Model 13
Question: Find the remainder when 43\times47\times49 is divided by 6.
Approach: Rather than multiply the three numbers first, we replace each by the remainder it leaves on division by 6, multiply those small remainders, and take the remainder of that. The remainder of a product equals the product of the remainders, reduced mod 6 once more.
43 \to 1,\quad 47 \to 5,\quad 49 \to 1 — Divided by 6, the three numbers leave remainders 1, 5 and 1:1 \times 5 \times 1 = 5 — Multiplying those small remainders gives 1\times5\times1=5:- Since 5 is already smaller than 6, no further reduction is needed and 5 is the answer:
Answer: The remainder is 5.
Model 14 — Remainder of a power (base ± 1 trick)
If the base is just one more than the divisor, every power leaves remainder 1. If it is just one less, an even power leaves 1 and an odd power leaves the divisor minus 1.
Model 14
Question: Find the remainder when 8^{25} is divided by 7.
Approach: The trick is to write the base one step from the divisor. Here 8 is exactly one more than 7, so it leaves remainder 1; and since 1 raised to any power stays 1, every power of 8 leaves remainder 1 on division by 7.
8 \equiv 1 \pmod{7} — Because 8 is one more than 7, it is congruent to 1 modulo 7:8^{25} \equiv 1^{25} = 1 \pmod{7} — Raising both sides to the 25th power, 1^{25} is still 1, so the remainder is 1:
Answer: The remainder when 8^{25} is divided by 7 is 1.
Shortcut: Example of the other case: 6^{20}\div7 — since 6\equiv-1, an even power gives (-1)^{20}=1.
9. Shortcuts & exam tips
10. At a glance — formula sheet
Quick-revision formula sheet| Topic | Key result |
|---|
| Repeating decimal | 0.\overline{ab}=\dfrac{ab}{99}, then reduce |
| Prime form | every prime > 3 is 6k\pm1 (reverse false) |
| Divisibility by 3 / 9 | digit-sum divisible by 3 / 9 |
| Divisibility by 11 | (odd-place sum) − (even-place sum) = 0 or 11 |
| Cyclicity | 0/1/5/6 → 1; 4/9 → 2; 2/3/7/8 → 4 |
| Number of factors | (x+1)(y+1)(z+1) |
| Product of factors | N^{\text{NOF}/2} |
| Division law | \text{Dividend}=Dq+R, R<D |
| Power remainder | (D+1)^n\to1, (D-1)^n\to(-1)^n |