Module 2: S4 Quant

Topic 7: Number System, LCM-HCF & Divisibility

1 Number System Classification & Traps

Before calculating, you must know exactly what classification a number falls into. UPSC uses definitional traps to eliminate candidates who rush.

The Real Number Tree

⚠️ Absolute Definitional Traps

0 is an EVEN integer. It is neither positive nor negative.
1 is NEITHER prime NOR composite. The smallest prime number is 2.
2 is the ONLY even prime number.
Co-Prime Numbers: Two numbers are co-prime if their HCF is 1. They do NOT have to be prime themselves! (e.g., 8 and 15 are co-prime).
$\pi$ vs $\frac{22}{7}$: $\pi$ is Irrational. $\frac{22}{7}$ is Rational.

2 Divisibility Rules (Speed Math Checks)

To quickly find factors or simplify equations, these rules must be second nature.

Divisor	The Rule (Check if...)
3 & 9	Sum of all digits is divisible by 3 or 9.
4	Last TWO digits are divisible by 4 (or are 00).
8	Last THREE digits are divisible by 8 (or are 000).
6	Number is divisible by BOTH 2 and 3.
11	(Sum of digits in Odd places) $-$ (Sum of digits in Even places) = $0$ or a multiple of $11$.

The 7, 11, 13 Mega-Hack (Triplets)

To check divisibility for 7, 11, or 13 on a massive number (e.g., 5,922):

Step 1: Form blocks of 3 digits from right to left.
Blocks: $\mathbf{005}$ and $\mathbf{922}$.
Step 2: Find the difference between alternate blocks.
$922 - 5 = \mathbf{917}$
Step 3: Check 917.
$917 \div 7 = 131$ (Divisible!)
Therefore, 5,922 is divisible by 7.

3 Unit Digit & The Concept of Cyclicity

UPSC loves asking for the unit digit of massive exponents like $3^{2026}$. You don't calculate this; you use the Cyclicity Rule. The unit digits of powers always repeat in a cycle of 4.

0, 1, 5, 6: Always end in the same digit. ($6^{999}$ ends in 6).
4, 9: Cycle of 2.
$4^{\text{odd}}$ ends in 4, $4^{\text{even}}$ ends in 6.
$9^{\text{odd}}$ ends in 9, $9^{\text{even}}$ ends in 1.
2, 3, 7, 8: Cycle of 4.
Rule: Divide the power by 4. Find the remainder ($R$).
If $R=1$, power is $1$. If $R=2$, power is $2$. If $R=3$, power is $3$.
If $R=0$, power is $4$ (DO NOT put 0!).

Cyclicity Clock for Base 3

Example: Find the unit digit of $3^{2026}$.
Step 1: Divide exponent 2026 by 4. (Speed hack: Just divide last two digits, $26 \div 4$).
Step 2: Remainder ($R$) is $2$.
Step 3: Calculate $3^2 = 9$. The unit digit is 9.

4 LCM & HCF: Rules and Word Problem Mastery

The Absolute Golden Rules

1. The Product Rule:
$$LCM(a,b) \times HCF(a,b) = a \times b$$ (Valid ONLY for two numbers!)

2. Fractions Rule:
$$LCM \text{ of fractions} = \frac{LCM \text{ of Numerators}}{HCF \text{ of Denominators}}$$ $$HCF \text{ of fractions} = \frac{HCF \text{ of Numerators}}{LCM \text{ of Denominators}}$$

Keyword Decoder (Word Problems)

UPSC rarely says "Find the LCM." They hide it in administrative scenarios.

Use LCM When:	Use HCF When:
Finding the "Smallest/Minimum"	Finding the "Greatest/Maximum"
Events repeating TOGETHER (Bells ringing, circular running)	Dividing items into equal groups of MAXIMUM size without mixing.

5 Admin Context: Practical Arithmetic

Scenario 1 (Bell Scheduling)

Three school bells ring at intervals of 12, 15, and 20 minutes respectively. If they ring together at 8:00 AM, when will they ring together next?

Logic: Keyword is "together next" ➔ LCM.
LCM of 12, 15, 20:
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$20 = 2^2 \times 5$
LCM = Highest powers = $2^2 \times 3 \times 5 = 60$ mins.
Answer: 60 mins after 8:00 AM = 9:00 AM.

Scenario 2 (Equal Grouping)

A Principal has 84 Arts students, 108 Science students, and 144 Commerce students. They must be seated in rooms so that every room has the same number of students and only one stream per room. Find the maximum students per room.

Logic: Keyword is "maximum... same number... without mixing" ➔ HCF.
HCF of 84, 108, 144:
Difference hack: $108 - 84 = 24$. HCF is 24 or a factor of 24.
24 does not divide 84.
Next factor of 24 is 12.
12 divides 84, 108, and 144.
Answer: 12 students per room.