Numbers, Divisibility, HCF/LCM and Progressions Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Numbers, Divisibility, HCF/LCM and Progressions notes for XAT: 38 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Numbers & Progressions

Chapter Roadmap

Numbers & Progressions

1. Divisibility, HCF & LCM

Core mechanics of factors and multiples. Essential for algebra and word problems.

2. Number Patterns & Progressions

Arithmetic and geometric sequences, sum of series.

3. Digit Puzzles & Number Logic

Analytical thinking with digits, remainders, and equations.

Goal: Deconstruct any number property problem and solve it using systematic, trap-free methods.

The Core of HCF and LCM

HCF

Greatest common sharing capacity. The largest tape measure that can measure both exactly.

LCM

Smallest common multiple. The first time two repeating events sync up.

The Golden Relationship

HCF always divides the numbers.
The numbers always divide the LCM.
HCF $\leq$ Smallest Number $\leq$ Largest Number $\leq$ LCM.

Prime Factorization Method

Break each number down into its prime building blocks.

For HCF

Take the lowest power of every common prime factor.

For LCM

Take the highest power of every prime factor present.

Example: 12 and 18
12=22×31
18=21×32
HCF =21×31=6
LCM =22×32=36

The Golden Formula for Two Numbers

The Golden Formula

For any two numbers $a$ and $b$ :

HCF (a, b) \times LCM (a, b) = a \times b

How to Use It

If you know any three of these four values (HCF, LCM, first number, second number), you can instantly find the fourth.

Strict Warning: This formula works strictly for two numbers. It is completely false for three or more numbers.

Numbers, Divisibility, HCF/LCM and Progressions: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

If the Highest Common Factor (HCF) of two numbers is 12 and their Least Common Multiple (LCM) is 180, what is the product of these two numbers?

A.
2160
B.
1920
C.
15
D.
216

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct application of the Golden Formula for two numbers, recognisable because it links HCF, LCM, and Product directly.

Step 1: Recall the fundamental relationship for any two positive integers $a$ and $b$ : $HCF (a, b) \times LCM (a, b) = a \times b$ .

Step 2: Substitute the given values into the formula: $Product = 12 \times 180$ .

Step 3: Calculate the product: $12 \times 180 = 2160$ .

Answer: The product of the two numbers is 2160.

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

Consider an arithmetic progression consisting of positive integers. The sum of the first $n$ terms is denoted by $S_{n}$ .

A student calculates the value of $\frac{S _{20}}{S _{10}}$ and finds it to be exactly $3$ .

Based on this information, what is the value of $\frac{S _{30}}{S _{10}}$ ?

A.
4
B.
5
C.
6
D.
Cannot be determined uniquely

Correct Answer:

Step-by-Step Solution

Key idea: This problem tests the relationship between sums of blocks in an AP. There is a powerful property: for any AP, the sums of consecutive blocks of equal size (e.g., first 10, next 10, next 10) themselves form an arithmetic progression. Alternatively, we can use the sum formula directly.

Step 1: Write the sum formula.

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .

Step 2: Express $S_{10}$ and $S_{20}$ .

$S_{10} = \frac{10}{2} [2 a + 9 d] = 5 (2 a + 9 d) = 10 a + 45 d$ .

$S_{20} = \frac{20}{2} [2 a + 19 d] = 10 (2 a + 19 d) = 20 a + 190 d$ .

Step 3: Apply the given condition $\frac{S _{20}}{S _{10}} = 3$ .

$S_{20} = 3 S_{10}$

$20 a + 190 d = 3 (10 a + 45 d)$

$20 a + 190 d = 30 a + 135 d$

$190 d - 135 d = 30 a - 20 a$

$55 d = 10 a$

$a = 5.5 d$ or $2 a = 11 d$ .

Step 4: Find $S_{30}$ .

$S_{30} = \frac{30}{2} [2 a + 29 d] = 15 (2 a + 29 d)$ .

Substitute $2 a = 11 d$ :

$S_{30} = 15 (11 d + 29 d) = 15 (40 d) = 600 d$ .

Step 5: Calculate the required ratio.

We need $\frac{S _{30}}{S _{10}}$ .

Express $S_{10}$ in terms of $d$ :

$S_{10} = 10 a + 45 d = 10 (5.5 d) + 45 d = 55 d + 45 d = 100 d$ .

Ratio $= \frac{600 d}{100 d} = 6$ .

Alternative elegant method (Block Sums):

Let $B_{1} = S_{10}$ , $B_{2} = S_{20} - S_{10}$ , $B_{3} = S_{30} - S_{20}$ .

$B_{1}, B_{2}, B_{3}$ are in AP.

Given $S_{20} / S_{10} = 3 ⟹ (B_{1} + B_{2}) / B_{1} = 3 ⟹ B_{1} + B_{2} = 3 B_{1} ⟹ B_{2} = 2 B_{1}$ .

Since $B_{1}, B_{2}, B_{3}$ are in AP and $B_{2} = 2 B_{1}$ , the common difference of this block-AP is $B_{2} - B_{1} = B_{1}$ .

So $B_{3} = B_{2} + B_{1} = 2 B_{1} + B_{1} = 3 B_{1}$ .

$S_{30} = B_{1} + B_{2} + B_{3} = B_{1} + 2 B_{1} + 3 B_{1} = 6 B_{1} = 6 S_{10}$ .

Ratio is 6.

Answer: 6

More notes in this unit

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Numbers, Divisibility, HCF/LCM and Progressions Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Numbers, Divisibility, HCF/LCM and Progressions notes for XAT: 38 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice ques

A question from this chapter

Question 1

If the Highest Common Factor (HCF) of two numbers is 12 and their Least Common Multiple (LCM) is 180, what is the product of these two numbers?

Question 2

Consider an arithmetic progression consisting of positive integers. The sum of the first n terms is denoted by S_{n} . A student calculates the value of \frac{S _{20}}{S _{10}} and finds it to be exactly 3 . Based on this information, what is the value of \frac{S _{30}}{S _{10}} ?

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