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

Numbers, Divisibility, HCF/LCM and Progressions short notes for XAT: 3 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Summary: The Divisibility and HCF/LCM Toolkit

Summary Toolkit

HCF: Lowest powers of common primes.

LCM: Highest powers of all primes.

Two Numbers:

HCF \times LCM = Product

Fractions:

HCF = \frac{HCF ( N )}{LCM ( D )}

Divisibility: Break into co-primes.

Word Problems: HCF for max, LCM for min.

Memory Hook: HCF divides the numbers, the numbers divide the LCM. HCF is for sharing, LCM is for syncing.

Summary: Progressions Revision Sheet

Progressions revision sheet

AP

$T_{n} = a + (n - 1) d$
$S_{n} = \frac{n}{2} (a + l)$
For 3 terms in AP, use $a - d, a, a + d$ .

GP

$T_{n} = a r^{n - 1}$
$S_{n} = a \frac{1 - r ^{n}}{1 - r}$ for $r \neq = 1$
$S_{\infty} = \frac{a}{1 - r}$ only when $|r|<1$

Pattern recognition

Fixed absolute change: AP
Fixed multiplier or percentage change: GP
Repeated cycles: LCM counting
Multiplicative choices: prime exponent tracking

Final check before solving

Is the question asking for a term or a sum?
Is the GP ratio greater than, less than, or equal to one?
Is the starting occasion included in the count?

Digit Puzzles: Final Toolkit

Summary

Digit Puzzles: Final Toolkit

Keep these core mechanics active in your memory for exam day.

Place value expansion

$N = 100 a + 10 b + c$ . This is the bridge between words and math.

Swap differences

Swap adjacent digits: difference is $9 \times ∣ a - b ∣$ .
Swap outer digits: difference is $99 \times ∣ a - c ∣$ .

Divisibility filtering

Convert sum conditions into modular arithmetic. Group available digits by remainders before attempting to arrange them.

Total sum trick

If the sum of a subset of digits must be divisible by $k$ , subtract that subset's sum from the total sum of all digits to find the constraint on the remaining digits.

Leading zeros

Mathematical numbers mean no leading zero. PINs and codes mean leading zero is allowed.

Negative questions

Disprove options using counter-examples rather than proving the negative universally.

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

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

Numbers, Divisibility, HCF/LCM and Progressions short notes for XAT: 3 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice

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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