CAT 2025 Number System: 5 Key Properties to Solve 70% of Questions in QA

Number System Topics

The CAT Number System topics from which most of the questions are asked in exams are:

• Properties of numbers

• Factorization

• Divisibility rule

• HCF and LCM

• Unit digit

• Last two digits

• Base system

Property 1: Divisibility Rules

In CAT, Divisibility rules are used

• To determine whether a given number is divisible by another number leaving no Remainder.

• To find the missing digit in a given number.

Key divisibility tests (2, 3, 5, 7, 11…)

Application in factor-based questions

Divisibility rules can be applied in finding the factors of a given number directly.

For example:

If a number has either 0 or 5 at unit place, it must have a factor 5.

If a number has an even digit at unit place, it must have a factor 2.

If the sum of the digits of the number is divisible by 3, it must have a factor 3.

Similarly, we can apply the divisibility rules of 6, 9, 11 etc.

If you need to find the prime factorization, use the divisibility rule of prime numbers.

Property 2: Factors & Multiples

In CAT, concept of factors and multiples are used to

• Determine the number of factors

• HCF and LCM

• CAT Word problems

Factor: A factor of a number divides evenly into that number, leaving no remainder.

Example: Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24

Multiple: A multiple of a number is the result of multiplying that number by another whole number.

Example: Multiples of 48 = 48, 96, 144, ……

Important formula for number of factors

1. Number of factors:

Concept/ Formula: For $N=a^p×b^q ×c^r×….$ Where a, b, and c are prime factors. p, q, r are positive integers.

Number of all factors = $(p+1)(q+1)(r+1)….$

$N =1280=2×2×2×2×2×2×2×2×5=2^8×5$

Number of factors = (8 + 1) × (1 + 1) = 18

2. Sum of all factors:

Concept/ Formula: For $N=a^p×b^q ×c^r×….$ Where a, b, and c are prime factors. p, q, r are positive integers.

Sum = $(1+a^1+a^2+⋯+a^p )(1+b^1+b^2+⋯+b^q )(1+c^1+c^2+⋯.+c^r )…..$

$N =1280=2×2×2×2×2×2×2×2×5=2^8×5$

Sum of all factors = $(1+2+2^2+⋯+2^8 )(1+5) =(2^9-1)/(2-1)×6=511×6=3066$

HCF & LCM in problem solving

Highest Common Factor (HCF):

The HCF of two or more numbers is the largest positive integer that divides them without leaving a remainder. It is also known as the Greatest Common Divisor (GCD). HCF is used to solve problems related to dividing objects into smaller equal groups.

Least Common Multiple (LCM):

The LCM of two or more numbers is the smallest multiple that is evenly divisible by all of the given numbers. LCM is useful when solving problems that involve finding the least number of repetitions required for multiple events to synchronise or repeat together.

HCF and LCM of fractions:

HCF (Highest Common Factor) of fractions

The HCF of two or more fractions is the largest fraction that can exactly divide each given fraction. To find the HCF of fractions, we need to find the HCF of the numerators and the LCM of the denominators. We’ll find HCF and LCM of the numbers using:

LCM (Least Common Multiple) of fractions

The LCM of two or more fractions is the smallest fraction that is divisible by each of the given fractions. To find the LCM of fractions, we need to find the LCM of the numerators and the HCF of the denominators.

Property 3: Even–Odd & Prime–Composite Behaviour

The concept of even and odd digits in CAT is useful to verify answers of number system questions.

The concept of prime and composite is useful for prime factorization, HCF and LCM questions.

Rules of operations with even & odd numbers

Classification of numbers into Even and Odd Numbers:

Non – negative integers are classified into even and odd numbers.

Odd Numbers: A whole number that is not exactly divisible by 2 is called an odd number.

For example: 3, 5, 7, 9, 11, 13, 15, are odd number.

Or a number having 1, 3, 5, 7, and 9 at its unit’s place is called an odd number.

Even Numbers: A whole number exactly divisible by 2 is called an even number.

For example: 0, 2, 4, 6, 8, 10, 12, 14, 16 .......................... are even number.

Or a number having 0, 2, 4, 6, 8 at its unit’s place is called an even number.

Important Results on Even and Odd Integers:

Prime number patterns in CAT questions

Classification of numbers into Prime and Composite Numbers:

Natural Numbers can be classified as

Prime Number: Prime numbers are those numbers which is greater than I, and have only two factors-1 and the number itself. Prime numbers are divisible only by the number 1 or itself.

Example: 2, 3, 5, 7, and 11 are the first few prime numbers.

Composite Number: Numbers having more than two factors are called composite numbers.

Example: 4, 6, 9, 10, and 15 are a few examples of composite numbers.

Note:

1. Co-Prime Numbers:

A pair of numbers that have exactly one factor in common.

Example:

16 and 25

Factors of 16: 1, 2, 4, 8, 16

Factors of 25: 1, 5, 25

Common factor: 1

2. All prime numbers are co-prime with each other.

3. 2 is the smallest prime number.

4. 2 is the only even prime number.

5. For quick answer, you should remember

Property 4: Remainder theorem (Short cut to find remainder)

The concept of remainders in CAT is used

• To determine the remainder of a large number divided by another number.

• To find the unit digit and last digit of a number

Basics of remainders

Basic Remainder Theorem to find the remainder when a large number is divided by small numbers:

It is better to understand this with the help of examples:

Question: Find the remainder when $6789^{12345}$ is divided by 8.

Solution:

Divide $6789^{12345}$ by 8, remainder will be of the form $5^{12345}$ ≡ $5×5^{12344}≡5×25^{6172}$

Now, divide it by 8, remainder will be 5.

Binomial theorem to find the remainder:

The Binomial Theorem states that for any positive integer n and any real numbers a and b:

[a+bn=k=0n knCan-kbk]

Example: Find the remainder when $27^{10}$ is divided by 5.

Solution:

$27^{10}= (25+2)^{10}$, if we divide this number by 5, the remainder will be $2^{10}$.

Now divide $2^{10}$ by 5 to find the final remainder.

So, when$2^{10}$ divided by 5, remainder = 4

Use in cyclicity and last digit questions

Concept of Cyclicity to find unit digit:

Unit digit cyclicity refers to the repetition pattern of the unit digit (the digit in the one’s place) of a number. To determine the unit digit cyclicity of a number, we need to analyse the cyclicity of its unit digit (also known as the ones digit) raised to various powers.

Formula for Unit Digit Cyclicity:

The unit digit of any number raised to a power follows a specific cyclicity pattern.

To find the unit digit of an where a is the unit digit and n is any natural number.

Example 1: $2^1=2,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64$ and so on. The unit digit repeats every four powers (2,4,8,6).

Example 2: $4^1=4,4^2=16,4^3=64,4^4=256$, and so on. The unit digit alternates between 4 and 6.

Rules to determine the last two digits of the number:

Rule 1: For the numbers of the form $x^2 /(50±x)^2 / (100±x)^2…, x^2$ will be the last two digits.

Example: Find the last two digits of $96^2$.

Solution: $96^2 = (100-4)^2$

Last two digits = $4^2=16$

Rule 2: For the numbers having last digit odd except 5.

For (odd)20k, last two digits = 01, k is positive integer.

Example: Find the last two digits of 2342.

Solution: $23^42=23^40×23^2$

Last two digits = 01 × 29 = 29, [Here 29 are the last two digits of 232]

Rule 3: For the numbers having last digit 5.

For (odd5)^odd, last two digits = 75, otherwise 25.

Example: Find the last two digits of 2542 and 7575.

Solution: 2542, last two digits 25, since the digit before 5 is even in 25.

7575, last two digits 75, since the digit before 5 is odd in 75 and the power is odd.

Rule 4: For the numbers having last digit even.

For (Even)^20k, last two digits = 76, k is positive integer.

Example: Find the last two digits of 4624.

Solution:

$46^24=46^20×46^2×46^2$

Writing the last two digits of each term

Last two digits ≡ 76 × 16 × 16

Last two digits ≡ 76 × 56

Last two digits ≡ 56

Property 5: Digit Sum Properties

Sum of the digits of a number till single digit.

Example: To find digit sum of 28976

Step 1: 2 + 8 + 9 + 7 + 6 = 32
Step 2: 3 + 2 = 5 (Ans)

Sum/product of digits tricks

Rule 1: If the digit-sum of a number is 9, then we can eliminate the 9 straight away and the digit- sum becomes ‘zero.’
Rule 2: While solving problems if the digit sum value comes negative then just add it to 9 & result will be the actual digit sum value.

Application in divisibility and missing digit problems

Application in Divisibility by 3 and 9:

Divisibility by 3:

A number is divisible by 3 if the sum of its digits is divisible by 3.

Example: Check whether 3987654 is divisible by 3.

Sum of the digits = 3 + 9 + 8 + 7 + 6 + 5 + 4 = 42

42 = 4 + 2 = 6 which is divisible by 3. So, the given number is divisible by 3.

Divisibility by 9:

A number is divisible by 9 if the sum of its digits is divisible by 9.

Example: Check whether 3987657 is divisible by 9.

Sum of the digits = 3 + 9 + 8 + 7 + 6 + 5 + 7 = 45

45 = 4 + 5 = 9 which is divisible by 9. So, the given number is divisible by 9.

Application in Missing Digit Problems:

This technique extends to solving for a missing digit by substituting it into the digit sum equation.

Example: In the number 2345x2 is divisible by 3, find possible values of x.

Digit sum = 2 + 3 + 4 + 5 + x + 2 = 16 + x

So, possible values of x are 2, 5, 8 so that sum of digits is divisible by 3 and hence number will be divisible by 3 for x = 2, 5, 8

Example: For the number 6523678pq, find the smallest possible sum of p and q so that the number is divisible by 9.

The sum of the known digits is 6 + 5 + 2 + 3 + 6 + 7 + 8 + p + q = 37 + p + q. For this number to be divisible by 9, the sum of all digits must be a multiple of 9. So, 37 + p + q must be a multiple of 9. The smallest multiple of 9 greater than 37 is 45, so p + q = 45 - 37 = 8.

Why These Properties Work

These properties work because most of the CAT numbers questions reduce to these 5 concepts.

CAT number system questions with solutions

Q.1) A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is:

A) 150

B) 140

C) 145

D) 135

Solution:-

Since the total number of students, when divided by either 9 or 10 or 12 or 25 each, gives a remainder of 4, the number will be in the form of LCM(9,10,12,25)k + 4 = 900k + 4.
It is given that the value of (900k + 4) is less than 5000.
Also, it is given that (900k + 4) is divided by 11.
It is only possible when k = 2 and total students = 1804.
So, the number of 12 students group = $\frac{1800}{12}$ = 150
Hence, the correct answer is option (1).

Q.2) When $10^{100}$ is divided by 7, the remainder is

A) 3

B) 4

C) 6

D) 7

Solution:-

We know that $10 \div 7$ leaves a remainder of $3$, so $10^{100}$ will leave the same remainder as $3^{100}$ when divided by $7$
Now, $3^3 = 27$ and $27 \div 7$ leaves remainder $-1$ (since $27 = 28 - 1$)
So, $3^{100} = (3^3)^{33} \times 3 = 27^{33} \times 3$
This becomes $(-1)^{33} \times 3 = -3$

So, the remainder is $-3$.

Since we want a positive remainder, add $7$: $-3 + 7 = 4$
The remainder is 4.

Hence, the correct answer is option 2.

Q.3) For some natural number n, assume that (15,000)! is divisible by n!. The largest possible value of n is

A) 4

B) 7

C) 6

D) 8

Solution:-

To find the largest possible value of n, we need to find the value of n such that n! is less than 15000.

$7!=5040$

$8!=40320 > 15000$
This implies 15000! is not divisible by 40320.
Therefore, the maximum value n can take is 7.
Hence, the correct answer is 7.

Q.4) Minu purchased a pair of sunglasses for Rs.1000 and sold them to Kanu for 20% profit. Then, Kanu sells it back to Minu at a 20% loss. Finally, Minu sells the same pair of sunglasses to Tanu. If the total profit made by Minu from all her transactions is Rs.500, then the percentage of profit made by Minu when she sold the pair of sunglasses to Tanu is

A) 35.42%

B) 31.25%

C) 52%

D) 54%

Solution:-

Initial Cost Price for Minu = 1000
Cost Price to Kanu = 120% of 1000 = 1200
Profit made by Minu = 1200 – 1000 = 200
Again Minu purchased at 80% of 1200 = 960
The total profit made by Manu is 500. So, profit made when she sold it to Tanu = 500 – 200 = 300
Final Selling Price for Minu when sold to Tanu = 960 + 300 = 1260
Profit percent = $\frac{300}{960}$ $\times$ 100 = 31.25%

Hence, the correct answer is option (2).

Q.5) Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of Rs.1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of Rs. 744. Then the amount, in rupees, that she had spent in buying almonds is:

A) 1680

B) 1176

C) 2520

D) 2530

Solution:-

It is given,7C = 30P = 9A and Ankita bought 4C, 14P and 6A.
Let 7C = 30P = 9A = 630k
C = 90k, P = 21k, and A = 70k
Cost price of 4C, 14P and 6A = 4(90k)+14(21k)+6(70k) = 1074k
Marked up price = 1074k + 1752
$\begin{aligned} & \text { S.P }=\frac{1}{6}(1074 k+1752)+\left(\frac{4}{5}\right)\left(\frac{5}{6}\right)(1074 k+1752)=\frac{5}{6}(1074 k+1752) \\ & \text { S.P-C.P }=\text { profit } \\ & 1460-\frac{1074 k}{6}=744 \\ & \frac{1074 k}{6}=716\end{aligned}$
k = 4
Money spent on buying almonds = 420k = 420 × 4 = Rs 1680
The correct answer is Rs. 1680.

Q.6) Amal buys 110 kg of syrup and 120 kg of juice, syrup being 20% less costly than juice, per kg. He sells 10 kg of syrup at 10% profit and 20 kg of juice at 20% profit. Mixing the remaining juice and syrup, Amal sells the mixture at Rs 308.32 per kg and makes an overall profit of 64%. Then, Amal’s cost price for syrup, in rupees per kg, is:

A) 160

B) 180

C) 220

D) 240

Solution:-

Let the price of juice be Rs. $x$ per kg.
Since the cost price of syrup is 20% less than the cost price of juice, the cost price of syrup is $0.8x$ per kg.
Total cost price of syrup $= 110 × 0.8x = 88x$
Total cost price of juice $= 120 × x = 120x$
Total cost price $= 88x + 120x = 208x$
Since the overall profit percentage is 64%,
Total profit $= 0.64(208x) = 133.12x$
Profit generated by selling 10 kg of syrup which costs
Rs $0.8x$ per kg at 10% profit $= 0.1 ×10 ×0.8x = 0.8x$
Profit generated on selling 20 kg of juice which costs Rs $x$ per kg at 20% profit $= 0.2 ×20x = 4x$
The remaining profit $(133.12x - (0.8x + 4x) = 128.32x)$ is generated by selling 100 kg of syrup and
100 kg of juice at Rs 308.32 per kg.
The total selling price of 100 kg of syrup and 100 kg of juice is 200 × 308.32 = 2×(30832)
The cost price of 100 kg of syrup $= 0.8x ×100 = 80x$
Cost price of 100 kg of juice = $100x$
Total cost price $= 80x + 100x = 180x$
Profit = Selling price – Cost price
⇒ $128.32x = 2×(30832) - 180x$
⇒ $308.32x = 2×(30832)$
⇒ $x = 200$
Cost price of syrup per kg = 0.8 × (200) = Rs 160
Hence, the correct answer is option (1).

Key Strategies for CAT number system preparation

We suggest you 5 step key strategy to prepare numbers for CAT:

Step 1: Understand the weightage and question type of number system in CAT. Then break it into core concepts.

Step 2: Focus on building basic concepts like

• Understanding of number properties

• Quick methods to find squares, cubes, square roots, and cube roots

• Classification of numbers

• HCF and LCM

• Factors and LCM

• Remainders etc

Step 3: Learn the application of above discussed concepts in CAT

Step 4: Practice with CAT type questions. Take help from standard books as mentioned in Study resources in later section.

Step 5: Revise and learn short tricks

Best resources & practice material for CAT Number System

1. Arun Sharma: A Quantitative Approach for CAT (7th Edition)
2. Quantitative Aptitude for CAT by Nishit K Sinha

CAT 2025 Preparation Resources by Careers360

The candidates can download the various CAT preparation resources designed by Careers360 using the links given below.