Dear CAT aspirants, if you are planning to prepare for the Number System, this article is created especially for you. Here, you will find the key properties of numbers that cover a major part of the syllabus and help you solve around 70% of the questions from this topic in the CAT Quantitative Aptitude section. In CAT 2025, you can expect 3-4 questions from Number System, and mastering these properties will give you a clear advantage. In this article, careers360 will provide you 5 most important properties of numbers for CAT that covers 70% of CAT number system questions.
A 99 percentile in the CAT exam puts you in strong contention for many IIMs, but doesn’t guarantee a call, especially from top IIMs like A, B, and C, which often require 99.5+ along with a strong overall profile and WAT/PI performance.
Sum the digits. If the result is divisible by 3, then the original number is divisible by 3.
4359 → 4 + 3 + 5 + 9 = 21 which is divisible by 3.
4
Examine the last two digits.
40864: 64 is divisible by 4. So, 40864 is divisible by 4.
5
The last digit is 0 or 5.
685: the last digit is 5.
4590: last digit is 0.
Both numbers are divisible by 5.
6
It is divisible by 2 and by 3.
1,458: 1 + 4 + 5 + 8 = 18, so it is divisible by 3 and the last digit is even, hence the number is divisible by 6.
7
A number is divisible by 7 if the difference between twice the unit digit and the number formed by the remaining digits is divisible by 7.
343: 34-2×3=28 which is divisible by 7. Hence, 343 is divisible by 7.
8
A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
34152: Examine divisibility of just 152: 19 × 8
So, the number is divisible by 8.
9
Sum the digits. If the result is divisible by 9, then the original number is divisible by 9.
435429 → 4 + 3 + 5 + 4 + 2 + 9 = 27 which is divisible by 9.
11
A number is divisible by 11 if the difference between the sum of digits at odd places (1st, 3rd, 5th, etc.) and the sum of digits at even places (2nd, 4th, 6th, etc.) is either 0 or divisible by 11.
918,082: 9 − 1 + 8 − 0 + 8 − 2 = 22.
So, the number is divisible by 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
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.
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
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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.
Digit
Cyclicity
Unit digit
Remark
1
1
1
2
4
The unit digit repeats every four powers (2,4,8,6).
3
4
The unit digit repeats every four powers (3,9,7,1)
4
2
4,6
4 raised to power any even number gives 6 otherwise 4.
5
1
5
6
1
6
7
4
The unit digit repeats every four powers (7,9,3,1)
8
4
The unit digit repeats every four powers (8,4,2,6)
9
2
9,1
9 raised to power any even number gives 1 otherwise 9.
0
1
0
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.
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
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.
Concept
Question Type
Divisibility & Factors
Prime factorization
HCF & LCM
Finding number of factors / sum of factors
Divisibility rules through digit sums (9, 11 tests etc.)
Remainders & Modular Arithmetic
Finding last digit / cyclicity
Remainder theorems
Divisibility-based remainder patterns
Base System & Units/Last Digits
Conversion between bases
Identifying units digit / last two digits
Place value logic
Even–Odd, Positives–Negatives & Digit Properties
Sum/product of digits properties
Solution check through even and odd digit properties
Factors and Multiples
Trailing zeros
Number of factors
Sum of factors
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
CAT does not restrict eligibility based on stream. Candidates from commerce, arts, science, engineering or any other background can appear for CAT and apply to MBA programmes offered by IIMs and other top B-schools.
The basic eligibility criteria for CAT are:
You must hold a bachelor’s degree in any discipline from a recognised university with at least 50 percent marks (45 percent for SC, ST and PwD categories). Final-year graduation students are also eligible to apply.
Being a commerce graduate can actually be an advantage in areas such as accounting, finance, economics and business studies, especially during MBA coursework in finance, marketing and operations.
After qualifying CAT, admission depends on multiple factors including CAT percentile, academic background, work experience (if any), performance in GD, WAT and personal interview rounds. Many IIMs and B-schools follow a diverse academic background policy, which means non-engineering candidates, including commerce graduates, often receive additional weightage.
Apart from IIMs, several reputed institutes such as FMS Delhi, SPJIMR Mumbai, MDI Gurgaon, IMT Ghaziabad, and many state and private universities also accept CAT scores for MBA admissions.
So, as a commerce graduate, you are fully eligible to appear for CAT and pursue an MBA, provided you meet the minimum academic requirements and perform well in the selection process.
If you want, I can also help you with realistic CAT percentile targets based on your academic profile or suggest suitable MBA colleges.
With a CAT 2025 raw score of 84 and strong sectionals, your expected percentile may fail in the high 98 range , depending on scaling. Your excellent academics and 2 years of work experience strengthen your profile .IIM Lucknow calls are possible but competitive for EWS engineers , while FMS Delhi looks mainly at CAT score and VARC , so you have a reasonable chance there . Final calls depend on cutoffs and composite score.
For MANAGE Hyderabad MBA , the OBC cutoff in CAT 2024 was generally around 85-90 percentile for shortlist consideration , tough exact figures vary by section and profile . For CAT 2025, if fewer candidates appeared , the cutoff might slightly adjust , but percentile based cutoffs dont change drastically year to year. A minor decrease is possible , but you should still aim for 90+ percentile to be competitive. Factors like VARC, QA-LRDI and overall profile influence the final shortlist.
For a raw score of 56 in CAT 2025 Slot 3, your expected overall percentile is likely to be in the range of the 90th-95th percentile. The exact percentile can vary slightly on the final normalization process and the process and performance of all test-takers. In this
article
you'll find more about the CAT result.
Hello,
With a projected CAT percentile of 87% but not clearing sectional cutoffs, your chances at top IIMs are limited because they require both overall percentile and sectional minimums. However, you still have a good shot at other reputed management institutes and non-IIM B-schools. Consider colleges like NMIMS, SPJIMR, IMT, TAPMI, Great Lakes, and other well-ranked private or state-level B-schools that accept CAT scores and weigh your profile holistically. Your academic record, BSc in Animation with 80%, and 5 years of work experience at Ubisoft India are strong points and may help in institutes that value work experience in their selection process. Also, explore institutes that accept XAT, MAT, or CMAT, where your profile can be competitive.
Hope this helps you.
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