Explore Exams
Login Signup
Sign in Open App
CAT Exam  >  CAT Test  >  Quantitative Aptitude (Quant)  >  Arun Sharma Test: Number System- 1 - CAT MCQ

Arun Sharma Test: Number System- 1 - CAT Prep Free MCQ with solutions


MCQ Practice Test & Solutions: Arun Sharma Test: Number System- 1 (15 Questions)

You can prepare effectively for CAT Quantitative Aptitude (Quant) with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Arun Sharma Test: Number System- 1". These 15 questions have been designed by the experts with the latest curriculum of CAT 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 25 minutes
  • - Number of Questions: 15

Sign up on EduRev for free to attempt this test and track your preparation progress.

*Answer can only contain numeric values
Arun Sharma Test: Number System- 1 - Question 1
Save

Find the highest power of 30 in 50!


Detailed Solution: Question 1

30=2×3×5.
Now 5 is the largest prime factor of 30, therefore, the powers of  5 in 50! will be less than those of 2 and 3. Therefore, there cannot be more 30 s than there are 5 in 50! So we find the highest power of 5 in 50! The highest power of 5 in 50! Hence the highest power of 30 in 50! = 12

*Answer can only contain numeric values
Arun Sharma Test: Number System- 1 - Question 2
Save

The number of even factors of S, where S = 13 + 23 + 33 +... 10003 is


Detailed Solution: Question 2

S = 13 + 23 + 33 +... 10003 is

We know that the sum of cubes of natural number starting from 1 upto n is given by: 

So, for even factors, we need at least one 2 among the four 2s available with S. We can select either one, two, three or four 2s. For the other prime factors, we also have the option to choose 0 out of the available number of 5s, 11s, 7s and 13s.

So, total even factors= 4×7×3×3×3 = 756.

Clear all your doubts with EduRev
Arun Sharma Test: Number System- 1 - Question 3
Save

If the number of factors of N is 8 and that of 5N is 16, then the number of factors of 2*N can be

Detailed Solution: Question 3

Let p1, p2, p3 be distinct prime numbers which are not equal to 5.

From the given information, we can conclude that N can be

Case 1: N = p1*p2*p3

Case 2: N = p13∗p2

Case 3: N = p17

Now, 2 can or cannot be one of p1,p2,p3

In case 1:

If 2 ≠ ​p1/p2/p3 then number of factors of 2*p1*p2*p3 = 16.

If 2 is equal to p1 or p2 or p3 then the number of factors of 2*N will be 12.

In case 2:

If 2 ≠  p1 or p2, then number of factors of 2*N=16.

If 2=p1, the number of factors of 2*N will be 10

If 2=p2, the number of factors of 2*N will be 12.

Case 3:

If 2 ≠  p, the number of factors of 2*N = 16.

If 2=p1, the number of factors of 2*N = 9

.'. The number of factors of 2N can be 9,10,12,16.

Arun Sharma Test: Number System- 1 - Question 4
Save

If A, B, C and D are distinct natural numbers, the sum of which adds up to 50. Then, the maximum value of their product is __?

Detailed Solution: Question 4

Given, A, B, C and D are 4 distinct integers and A+B+C+D= 50.
The products of four numbers will be maximum if the numbers are as close to each other as possible.
The A.M. of the four numbers= 
.'. The four numbers are around 12 and continuous. A, B, C and D are 11, 12, 13 and 14.
And hence, the maximum value of the products of the four number=11×12×13×14=24024. 

*Answer can only contain numeric values
Arun Sharma Test: Number System- 1 - Question 5
Save

For natural numbers x,y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is


Detailed Solution: Question 5

It is given, y(x + z) = 19

y cannot be 19. 

If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.

Therefore, y = 1 and x + z = 19

It is given, z(x + y) = 51

z can take values 3 and 17

Case 1:

If z = 3, y = 1 and x = 16

xyz = 3*1*16 = 48

Case 2:

If z = 17, y = 1 and x = 2

xyz = 17*1*2 = 34

Minimum value xyz can take is 34.

*Answer can only contain numeric values
Arun Sharma Test: Number System- 1 - Question 6
Save

Let A be the largest positive integer that divides all the numbers of the form 3k + 4k + 5k and B be the largest positive integer that divides all the numbers of the form 4k + 3(4k) + 4k+2, where k is any positive integer. Then (A + B) equals


Detailed Solution: Question 6

A is the HCF of 3k + 4k + 5k for different values of k.

For k = 1, value is 12

For k = 2, value is 50

For k = 3, value is 216

HCF is 2. Therefore, A = 2

HCF of the values is when k = 1, i.e. 5*16 = 80

Therefore, B = 80

A + B = 82

Arun Sharma Test: Number System- 1 - Question 7
Save

On day one, there are 100 particles in a laboratory experiment. On day n, where n ≥ 2, one out of every n particles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals

Arun Sharma Test: Number System- 1 - Question 8
Save

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

Arun Sharma Test: Number System- 1 - Question 9
Save

Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of x km till then, where x is a whole number and is palindromic, i.e., X remains unchanged when its digits are reversed. At the end of the trip, the car had travelled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/h, then the greatest possible average speed at which she drove during the trip, in km/h, was

Detailed Solution: Question 9

The distance travelled before the trip (x in km.) is a palindrome.
Let the average speed of Brishti be s kmph.
The total distance covered by Brishti after the trip is 26862 km.
The distance covered by Brishti during the trip is 8×s = 8s km.
x = 26862 – 8s
For x to be a palindrome, of the given options, s can only be 100.
Therefore, the average speed of Brishti during the trip is 100 kmph.

Arun Sharma Test: Number System- 1 - Question 10
Save

For any natural numbers m, n, and k, such that k divides both m + 2n and 3m + 4n, k must be a common divisor of

Detailed Solution: Question 10

It is given that k divides m+2n and 3m+4n.

Since k divides (m+2n), it implies k will also divide 3(m+2n). Therefore, k divides 3m+6n.

Similarly, we know that k divides 3m+4n.

We know that if two numbers a, and b both are divisible by c, then their difference (a-b) is also divisible by c.

By the same logic, we can say that {(3m+6n)-(3m+4n)} is divisible by k. Hence, 2n is also divisible by k.

Now, (m+2n) is divisible by k, it implies 2(m+2n) =2m+4n is also divisible by k.

Hence, {(3m+4n)-(2m+4n)} = m is also divisible by k.

Therefore, m, and 2n are also divisible by k.

The correct option is C

*Answer can only contain numeric values
Arun Sharma Test: Number System- 1 - Question 11
Save

In a 3-digit number N, the digits are non-zero and distinct such that none of the digits is a perfect square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of N is


Detailed Solution: Question 11

According to question, in the given 3-digit number N, the digits are non-zero and distinct.
So, the possible digits in 3-digit number = 2,3,5,6,7,8
It is also given that only one of the digits is a prime number.
So, the minimum possible value of N = 268
(2 is the smallest prime digit, and the non-prime digits has to be 6 and 8)
Now, 
268 = 4 × 67 = 22 × 67
So, the number of factors =  (2 + 1)(1 + 1) = 3 × 2 = 6

*Answer can only contain numeric values
Arun Sharma Test: Number System- 1 - Question 12
Save

Suppose a,b,c are three distinct natural numbers, such that 3ac = 8(a + b). Then, the smallest possible value of 3a + 2b + c is


Detailed Solution: Question 12

Our task is to minimise 3a +2b + c.

Here, the coefficient for cc is the minimum. 

3ac = 8(a + b)

We know that a, b, and c are natural numbers. So, the product acac should definately be a multiple of 8.

Case 1: a = 1, c = 8 and b = 2 ⇒ 3a+2b+c = 15

Case 2: a = 2, c = 4 and b = 1 ⇒ 3a+2b+c = 12

So, 12 is the correct answer.

Arun Sharma Test: Number System- 1 - Question 13
Save

The number of divisors of (26 × 35 × 53 × 72), which are of the form (3r + 1), where r is a non-negative integer, is

Detailed Solution: Question 13

The divisors of the given number will have the form 2a∗3b∗5c∗7d with 0 ≤ a ≤ 6, 0 ≤ b ≤ 5, 0 ≤ c ≤ 3, 0 ≤ d ≤ 2

Because the divisors should be in the form 3r + 1, it cannot be divisible by 3, so (b = 0). 

Reduce modulo 3: 2 mod 3 = 2,5  mod 3 = 2,7  mod 3 = 1

Hence,  2a5c7d ≡ 2a+c⋅1d ≡ 2a+c(mod 3)

2k will be in the form 3r+1 only when K is even. So, we need a+c to be even

a ∈ 0,…,6 has 4 even, 3 odd values 

c ∈ 0,…,3 has 2 even, 2 odd

Number of (a,c) with (a+c) even is 4⋅2 + 3⋅2 = 8+6 = 14

For each such pair, there are 3 choices for d = 0,1,2. Thus, total divisors in the form (3r+1) equals 14 × 3 = 42.

*Answer can only contain numeric values
Arun Sharma Test: Number System- 1 - Question 14
Save

The sum of digits of the number (625)65 × (128)36 is


Detailed Solution: Question 14

Thus, the number equals 58 followed by 252 zeros. Now (58 = 390625), and the sum of the digits of (390625) is
3 + 9 + 0 + 6 + 2 + 5 = 25.
Zeros add nothing, so the required sum of digits is 25.

Arun Sharma Test: Number System- 1 - Question 15
Save

 How many pairs of integers (x, y) exist such that the product of x, y and HCF(x, y) = 1080?

Detailed Solution: Question 15

  1. Let g = HCF(x, y). Write x = g·d, y = g·h with gcd(d, h) = 1.
    Then x·y·HCF(x, y) = (g d)(g h)g = g³ d h = 1080.

  2. Prime factorization: 1080 = 2³ · 3³ · 5.
    For g³ | 1080, g can only use primes with exponents ≤ 1 (since exponents triple):
    g ∈ {1, 2, 3, 6}.

  3. For each g, N = 1080 / g³ must be split as d·h with gcd(d, h) = 1.
    If N has k distinct primes, the number of ordered coprime splits (d, h) is 2^k
    (each prime-power goes wholly to d or to h).

  • g = 1 ⇒ N = 1080 (primes {2,3,5}, k=3) ⇒ 2³ = 8 ordered pairs

  • g = 2 ⇒ N = 135 = 3³·5 (primes {3,5}, k=2) ⇒ 4 ordered pairs

  • g = 3 ⇒ N = 40 = 2³·5 (primes {2,5}, k=2) ⇒ 4 ordered pairs

  • g = 6 ⇒ N = 5 (primes {5}, k=1) ⇒ 2 ordered pairs

Total ordered pairs = 8 + 4 + 4 + 2 = 18.
Since d ≠ h in all cases (N ≠ 1), unordered pairs = 18 / 2 = 9.

Hence, 9 pairs.

136 videos|255 docs|86 tests
Join Course
Information about Arun Sharma Test: Number System- 1 Page
In this test you can find the Exam questions for Arun Sharma Test: Number System- 1 solved & explained in the simplest way possible. Besides giving Questions and answers for Arun Sharma Test: Number System- 1, EduRev gives you an ample number of Online tests for practice
136 videos|255 docs|86 tests
Join Course
Download as PDF
;

About this CAT Practice Test

This test contains 15 MCQs on Arun Sharma Test: Number System- 1 with detailed solutions for each question. It is part of the Quantitative Aptitude (Quant) course on EduRev, designed to align with the current CAT syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

Explore more CAT Study Resources

If you found this Arun Sharma Test: Number System- 1 MCQ Test helpful, you can further enhance your CAT preparation with in-depth courses on EduRev, for all the subjects and topics available in Quantitative Aptitude (Quant) course. The course offers:
  • Video Lectures: covering all the topics of Quantitative Aptitude (Quant)
  • Notes for Revision: offering revision notes, important questions, summaries, flashcards, cheatsheets & more for CAT preparation
  • MCQ Test: variety of questions as per the pattern of CAT exam
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily