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CAT Exam  >  CAT Test  >  Quantitative Aptitude (Quant)  >  Test: Arun Sharma Based Level 2: Remainder & Divisibility - CAT MCQ

Arun Sharma Based Level 2: Remainder & Divisibility - Free MCQ Practice


MCQ Practice Test & Solutions: Test: Arun Sharma Based Level 2: Remainder & Divisibility (10 Questions)

You can prepare effectively for CAT Quantitative Aptitude (Quant) with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Arun Sharma Based Level 2: Remainder & Divisibility". These 10 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: 15 minutes
  • - Number of Questions: 10

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Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 1
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Find the remainder when 73 + 75 + 78 + 57 + 197 is divided by 34.

Detailed Solution: Question 1

The remainder would be given by: (5 + 7 + 10 + 23 + 27)/34 = 72/34
-> remainder = 4. Option (b) is correct.

Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 2
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Find the remainder when 51203 is divided by 7.

Detailed Solution: Question 2

51203/7 -> 2203/7 = (23)67 X 22/7 = 867 X 4/7 -> remainder = 4. Option (a) is correct

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Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 3
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Find the remainder when 21875 is divided by 17.

Detailed Solution: Question 3

21875/17 -> 4875/17 = (44)n X 43/17 = 256n X 64/17 -> 1n X 13/17 -> remainder =13. Option (b) is correct.

Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 4
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Find the number of consecutive zeroes at the end of:

57 X 60 X 30 X 15625 X 4096 X 625 X 875 X 975

Detailed Solution: Question 4

The given expression has fifteen 2’s and seventeen 5’s. The number of zeroes would be 15 as the number of 2’s is lower in this case. Option (d) is correct.

Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 5
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Find the number of consecutive zeroes at the end of the following numbers. -  100! x 200! 

Detailed Solution: Question 5

The number of zeroes would depend on the number of 5’s in the value of the factorial.

100! would end in 20 + 4 = 24 zeroes

200! Would end in 40 + 8 + 1 = 49 zeroes.

When you multiply the two numbers (one with 24 zeroes and the other with 49 zeroes at it’s end), the resultant total would end in 24 + 49 = 73 zeroes. Option (b) is correct.

Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 6
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Find the number of consecutive zeroes at the end of the following numbers. 72!

Detailed Solution: Question 6

The number of zeroes would depend on the number

of 5’s in the value of the factorial. 72! -> 14 + 2 = 16. Option (d) is correct.

Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 7
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Find the maximum value of n such that 50! is perfectly divisible by 2520". 

Detailed Solution: Question 7

2520=7 X 32 X 23 X 5.

The value of n would be given by the value of the number of 7s in 50! This Value Is Equal To[50/7] + [50/49] = 7 + 1 = 8 Option (b) is correct.

Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 8
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Find the maximum value of n such that 570 X 60 X 30 X 90 X 100 X 500 X 700 X 343 X 720 X 81 is perfectly divisible by 30n.

Detailed Solution: Question 8

Checking for the number of 2’s, 3’s and 5’s in the given expression you can see that the minimum is for the number of 3’s (there are 11 of them while there are 12 5’s and more than 11 2’s) Hence, option (b) is correct.

Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 9
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Find the maximum value of n such that 77! is per-fectly divisible by 720n.

Detailed Solution: Question 9

720 = 24 X 51 X 32

In 77! there would be 38 + 19 + 9 + 4 + 2 + 1 = 73 twos Æ hence [73/4] = 18 24s

In 77! there would be 25 + 8 + 2 = 35 threes -> hence [35/2] = 17 32s

In 77! there would be 15 + 3 = 18 fives

Since 17 is the least of these values, option (c) is correct.

Test: Arun Sharma Based Level 2: Remainder & Divisibility - Question 10
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Find the number of consecutive zeroes at the end of the following numbers. 100! + 200! 

Detailed Solution: Question 10

The number of zeroes would depend on the number of 5’s in the value of the factorial. 100! wouldendin 20 + 4 = 24zeroes 200!Wouldendin 40 + 8 + 1 = 49zeroes.

When you add the two numbers (one with 24 zeroes and the other with 49 zeroes at it’s end), the resultant total would end in 24 zeroes. Option (b) is correct

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About this CAT Practice Test

This test contains 10 MCQs on Arun Sharma Based Level 2: Remainder & Divisibility 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.

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If you found this Arun Sharma Based Level 2: Remainder & Divisibility 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:
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