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Test: Arun Sharma Based Level 3: Remainder & Divisibility


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10 Questions MCQ Test Quantitative Aptitude (Quant) | Test: Arun Sharma Based Level 3: Remainder & Divisibility

Test: Arun Sharma Based Level 3: Remainder & Divisibility for CAT 2022 is part of Quantitative Aptitude (Quant) preparation. The Test: Arun Sharma Based Level 3: Remainder & Divisibility questions and answers have been prepared according to the CAT exam syllabus.The Test: Arun Sharma Based Level 3: Remainder & Divisibility MCQs are made for CAT 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Arun Sharma Based Level 3: Remainder & Divisibility below.
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Test: Arun Sharma Based Level 3: Remainder & Divisibility - Question 1

Find the remainder when 73 + 75 + 78 + 57 + 197 is divided by 34.

Detailed Solution for Test: Arun Sharma Based Level 3: Remainder & Divisibility - 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 3: Remainder & Divisibility - Question 2

Find the remainder when 51203 is divided by 7.

Detailed Solution for Test: Arun Sharma Based Level 3: Remainder & Divisibility - Question 2

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

Test: Arun Sharma Based Level 3: Remainder & Divisibility - Question 3

Find the remainder when 21875 is divided by 17.

Detailed Solution for Test: Arun Sharma Based Level 3: Remainder & Divisibility - 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 3: Remainder & Divisibility - Question 4

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 for Test: Arun Sharma Based Level 3: Remainder & Divisibility - 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 3: Remainder & Divisibility - Question 5

Find the number of consecutive zeroes at the end of the following numbers. -  100! x 200! 

Detailed Solution for Test: Arun Sharma Based Level 3: Remainder & Divisibility - 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 3: Remainder & Divisibility - Question 6

Find the number of consecutive zeroes at the end of the following numbers. 72!

Detailed Solution for Test: Arun Sharma Based Level 3: Remainder & Divisibility - 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 3: Remainder & Divisibility - Question 7

Find the maximum value of n such that 50! is perfectly divisible by 2520". 

Detailed Solution for Test: Arun Sharma Based Level 3: Remainder & Divisibility - 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 3: Remainder & Divisibility - Question 8

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 for Test: Arun Sharma Based Level 3: Remainder & Divisibility - 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 3: Remainder & Divisibility - Question 9

Find the maximum value of n such that 77! is per-fectly divisible by 720n.

Detailed Solution for Test: Arun Sharma Based Level 3: Remainder & Divisibility - 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 3: Remainder & Divisibility - Question 10

Find the number of consecutive zeroes at the end of the following numbers. 100! + 200! 

Detailed Solution for Test: Arun Sharma Based Level 3: Remainder & Divisibility - 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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