Test: Arun Sharma Based Level 3: Remainder & Divisibility - CAT MCQ

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

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

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

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.

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.

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.

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.

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.

720 = 2⁴ X 5¹ X 3²

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

In 77! there would be 25 + 8 + 2 = 35 threes -> hence [35/2] = 17 3^2s

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

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

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