# DIVISIBILITY

## DIVISIBILITY

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Question 1 |

Find the number of divisors of 1728. ?

A | 18 |

B | 30 |

C | 28 |

D | 20 |

Question 2 |

Find the number of divisors of 544 excluding 1 and 544. ?

A | 12 |

B | 18 |

C | 11 |

D | 10 |

Question 3 |

Find the number of divisors of 1080 excluding the divisors, which are perfect squares ?

A | 28 |

B | 29 |

C | 30 |

D | 31 |

Question 4 |

Find the number of divisors of 544 excluding 1 and 544 ?

A | 12 |

B | 18 |

C | 11 |

D | 10 |

Question 5 |

Find the number of divisors 544 which are greater than 3. ?

A | 15 |

B | 10 |

C | 12 |

D | 11 |

Question 6 |

Find the sum of divisors of 544 excluding 1 and 544. ?

A | 1089 |

B | 545 |

C | 589 |

D | 1134 |

Question 7 |

Find the sum of divisors of 544 which are perfect squares. ?

A | 32 |

B | 64 |

C | 42 |

D | 21 |

Question 8 |

Find the sum of odd divisors of 544. ?

A | 18 |

B | 34 |

C | 68 |

D | 36 |

Question 9 |

Find the sum of even divisors of 4096 ?

A | 8192 |

B | 6144 |

C | 8190 |

D | 6142 |

Question 10 |

Find the sum the sums of divisors of 144 and 160 ?

A | 589 |

B | 781 |

C | 735 |

D | None of these |

Question 11 |

Find the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600 ?

A | 639 |

B | 735 |

C | 651 |

D | 589 |

Question 12 |

Find the maximum value of n such that 157! is perfectly divisible by 10^n ?

A | 37 |

B | 38 |

C | 16 |

D | -1.15 |

Question 12 Explanation:

[157/5] = 31. [31/5] = 6. [6/5] = 1. 31 + 6 + 1 = 38.

Question 13 |

Find the maximum value of n such that 157! is perfectly divisible by 12^n ?

A | 75 |

B | 76 |

C | 77 |

D | 78 |

Question 13 Explanation:

No of 2’s in 157! = [157/2] + [157/4] + [157/8]… + [157/128] = 78 + 39 + 19 + 9 + 4 + 2 + 1 =152. Hence, the number of 2^2 s would be [152/2] = 76.
Number of 3’s in 157! = 52 + 17 + 5 +1 = 75.
The answer would be given by the lower of these values. Hence, 75

Question 14 |

Find the maximum value of n such that 157! is perfectly divisible by 18^n ?

A | 37 |

B | 38 |

C | 39 |

D | 40 |

Question 14 Explanation:

From the above solution:Number of 2’s in 157! = 152
Number of 3^2's in 157! = [75/2]=37

Question 15 |

Find the maximum value of n such that 570 × 60 × 30 × 90 × 100 × 500 × 700 × 343 × 720 × 81 is perfectly divisible by 30^n ?

A | 11 |

B | 12 |

C | 13 |

D | 14 |

Question 15 Explanation:

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)

Question 16 |

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

A | 17 |

B | 9 |

C | 8 |

D | 16 |

Question 16 Explanation:

The number of zeroes would depend on the number of 5’s in the value of the factorial. 72! then 14 + 2 = 16.

Question 17 |

Find the number of consecutive zeroes at the end of the following numbers in 77! × 42! ?

A | 24 |

B | 9 |

C | 27 |

D | 18 |

Question 17 Explanation:

The number of zeroes would depend on the number of 5’s in the value of the factorial.
77! × 42! Æ 15 + 3 = 18 (for 77!) and 8 + 1 = 9 (for 42!).
Thus, the total number of zeroes in the given expression would be 18 + 9 = 27

Question 18 |

Find the number of consecutive zeroes at the end of the following numbers in

**100! + 200!**?A | 73 |

B | 24 |

C | 11 |

D | 22 |

Question 18 Explanation:

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 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.

Question 19 |

Find the number of consecutive zeroes at the end of the following numbers in

**57 × 60 × 30 × 15625 × 4096 × 625 × 875 × 975 ?**A | 6 |

B | 16 |

C | 17 |

D | 15 |

Question 19 Explanation:

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.

Question 20 |

Find the number of consecutive zeroes at the end of the following numbers in

**1! × 2! × 3! × 4! × 5! × - - - - - - - × 50! ?**A | 235 |

B | 12 |

C | 262 |

D | 105 |

Question 20 Explanation:

1! to 4! would have no zeroes while 5! to 9! All the values would have 1 zero. Thus, a total of 5 zeroes till 9!. Going further 10! to 14! would have two zeroes each – so a total of 10 zeroes would
come out of the product of 10! × 11! × 12! × 13! × 14!.
Continuing this line of thought further we get:
Number of zeroes between 15! × 16!... × 19! = 3 + 3 + 3 + 3 + 3 = 3 × 5 = 15
Number of zeroes between 20! × 21!... × 24! = 4 × 5 = 20
Number of zeroes between 25! × 26!... × 29! = 6 × 5 = 30
Number of zeroes between 30! × 31!... × 34! = 7 × 5 = 35
Number of zeroes between 35! × 36!... × 39! = 8 × 5 = 40
Number of zeroes between 40! × 41!... × 44! = 9 × 5 = 45
Number of zeroes between 45! × 46!... × 49! = 10 × 5 = 50
Number of zeroes for 50! = 12
Thus, the total number of zeroes for the expression 1! × 2! × 3! …. × 50! = 5 + 10 +15 + 20 + 30 +
35 + 40 + 45 + 50 + 12 = 262 zeroes.

Question 21 |

Find the number of consecutive zeroes at the end of the following numbers in

**1^1 × 2^2 × 3^3 × 4^4 × 5^5 × 6^6 × 7^7 × 8^8 × 9^9 × 10^10**.A | 25 |

B | 15 |

C | 10 |

D | 29 |

Question 21 Explanation:

The number of 5’s is `15 while the number of 2’s is much more.

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