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## Number System: Factorials & No. of Zeros Test-1

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Question 1 |

What will be the highest power of 7 that divides the 39!

8 | |

3 | |

4 | |

5 |

Question 1 Explanation:

Since 7is prime number so we will find only number of sevens in 39!

So when we divide the 39 with 7 we get 5 as quotient and 4 as the remainder ,

since now quotient is smaller than the divisor and cannot be be divided further so the total number of sevens in 39! are 5

That means the highest power of 7 that divides the 39! Is 7

so (d) is the right answer

So when we divide the 39 with 7 we get 5 as quotient and 4 as the remainder ,

since now quotient is smaller than the divisor and cannot be be divided further so the total number of sevens in 39! are 5

That means the highest power of 7 that divides the 39! Is 7

^{5},so (d) is the right answer

Question 2 |

What will be the highest power of 8 that divides 88!

28 | |

33 | |

24 | |

25 |

Question 2 Explanation:

Since 8 can be written as 2 x 2 x 2 =2

So we will make pairs of 2

Now the number of two’s in 88! are 85

And the number of pairs of 2

so the maximum pair that can be made are 28 and finally the maximum power of 8

that can divide the 88! So the right choice for this question is (a)

^{3}So we will make pairs of 2

^{3}from the number of 2’s in 88!Now the number of two’s in 88! are 85

And the number of pairs of 2

^{3 }are 85/3= 28 with one 2 remainderso the maximum pair that can be made are 28 and finally the maximum power of 8

that can divide the 88! So the right choice for this question is (a)

Question 3 |

What is the highest power of 24 that can divide 80!

32 | |

34 | |

26 | |

28 |

Question 3 Explanation:

24 can be written as 2 x 2 x 2 x 3 = 2

So we will make pairs of 2

The number of 2’s in 80! are 78 ;

Number of 8’s in 80 ! is 78/3 = 26

The number of 3’s in 80 ! are 36

So the limiting power in 24 = power of 8 = 26

the highest power of 24 that can divide the 80! is 26.

^{3}x 3So we will make pairs of 2

^{3}x 3 from the number of two’s and number of threes ‘s in 80!The number of 2’s in 80! are 78 ;

Number of 8’s in 80 ! is 78/3 = 26

The number of 3’s in 80 ! are 36

So the limiting power in 24 = power of 8 = 26

the highest power of 24 that can divide the 80! is 26.

Question 4 |

Which of the following cannot be the number of zeros at the end of any factorial?

24 | |

27 | |

29 | |

31 |

Question 4 Explanation:

To solve this question we have to remember some points that the number of zeros are depends upon the number of pairs of 2 x 5
We know that number of zero in 5! = 1

5! to 9! = 1

10! to 14 ! = 2

15! to 19 ! = 3

20! to 24 ! = 4

25! to 29 ! = 6

Here the order of zero shifted to 4 to 6 because in 25! ,

25 it self is square of 5 i.e. 25 = 5 x5 therefore there are six 5’s in 25!

so the number of zeros are 6 . Come to the question

So we know that the number of zeros in 100! are 24

101! to 104 ! = 24

105! to 109 ! = 25

110! to 114 ! = 26

115! to 119 ! = 27

120! to 124 ! = 28

125! to 129 ! = 31

From this 125 is a multiple of 25 and can be written as 5 x 5x 5

So there will be the addition of 3 more zeros in it so numbers of zeros will shifted to 28 to 31.

Now see the options , option number (c) is not possible in any case

5! to 9! = 1

10! to 14 ! = 2

15! to 19 ! = 3

20! to 24 ! = 4

25! to 29 ! = 6

Here the order of zero shifted to 4 to 6 because in 25! ,

25 it self is square of 5 i.e. 25 = 5 x5 therefore there are six 5’s in 25!

so the number of zeros are 6 . Come to the question

So we know that the number of zeros in 100! are 24

101! to 104 ! = 24

105! to 109 ! = 25

110! to 114 ! = 26

115! to 119 ! = 27

120! to 124 ! = 28

125! to 129 ! = 31

From this 125 is a multiple of 25 and can be written as 5 x 5x 5

So there will be the addition of 3 more zeros in it so numbers of zeros will shifted to 28 to 31.

Now see the options , option number (c) is not possible in any case

Question 5 |

What will be the number of zeros in the end of (45!)

^{4!}10 | |

40! | |

40 | |

240 |

Question 5 Explanation:

Initially 45 can be written as = 3 x 3 x 5 ,So we need pair of 3

For this lets see the number of 3’s and 5’s in 45!

The number of 3’s in 45! are 21

Number of 5’s in 45 ! are 10

So we can make only 10 maximum pairs of 3

So the number of zeros in 45! are 10

Now we are asked for the number of zeros in (45!)

Then in 24 , 45! There would be 240 zeros

So the best answer for this question is option (d).

^{2}x 5For this lets see the number of 3’s and 5’s in 45!

The number of 3’s in 45! are 21

Number of 5’s in 45 ! are 10

So we can make only 10 maximum pairs of 3

^{2}x 5So the number of zeros in 45! are 10

Now we are asked for the number of zeros in (45!)

^{4!}So if in one 45! There are 10 zeroThen in 24 , 45! There would be 240 zeros

So the best answer for this question is option (d).

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