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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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