- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.

## Number System: Factorials & No. of Zeros Test-2

Congratulations - you have completed

*Number System: Factorials & No. of Zeros Test-2*.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
Your answers are highlighted below.

Question 1 |

What will be the number of trailing zeros in 135! + 100!

34 | |

24! | |

24 | |

none |

Question 1 Explanation:

135! + 100!

= 135 x 134 x 132 xâ€¦â€¦â€¦â€¦101 x 100! + 100!

= 100!( 135 x 134 x 132 xâ€¦â€¦â€¦â€¦101 + 1)

Now the number of zeros in 100! are 24 .

And if you see the last digit of this term ( 135 x 134 x 132 xâ€¦â€¦â€¦â€¦101 + 1) is 1

so there is no any zero I the end of this term so the number of zeros in 135! + 100! are 24

= 135 x 134 x 132 xâ€¦â€¦â€¦â€¦101 x 100! + 100!

= 100!( 135 x 134 x 132 xâ€¦â€¦â€¦â€¦101 + 1)

Now the number of zeros in 100! are 24 .

And if you see the last digit of this term ( 135 x 134 x 132 xâ€¦â€¦â€¦â€¦101 + 1) is 1

so there is no any zero I the end of this term so the number of zeros in 135! + 100! are 24

Question 2 |

What will be the highest power of 6 that can divide 37! -36!

18 | |

18! | |

19 | |

cannot be determined |

Question 2 Explanation:

37! -36!

= 37 x 36! â€“ 36!

= 36!(37 â€“ 1)

= 36 ! (36)

In 36! , the number of 2â€™s are 34 and the number of threeâ€™s are 17.

So the number of pairs that can formed from this are 17.

Remember, the additional 36 is also here.

So we need two more 6â€™s.

So the maximum power that can divide the 37! -36! is 19.

= 37 x 36! â€“ 36!

= 36!(37 â€“ 1)

= 36 ! (36)

In 36! , the number of 2â€™s are 34 and the number of threeâ€™s are 17.

So the number of pairs that can formed from this are 17.

Remember, the additional 36 is also here.

So we need two more 6â€™s.

So the maximum power that can divide the 37! -36! is 19.

Question 3 |

If (n+1)! has 2 more zeros at the end as compared to n!, how many two digit values (n+1) can be assume?

4 | |

3 | |

2 | |

none |

Question 3 Explanation:

We know that the number of zero depends on the pairs of 2 x 5.

So when there is increase in the number of fives, then a zero is increased.

So letâ€™s calculate the number of values:

5! to 9! = 1

10! to 14 ! = 2

15! to 19 ! = 3

20! to 24 ! = 4

25! to 29 ! = 6

That means on every multiplier of 25, there is an increase by 2 in the number of zeros.

This means that the value of (n+1) will be a factor of five square.

There are three such two-digit values possible:

25 x 1 = 25

25 x 2 = 50

25 x 3 = 75

So when there is increase in the number of fives, then a zero is increased.

So letâ€™s calculate the number of values:

5! to 9! = 1

10! to 14 ! = 2

15! to 19 ! = 3

20! to 24 ! = 4

25! to 29 ! = 6

That means on every multiplier of 25, there is an increase by 2 in the number of zeros.

This means that the value of (n+1) will be a factor of five square.

There are three such two-digit values possible:

25 x 1 = 25

25 x 2 = 50

25 x 3 = 75

Question 4 |

By what least number should 36! be divided such that the quotient is not multiple of 4?

3 ^{4} | |

2 ^{17} | |

12 | |

18 |

Question 4 Explanation:

The question has asked for the highest power of 4 that can divide 36

The quotient power should not be the multiple of 4.

Now the number 4 = 2 x 2

Therefore we need the pairs of 2 x2

So the number of 2â€™s in 36! ,are 34

So we can make 17 pairs of 2

So the least number is 2

is divided, the quotient is not multiple of 4

The quotient power should not be the multiple of 4.

Now the number 4 = 2 x 2

Therefore we need the pairs of 2 x2

So the number of 2â€™s in 36! ,are 34

So we can make 17 pairs of 2

So the least number is 2

^{17}by which when 36!is divided, the quotient is not multiple of 4

Question 5 |

What will be the number of zeros in the end of 625!

150 | |

156 | |

158 | |

15! |

Question 5 Explanation:

This is very simple question in the end of the test

Number of zeros always depends on the 2 x 5 pair so number of twoâ€™s in 625 ! are 520

And the number of fives are 156

Therefore maximum pair can be formed are 156

So the zeros at the end of 625! Will be 156 .

Number of zeros always depends on the 2 x 5 pair so number of twoâ€™s in 625 ! are 520

And the number of fives are 156

Therefore maximum pair can be formed are 156

So the zeros at the end of 625! Will be 156 .

Once you are finished, click the button below. Any items you have not completed will be marked incorrect.

There are 5 questions to complete.

List |