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

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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 x…………101 x 100! + 100!

= 100!( 135 x 134 x 132 x…………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 x…………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 x…………101 x 100! + 100!

= 100!( 135 x 134 x 132 x…………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 x…………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! – 36!

= 36!(37 – 1)

= 36 ! (36)

In 36!, the number of 2’s are 34 and the number of three’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 6’s.

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

= 37 x 36! – 36!

= 36!(37 – 1)

= 36 ! (36)

In 36!, the number of 2’s are 34 and the number of three’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 6’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 let’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 let’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 2’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 2’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 two’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 two’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 .

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