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## 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%%
 Question 1
What will be the number of trailing zeros in 135! + 100!
 A 34 B 24! C 24 D 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
 Question 2
What will be the highest power of 6 that can divide 37! -36!
 A 18 B 18! C 19 D 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.
 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?
 A 4 B 3 C 2 D 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
 Question 4
By what least number should 36! be divided such that the quotient is not multiple of 4?
 A 34 B 217 C 12 D 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 217 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!
 A 150 B 156 C 158 D 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 .
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