This set of Aptitude Questions and Answers (MCQs) focuses on “Factorials”.
1. Find the highest power of a prime number 2 in 50!.
a) 47
b) 50
c) 45
d) 42
View Answer
Explanation: Highest power of a prime number 2 in 50! = \(\frac{50}{2}+\frac{50}{2^2}+\frac{50}{2^3}+\frac{50}{2^4}+\frac{50}{2^5}\)
= 25 + 12 + 6 + 3 + 1 = 47
2. Find the number of zeroes at the end of the product of the numbers from 1 to 75.
a) 24
b) 20
c) 18
d) 22
View Answer
Explanation: Only multiples of 2 and 5 yield zeroes on multiplication.
In the product power of 5 is less than power of 2. So, we shall find power of 5 in the product.
Highest power of 5 in the product = \(\frac{75}{5}+\frac{75}{5^2}\) = 15 + 3 = 18.
Therefore, the number of zeroes at the end of the product of the numbers from 1 to 75 is 18.
3. Find the highest power of 11 that completely divides 130!.
a) 8
b) 14
c) 12
d) 10
View Answer
Explanation: The highest power of 11 that completely divides 130! is,
= \(\frac{130}{11}+\frac{130}{11^2}\)=11+1=12.
4. Find the value of \(\frac{10!}{7!}\).
a) 720
b) 810
c) 630
d) 1000
View Answer
Explanation: \(\frac{10!}{7!}=\frac{10*9*8*7!}{7!}\)=720.
5. What is the highest power of 8 in 98!?
a) 29
b) 35
c) 31
d) 27
View Answer
Explanation: The highest power of 8 i.e., 23 in 98! is,
= \(\Big(\frac{98}{2}+\frac{98}{2^2}+\frac{98}{2^3}+\frac{98}{2^4}+\frac{98}{2^5}+\frac{98}{2^6} \Big)\)/3=(49+24 +12+6+3+1)/3=31.
6. Which of the following cannot be the number of zeroes at the end of any factorial?
a) 29
b) 27
c) 24
d) 31
View Answer
Explanation: We know that number of zeroes in 100! is 24.
i.e., from 100! to 104!, there are 24 zeroes.
From 105! to 109!, there are 25 zeroes.
From 110! to 114!, there are 26 zeroes.
From 115! to 119!, there are 27 zeroes.
From 120! to 124!, there are 28 zeroes.
From 125! to 129!, there are 31 zeroes, because there are three 5’s in 125 i.e., 53.
Therefore, 29 cannot be the number of zeroes at the end of any factorial.
7. What is the value of \(\frac{0!}{(2!)^0}\)?
a) 1
b) 1/2
c) 0
d) ∞
View Answer
Explanation: We know that, 0! = 1 and a0 = 1 i.e., (2!)0 = 1
Therefore, \(\frac{0!}{(2!)^0}\) = 1.
8. Find the value of \(\frac{(n-1)!}{(n+1)!}\).
a) \(\frac{1}{n^2-n} \)
b) \(\frac{n^2-n}{n^4-n^2} \)
c) \(\frac{1}{n+1} \)
d) \(\frac{n+1}{n} \)
View Answer
Explanation: \(\frac{(n-1)!}{(n+1)!}=\frac{1*2*3……(n-1)}{1*2*3……(n-1)*n*(n+1)}=\frac{1}{n^2+n}\).
\(\frac{(n-1)!}{(n+1)!}=\frac{1}{n^2+n}=\frac{n^2-n}{(n^2+n)*(n^2-n)}=\frac{n^2-n}{n^4-n^2} \).
9. Find the number of factors of \(\frac{6!}{2!*4!}\).
a) 4
b) 8
c) 2
d) 12
View Answer
Explanation: \(\frac{6!}{2!*4!}=\frac{1*2*3*4*5*6}{1*2*1*2*3*4}\)=15=31*51.
Therefore, the number of factors are (1+1)*(1+1) = 4.
10. Which of the following is/are true?
I. The factorial value of all number greater than 4 is divisible by 10.
II. The factorial value of all number is even except 1!.
a) Only I
b) Only II
c) Both I & II
d) Neither I nor II
View Answer
Explanation: We know that factorial value of all number greater than 4 contains zeroes at the end. So, they are divisible by 10. Hence, statement I is true.
The factorial value of all number is even except 1! and 0! i.e., 0! = 1! = 1. Hence, statement II is false.
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