This set of Aptitude Questions and Answers (MCQs) focuses on “Factorials – Set 2”.
1. Find the highest power of a number 18 in 100!.
a) 12
b) 28
c) 18
d) 24
View Answer
Explanation: 18 on prime factorization yields 2*32.
In 100!, power of 32 is less than power of 2. So, we shall find power of 32 i.e., 9 in 100!.
Highest power of 9 i.e., 32 in 100! = \(\Big(\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}+\frac{100}{3^4}\Big)\)/2 = (33 + 11 + 3 + 1)/2 = 24.
Therefore, highest power of a number 18 in 100! Is 24.
2. N! is having 37 zeroes at its end. How many values of N is/are possible?
a) 1
b) 2
c) 5
d) 6
View Answer
Explanation: The number of zero is determined by number of 5’s and 2’s, whichever is less.
In most of the cases, number of 5’s is less than number of 2’s.
\(\frac{N}{5}+\frac{N}{5^2}+\frac{N}{5^3}+\frac{N}{5^4}\)+⋯=37.
This gives the value of N = 150, 151, 152, 153, 154 i.e., 5 values.
3. Find the number of zeroes at the end of (55!)4!.
a) 312
b) 282
c) 318
d) 324
View Answer
Explanation: The number of zero is determined by number of 5’s and 2’s, whichever is less.
In most of the cases, number of 5’s is less than number of 2’s.
Highest power of 5 in 55! = \(\Big(\frac{55}{5}+\frac{55}{5^2}\Big)\) = 11 + 2 = 13.
4! = 4*3*2*1 = 24.
Therefore, the number of zeroes at the end of (55!)4! = 24 * 13 = 312.
4. (165!)/20n, is an integer. Find the highest possible value of n for this to be true.
a) 40
b) 50
c) 52
d) 32
View Answer
Explanation: The number of zero is determined by number of 5’s and 2’s, whichever is less.
In most of the cases, number of 5’s is less than number of 2’s.
20 = 5*22.
Highest power of 5 in the product = \(\frac{165}{5}+\frac{165}{5^2}+\frac{165}{5^3}\) = 33 + 6 + 1 = 40.
Therefore, the highest possible value of n for this to be true is 40.
5. What is the highest power of 3 available in the expression 58! – 38!?
a) 17
b) 18
c) 19
d) 16
View Answer
Explanation: 58! – 38! = 38! (58*57*56……39 – 1)
It depends on highest power of 3 in 38!.
highest power of 3 in 38! = \(\frac{38}{3}+\frac{38}{9}+\frac{38}{27}\) = 12 + 4 + 1 = 17.
6. Find the number of factors of 7!.
a) 60
b) 72
c) 48
d) 84
View Answer
Explanation: 7! = 7*6*5*4*3*2*1 = 24*32*5*7.
Therefore, the number of factors = (4+1)*(2+1)*(1+1)*(1+1) = 60.
7. Find the number of positive integers which divides (24)!.
a) 6536
b) 7536
c) 3576
d) 5376
View Answer
Explanation: (24)! = 16! = 215*36*53*72*11*13
Therefore, the number of positive integers which divides (24)! = (15+1)(6+1)(3+1)(2+1)(1+1)(1+1) = 5376.
8. How many factors of 19! are there, whose unit digit is 5?
a) 1296
b) 1496
c) 996
d) 1096
View Answer
Explanation: We need to find the odd factors of 19!
19! = 216*38*53*72*111*131*171*191.
The required answer is 9*3*3*2*2*2*2 = 1296.
9. N = 1! – 2! + 3! – 4! + …… + 47! – 48! + 49! – 50! + 51!. Then what is the unit digit of NN?
a) 1
b) 0
c) 5
d) 2
View Answer
Explanation: From 5! onwards, each of the numbers would have unit digit of 0. Hence, the unit digit of the given number depends on unit digit of 1! – 2! + 3! – 4!, which would be 1-2+6-4 = 1. Since N is a number that has units digit of 1, when it is raised to any power, the units digit would not change.
Therefore, the unit digit of NN is 1.
10. (3132!)10 = (x)34, then what will be the number of consecutive zeroes at the end of x?
a) 124
b) 194
c) 164
d) 144
View Answer
Explanation: In base 34, 10 means 34. In base 10, 10 is obtained by multiplying 2 and 5. In base 34, it is obtained by multiplying 2 and 17. Number of consecutive zeroes in base 34 at the end of the number is same as the number of 2’s and 17’s in 3132!. Since, the number of 2’s is much more than number of 17’s, so we have to count number of 17’s in 3132!.
Maximum power of 17 in 3132! = \(\frac{3132}{17}+\frac{3132}{17^2}\)=184+10=194.
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