Understanding the concepts of factorial and trailing zeroes is crucial for excelling in competitive exams like CAT, SSC, Banking, GRE, and GMAT. Factorials, represented by “!”, involve multiplying positive integers, while trailing zeroes are significant when dealing with permutations, combinations, and number theory. Mastering these concepts is essential for tackling quantitative aptitude questions with confidence.
Multiple Choice Questions
Que 1: How many trailing zeros do 60! have?
(a) 14
(b) 12
(c) 10
(d) 8
Solution: 60/5=12
12/5=2
12+2=14
Option (a) is correct.
Que 2: What is the smallest positive integer value of n that satisfies the conditions: the factorial of n, denoted as n! When written in base 6, ends with 10 zeros, and when written in base 8, ends with 7 zeros. Additionally, how many different values of n satisfy these conditions?
(a) 24 and 4
(b) 25 and 3
(c) 24 and 3
(d) 25 and 4
Solution: n! When written in base 6, it ends with 10 zero. This means that it is divisible by 610 so it is also divisible by 310 and not divisible by 311.
From options, only 24! and 25! satisfies this condition.
Possible values of n= 24, 25, and 26
n! When written in base 8, ends with 7 zero. This means that it is divisible by 87 so it is also divisible by 221 and not divisible by 224.
24! and 25! both satisfy this condition.
Possible values of n= 24, 25, 26, and 27
So, the Smallest value of n=24
n can hold three values={24, 25, 26}
Option (c) is correct.
Que 3: If N is a positive integer less than 31, how many different values can n take such that (n + 1) is a factor of n!
(a) 18
(b) 16
(c) 12
(d) 20
Solution: If n+1 is a prime then it can’t be the factor of n!
Remove all (Prime-1)’s
1,2,4,6,10,12,16,18,22,28⇒10 numbers …….(i)
4 can not be the factor of 3!
31 can not be the factor of 30! ………..Exceptions
So remove 4 and 30⇒2 numbers …….(i)
Remaining numbers= 30-(10+2)=18
Option (a) is correct.
Que 4: How many different values can the natural number n take such that the n!, is a multiple of 76 but not a multiple of 79.
(a) 7
(b) 21
(c) 14
(d) 12
Solution: n=42, 43, 44, 45, 46, 47,…..55
42/7=6
55/7=7, 7/7=1…… Total=7+1=8
Option (c) is correct.
Que 5: How many different values can the natural number n take such that the n!, is a multiple of 220 but not a multiple of 320.
(a) 11
(b) 21
(c) 16
(d) 5
Solution: By hit and trial
24! is the smallest multiple of 222 and 310
From options
44! is the multiple of 241 and 319
Option (b) is correct.
Que 6: What is the highest power of 12 that divides the factorial of 54.
(a) 25
(b) 26
(c) 30
(d) 4
Solution: 2’s in 54!⇒ 27+13+6+3+1=50
So, the total 4’s in 54!= 25
3’s in 54!= 18+6+2=26
Total 25 pairs of 12 can be formed
Option (a) is correct.
Que 7: When the factorial of 40, denoted as 40!, is expressed in base 8 form, what is the last non-zero digit in the base 8 expansion?
(a) 2
(b) 4
(c) 4 or 5
(d) 2 and 6
Solution: To check the number of trailing zeroes in base 8, we need to calculate 8n=(23)n
2’s in 40!⇒ 20+10+5+2+1=38
Number of 8’s⇒ (23)12⇒ 12
So, the number of trailing zeroes is 12, and the rest non-zero part is the multiple of the remaining 22, which is 4.
Non-Zero part= 8k+4 (Divisible by 4 but not by 8)
So, the unit digit= 4
Option (b) is correct.
Que 8: What is the number of factors of (K – 1), where K is the largest number with exactly 3 factors that can divide 25!?
(a) 16
(b) 12
(c) 9
(d) 14
Solution: K has exactly three factors which mean K=p2, Here p is a prime number.
For the largest K, p=11
K=112=121
K-1=120=23×3×5
Number of factors= 4×2×2=16
Option (a) is correct.
Que 9: How many zeroes are there in the series of numbers from 1 to 1000?
(a) 189
(b) 190
(c) 192
(d) 195
Solution: 1 to 99⇒9
100 to 999⇒(11+9)×9=180
1000⇒3
Total zeroes= 9+180+3=192
Option (c) is correct.
Note this pattern:
Number of zeros in 1 to 99= 9
Number of zeros in 1 to 999= 189
Number of zeros in 1 to 9999= 2889
Number of zeros in 1 to 99999= 38889
Number of zeros in 1 to 999999= 488889…..
Que 10: How many zeroes are there in the product of all even numbers with less than 3 digits?
(a) 11
(b) 10
(c) 13
(d) 12
Solution: 2×4×6×8×……×98
=249×[1×2×3×…..×49]
=249×49!
Number of zeroes in 49!= 9+1=10
Option (b) is correct.
The Sum and Product of the Divisors of a Number: Click Here