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Question 1: What is the remainder when 96! is divided by 291 ?
a) 86
b) 96
c) 76
d) 66
Question 2: What is the remainder when $120!$ is divided by $120^{24}$ ? ?
a) 2
b) 1
c) 0
d) 3
Question 3: Which digit from the right end of 933! is the first non-zero digit?
a) 300
b) 232
c) 400
d) 320
Question 4: In which position from the right is the first non-zero digit present in 334!?
a) 81st
b) 82nd
c) 83rd
d) 84th
Question 5: Find the number of zeroes at the end of 179!.
a) 32
b) 51
c) 43
d) 28
Answers & Solutions:
1) Answer (B)
Using Wilson’s theorem, 96! + 1 is divisible by 97.
Let 96! + 1 be equal to 97k for some natural number k.
Note that 291 = 97*3.
Hence, 96! + 1 is either 291p or 291p + 97 or 291p + 194 for some natural number p. This is because 96! + 1 is a multiple of 97.
We know that 96! is divisible by 3. So, 96! + 1 leaves a remainder of 1 when divided by 3. Using this, we can conclude that 96! + 1 is of the form 291p + 96.
Hence, 96! leaves a remainder of 96 when divided by 291.
2) Answer (C)
The product of any five consecutive numbers is divisible by 5!
So, 1*2*3*4*5 is divisible by 120.
Similarly, 6*7*8*9*10 is also divisible by 120.
120! can be written as (1*2*3*4*5)*(6*7*8*9*10)*…*(116*117*118*119*120)
Each of the 24 terms is divisible by 120
Hence, the product is divisible by
So, the remainder when 120!120! is divided by
is 0.
3) Answer (B)
Number of zeros in 933! = $\left \lfloor \frac{933}{5} \right \rfloor + \left \lfloor \frac{933}{25} \right \rfloor + \left \lfloor \frac{933}{125} \right \rfloor + \left \lfloor \frac{933}{625} \right \rfloor + \left \lfloor \frac{933}{3125} \right \rfloor + ..$
where $\left \lfloor x \right \rfloor$ means greatest integer less than or equal to x
= 186 + 37 + 7 + 1 + 0
= 231
=> 232nd digit from right end is the first non-zero digit.
4) Answer (B)
Number of zeros = $\left \lfloor \frac{334}{5} \right \rfloor$ + $\left \lfloor \frac{334}{25} \right \rfloor$ + $\left \lfloor \frac{334}{125} \right \rfloor$ + $\left \lfloor \frac{334}{625} \right \rfloor$
= 66 + 13 + 2 + 0
= 81
=> 82nd digit is the first non-zero digit
5) Answer (C)
Number of zeroes in 179! = $\left \lfloor \frac{179}{5} \right \rfloor + \left \lfloor \frac{179}{5^2} \right \rfloor + \left \lfloor \frac{179}{5^3} \right \rfloor + \left \lfloor \frac{179}{5^4} \right \rfloor$
= 35 + 7 + 1 + 0
= 43
