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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