Question

What is the remainder when (100!)^100 is divided by 23?

Answer 1

Answer:

0

Step-by-step explanation:

Formula : (a^n) mod m=(a mod m)^n

100! ^100 mod 23

=>(100! mod 23)^100

=>0^100

=>0

as 23 comes inside 100! if it cancelled the outcome will be 0.

Answer 2

Answer

When $(100!)^{100}$ is divided 23, remainder will be zero.

Step-by-step explanation:

TIP:

For this question we have to first notice if 23 is divided to whole expression of 100! to the power of 100 or 23 is divided to 100! which is then powered to 100. But in both the situation the remainder will remain 0. For both type of question the solution given below can be used. Only in second type of question we don't have to separate 100! instead we have to take 23 inside the bracket having power to 100.

SOLUTION:

$(100!)^{100}$ can also be written as  $(100!).(100!)^{99}$.

100! is a multiplication of all the number from 1 to 100 which includes the number 23.

So,

$\frac{(1.2......23.......99.100).(100!)^{99} }{23}$

The remainder of the above expression is 0.

To refer similar questions,

https://brainly.in/question/46583635

https://brainly.in/question/14910933

Thank you.