(n-1)!/n

For 32,

31!/32

32 = 4 x 8 --> both of these numbers exist in 31!.. remainder should be 0

What is the OA? Also, please post the answer choices along with Questions

Thanks OA is 0

CSASHISHPANDAY wrote:Let n~ be defined for all positive integers n as the remainder when (n - 1)! is divided by n.

What is the value of 32~ ?
A. 0
B. 1
C. 2
D. 8
E. 31

32~ = (32 - 1)!/32 = 31!/32 = 31!/2^5

Since we can safely say that there are at least five 2s in 31! (for example, 31! has the factors 16 = 2^4 and 8=2^3), the remainder is zero.

Alternate Solution:

32~ = (32 - 1)!/32 = 31!/32 = 31!/2^5

We want to know the remainder when 31! is divided by 2^5. If we can establish that there are at least 5 factors of 2 in 31!, then we will know that 2^5 evenly divides into 31!, which means that the remainder would be 0. Let's determine if we can find at least 5 twos in 31!:

31! = 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x ... x 1

31! = 31 x (2 x 15) x 29 x (2 X 2 x 14) x 27 x (2 x 13) x 25 x (2 x 2 x 2 x 3) x ... x 1

Notice that we have found seven 2s, which is two more than what we needed. Thus, we know that 2^5 will evenly divide into 31!, leaving a remainder of 0.

Answer: A