The remainder when 98! is divided by 101 is equal to ____________________________Correct answer is between '49.9,50.1'. Can you explain this answer?

Question

Answer

Solution:

To find out the remainder when 98! is divided by 101, we can use Wilson's theorem.

Wilson's theorem: If p is a prime number, then (p-1)! ≡ -1 (mod p)

Here, p = 101, so (p-1)! ≡ -1 (mod p)

Now, we can write 98! as (98 x 97 x 96 x ... x 2 x 1)

We can group the terms in pairs such that each pair multiplies to give 101. For example, (2 x 51), (3 x 34), (4 x 26), etc.

There are 48 such pairs, and we are left with a single term 50.

So, 98! ≡ (50 x (product of 48 pairs)) (mod 101)

Now, we need to find the product of the 48 pairs.

Each pair multiplies to give 101, so their product is 101^48.

So, 98! ≡ (50 x 101^48) (mod 101)

Using Wilson's theorem, we know that (101-1)! ≡ -1 (mod 101)

So, 100! ≡ -1 (mod 101)

Dividing both sides by 101, we get:

(100 x 99 x 98! ≡ -1 (mod 101)

Substituting the value of 98! from above, we get:

(100 x 99 x 50 x 101^48) ≡ -1 (mod 101)

Simplifying this expression, we get:

(99 x 50) ≡ -1 (mod 101)

4950 ≡ -1 (mod 101)

Dividing both sides by 101, we get:

49 ≡ 50^-1 (mod 101)

So, the remainder when 98! is divided by 101 is 50^-1, which is approximately 0.02 less than 50. Therefore, the correct answer is between 49.9 and 50.1.

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