# Wilson’s Theorem for CAT PDF

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Wilson’s Theorem for CAT PDF gives the clear explanation and example questions for Wilson’s Theorem. This an very important Remainder Theorem for CAT. Remainder theorem comes under the topic of Number systems for CAT. This theorem is easy to remember the questions will be generally asked on the application of this theorem.

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Wilson’s Theorem for CAT

According to Wilson’s theorem for prime number ‘p’, [(p-1)! + 1] is divisible by p.

In other words, (p-1)! leaves a remainder of (p-1) when divided by p.

Thus, (p-1)! mod p = p-1

For e.g.

4! when divided by 5, we get 4 as a remainder.

6! When divided by 7, we get 6 as a remainder.

10! When divided by 11, we get 10 as a remainder.

If we extend Wilson’s theorem further, we get an important corollary

(p-2)! mod p = 1

As from the Wilson’s theorem we have, (p-1)! mod p = (p-1)

Thus, [(p-1)(p-2)!] mod p = (p-1)

This will be equal to [(p-1) mod p] * [(p-2)! mod p] = (p-1)

For any prime number ‘p’, we observe that (p-1) mod p = (p-1). For e.g. 6 mod 7 will be 6.

Thus, (p-1) * [(p-2)! mod p] = (p-1)

Thus for RHS to be equal to LHS,

(p-2)! mod p = 1

Hence,

5! mod 7 will be 1

51! mod 53 will be 1

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Examples for Wilson’s Theorem for CAT:

Q.1) What will be the remainder when 568! Is divided by 569?

Solution:

According to Wilson’s theorem we have,

For prime number ‘p’, (p-1)! mod p = (p-1)

In this case 569 is a prime number.

Thus, 568! mod 569 = 568.

Hence, when 568! is divided by 569 we get 568 as remainder.

Q.2) What will be the remainder when 225! Is divided by 227?

Solution:

We know that for prime number ‘p’, (p-2)! mod p = 1.

In this case, 227 is a prime number.

Thus, 225! mod 227 will be equal to 1.

In other words, when 225! Is divided by 227 we get remainder as 1.

Q.3) What will be the remainder when 15! is divided by 19?

Solution:

19 is a prime number.

From corollary of Wilson’s theorem, for prime number ‘p’,  (p-2)! mod p = 1

Thus,

17! mod 19 = 1

[17*16*15!] mod 19 = 1

[17 mod 19]* [16 mod 19]* [15! mod 19] = 1

[-2]*[-3]* [15! mod 19] = 1

[6 * 15!] mod 19 = 1

Multiplying both sides by 3, we get

[18*15!] mod 19 = 3

[-1*15!] mod 19 = 3

Multiplying both sides by ‘-1’, we get

15! mod 19 = -3

Remainder of ‘-3’ when divided by 19 is same as remainder of ‘16’ when divided by 19.

Thus 15! mod 19 = 16

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Q.4) What will be the remainder when (23!)2 is divided by 47?

Solution:

47 is a prime number.

From corollary of Wilson’s theorem, for prime number ‘p’,  (p-2)! mod p = 1

Thus, 45! mod 47 = 1

[45*44*43*42*…*25*24*23!] mod 47 = 1

[(-2)*(-3)*(-4)*(-5)*…*(-22)*(-23) * 23!] mod 47 = 1

We see that, there are even number of terms from ‘-2’ to ‘-23’. Thus, negative sign cancels off.

We get,

[23!*23!] mod 47 = 1

Thus, (23!)2 mod 47 =1

Hence, when (23!)2 is divided 47, we get 1 as a remainder.