MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Definition of Sets

If p is pri...

Question

If $p$ is prime, show that $2 (p - 3)! + 1$ is a multiple of $p$ .

Open in App

Solution

According to Wilson's theorem if $p$ is a prime, then it would be divisible by $(p - 1)! + 1 = (p - 1) (p - 2) (p - 3)! + 1$
If $p - 1$ is divided by $p$ remainder is $- 1$
If $p - 2$ is divided by $p$ remainder is $- 2$
Hence, $p$ would be divisible by $- 1 \times - 2 \times (p - 3)! + 1 = 2 (p - 3)! + 1$

Suggest Corrections

thumbs-up

0

Similar questions

p

is a prime number, and a prime to

p

, and if a square number

c^{2}

can be found such that

c^{2} - a

is divisible by

p

a^{\frac{1}{2} (p - 1)} - 1

is divisible by

p

n

is a prime number greater than

3

n^{2} - 1

is a multiple of

24

p

is a prime number, show that the coefficients of the terms of

(1 + x)^{p - 1}

are alternately greater and less by unity than some multiple of

p

p

is a prime, show that the sum of the

(p - 1)^{t h}

p

numbers in arithmetical progression, wherein the common difference is not divisible by

p

1

than a multiple of

p

p

is a prime, and

x

p

x^{p^{r}} - p^{r - 1} - 1

is divisible by

p^{r}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Definition of Sets

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Definition of Sets

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon