MyQuestionIcon

1

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

Quantitative Aptitude

Fermat Theorem

Given that p ...

Question

Given that p is a prime number, $p^{2} (p^{27} - \frac{1}{p})$ is always divisible by?

A

25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

23

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

27

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

none of these

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D none of these
According to Fermat's theorem "If M is a prime number, then for any integer N, $N^{M} - N$ is divisible by M."

Open the brackets, $p^{2} (p^{27} - \frac{1}{p}) = p^{29} - p$ will always be divisible by 29, as it is prime.

Suggest Corrections

thumbs-up

0

Similar questions

Q. Given that p is a prime number. p²[p²⁷-

] is always divisible by?

p

is a prime number,

x^{p} - x

is divisible by

p

1

p

is a prime number

(p \neq 2)

[(2 + \sqrt{5})^{p}] - 2^{p + 1}

is always divisible by

p

[.]

denotes the greatest integer function).

Statement

2

n

is a prime, then

^{n} C_{1},^{n} C_{2},^{n} C_{3}, \dots,^{n} C_{n - 1}

must be divisible by

n

.

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

P

is a prime number then

n^{p} - n

is divisible by

p

n

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Factorising Denominator

QUANTITATIVE APTITUDE

Watch in App

explore_more_icon

Explore more

Other Quantitative Aptitude

Join BYJU'S Learning Program

CrossIcon