###### UGC NET JRF November 2020 Paper-2

October 12, 2023###### UGC NET JRF November 2020 Paper-2

October 12, 2023# UGC NET JRF November 2020 Paper-2

Question 16 |

Using ‘RSA’ public key cryptosystem, if p=3, q=11 and d=7, find the value of e and encrypt the number ’19’

20,19 | |

33,11 | |

3,28 | |

77,28 |

Question 16 Explanation:

n= P*q

n= 3*11

n=33

ф(n) = (p-1)(q-1)

ф(n) = 2*10

ф(n) = 20

E * d=1 mod ф(n)

e*7=1 mod 20

Now find the value of “e” in such a way that if we multiply the value of “e” with 7 then by dividing it with 20 mod value should come as “1”.

So if e=3 then the above condition can be satisfied hence option C is the right answer.

Encrypted value = (Message)

= (19)

= 28

n= 3*11

n=33

ф(n) = (p-1)(q-1)

ф(n) = 2*10

ф(n) = 20

E * d=1 mod ф(n)

e*7=1 mod 20

Now find the value of “e” in such a way that if we multiply the value of “e” with 7 then by dividing it with 20 mod value should come as “1”.

So if e=3 then the above condition can be satisfied hence option C is the right answer.

Encrypted value = (Message)

^{e}mod n= (19)

^{3}mod 33= 28

Correct Answer: C

Question 16 Explanation:

n= P*q

n= 3*11

n=33

ф(n) = (p-1)(q-1)

ф(n) = 2*10

ф(n) = 20

E * d=1 mod ф(n)

e*7=1 mod 20

Now find the value of “e” in such a way that if we multiply the value of “e” with 7 then by dividing it with 20 mod value should come as “1”.

So if e=3 then the above condition can be satisfied hence option C is the right answer.

Encrypted value = (Message)

= (19)

= 28

n= 3*11

n=33

ф(n) = (p-1)(q-1)

ф(n) = 2*10

ф(n) = 20

E * d=1 mod ф(n)

e*7=1 mod 20

Now find the value of “e” in such a way that if we multiply the value of “e” with 7 then by dividing it with 20 mod value should come as “1”.

So if e=3 then the above condition can be satisfied hence option C is the right answer.

Encrypted value = (Message)

^{e}mod n= (19)

^{3}mod 33= 28

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