UGC NET JRF November 2020 Paper-2
October 12, 2023UGC NET JRF November 2020 Paper-2
October 12, 2023UGC NET JRF November 2020 Paper-2
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Question 16
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Using ‘RSA’ public key cryptosystem, if p=3, q=11 and d=7, find the value of e and encrypt the number ’19’
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20,19
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33,11
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3,28
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77,28
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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)e mod n
= (19)3 mod 33
= 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)e mod n
= (19)3 mod 33
= 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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