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Nimfa-mama [501]
3 years ago
11

Suppose to build RSA crypto system you picked primes "p" and "q" as 3 and 7 and "e" as 5 what are the public and private keys? W

hat is the encryption of a message whose integer equivalent is 12. Show that the decryption gives back the message 12.
Engineering
1 answer:
nikdorinn [45]3 years ago
7 0

Answer:

1) Public key of the receiver is (e, n) is (5, 21)  and Private key of the receiver (d, n)   is (5 , 21) ,

2) the encryption of a message whose integer equivalent is 12 is 3

3) Decryption of the message ⇒ P = C^d mod n  

⇒ P = 3⁵ mod 21

⇒ P = 243 mod 21

⇒ 12

Explanation:

Given that,

p = 3

q = 7

e = 5

1)

Now, n = pq = 3 × 7 = 21

Ø(n) = (p-1) × (q-1)  = 2 × 6 = 12

Public key of the receiver is (e, n) is (5, 21)

and private key of the receiver is (d, n)

we have to find 'd' by using the expression

ed = 1 + kmodØ(n)

d = 1 + kmodØ(n) / e

now to get 'd' , we need to choose the least positive integer 'k', by substituting different values of ‘k’ from 0,

so for k =0 , d = (1+0) / 5 = 0.2 not an integer.

for k =1 , d = (1+12) / 5 = 13/5 = 2.6 not an integer.

for k =2 , d = (1+24) / 5 = 5 , now 5 is an integer

So k = 2 and d = 5

Private key of the receiver (d, n)   is (5 , 21)

2)

Now the encryption of a message whose integer equivalent is 12?

Encryption of the message ⇒ C = P^e mod n

⇒ C = 12⁵ mod 21

⇒ 248832 mod 21

⇒ 3

3)

Also the decryption gives back the message 12.

Decryption of the message ⇒ P = C^d mod n

⇒ P = 3⁵ mod 21

⇒ P = 243 mod 21

⇒ 12

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This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

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Explanation:

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Now, Using Fick's second law inform of diffusion

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Answer:

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{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}, here

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\mu _0=4\pi \times 10^{-7}

T_0 = 291.15

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