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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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putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

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12.5X10⁻⁶ *115-15 * 20

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3 0
3 years ago
What is the basic formula for actual mechanical advantage?
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Answer:

Mechanical Advantage Formula

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Explanation:

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Pressure drop is given by hagen poiseuille equation

\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

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4 0
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Answer:

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The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

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ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

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h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

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Answer:

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6 0
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