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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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harkovskaia [24]

Answer:

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Consider checking attachment for the step by step solution.

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3 years ago
Tensile testing provides engineers with the ability to verify and establish material properties related to a specific material.
Sedbober [7]

Answer:

True

Explanation:

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7 0
3 years ago
A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027
Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = \frac{d}{2} = 3 inches

density,  \rho} = 493 lbm/ ft^{3}

S.G = 1.0027

g = 9.8 m/ m^{2} = 386.22 inch/ s^{2}

Solution:

Using the formula for terminal velocity,

v_{T} = \sqrt{\frac{2V\rho  g}{A \rho C_{d}}}              (1)

(Since, m = V\times \rho)

where,

V = volume of sphere

C_{d} = drag coefficient

Now,

Surface area of sphere, A = 4\pi r^{2}

Volume of sphere, V = \frac{4}{3} \pi r^{3}

Using the above formulae in eqn (1):

v_{T} = \sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho  g}{4\pi r^{2} \rho C_{d}}}

v_{T} = \sqrt{\frac{2gr}{3C_{d}}}  

v_{T} = \sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}

Therefore, terminal velcity is given by:

v_{T} = \frac{27.79}{\sqrt{C_d}} inch/sec

3 0
3 years ago
A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces
nika2105 [10]

Answer:

a) volume flow rate of air at the inlet is 0.0471 m³/s

b) the velocity of the air at the exit is  8.517 m/s

Explanation:

Given that;

The electrical power Input W_elec = -1400 W = -1.4 kW

Inlet temperature of air T_in = 22°C

Inlet pressure of air p_in = 100 kPa

Exit temperature T_out = 47°C

Exit area of the dyer is A_out = 60 cm²= 0.006 m²

cp = 1.007 kJ/kg·K

R = 0.287 kPa·m3/kg·K

Using mass balance

m_in = m_out = m_air

W _elec = m_air ( h_in - h_out)

we know that h = CpT

so

W _elec = m_air.Cp ( T_in - T_out)

we substitute

-1.4 = m_air.1.007 ( 22 - 47 )

-1.4 =  - m_air.25.175

m_air = -1.4 / - 25.175

m_ air = 0.0556 kg/s

a) volume flow rate of air at the inlet

we know that

m_air = P_in × V_in

now from the ideal gas equation

P_in = p_in / RT_in

we substitute our values

= (100×10³) / ((0.287×10³)(22+273))

= 100000 / 84665

P_in = 1.18 kg/m³

therefore inlet volume flowrate will be;

V_in = m_air / P_in

= 0.0556 / 1.18

= 0.0471 m³/s

the volume flow rate of air at the inlet is 0.0471 m³/s

b) velocity of the air at the exit

the mass flow rate remains unchanged across the duct

m_ air = P_in.A_in.V_in = P_out.A_out.V_out

still from the ideal gas equation

P_out = p_out/ RT_out   ( assume p_in = p_out)

P_out = (100×10³) / ((0.287×10³)(47+273))

P_out  = 1.088 kg/m³

so the exit velocity will be;

V_out = m_air / P_out.A_out

we substitute our values

V_out = 0.0556 / ( 1.088 × 0.006)

= 0.0556 / 0.006528

= 8.517 m/s

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6 0
3 years ago
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7nadin3 [17]

Answer:

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Explanation:

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3 0
3 years ago
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