Question

1011 x2 left to right multiplication​

Answer 1

Answer:

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Step-by-step explanation:

a + b + c = 10

⇒ (a + b + c)2 = (10)2

⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 100

⇒ 38 + 2(ab + bc + ca) = 100

⇒ 2(ab + bc + ca) = 62

⇒ 2(ab + bc + ca) = 62

⇒ (ab + bc + ca) = 62/2

⇒ ab + bc + ca = 31

Alternative Method

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

⇒ (10)2 = 38 + 2(ab + bc + ca)

⇒ 100 = 38 + 2(ab + bc + ca)

⇒ 100 - 38 + 2(ab + bc + ca)

⇒ 62 = 2(ab + bc + ca)

⇒ 62/2 = ab + bc + ca

⇒ 31 = ab + bc + ca

HEYA FRIEND HAVE A GOOD DAY

∴ ab + bc + ca = 31

Answer 2

Answer:

Explanation: You do basically written multiplication. You shift the a to the right and always look at the last bit. If it is 0, you add nothing if it is 1 you xor with b. Since xor is not exactly addition on integers, this is not just a*b. You can think about why the addition of two polynomials can be done by xor. Since we shift b to the left, it always is multiplied with the current monom from a.

Step-by-step explanation:

unsigned int multiply_poly(unsigned int a, unsigned int b)

{

unsigned int ret = 0;

while(a)

{

if(a & 1)

{

ret ^= b;

}

a >>= 1;

b <<= 1;

}

return ret;

}