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Question 2 For each of the following, find b if possible. If not possible, say why. b) 23₁0=25 c) b2,=2b1 Question 3 a) 111, +101,+ 110,

User Rob Cole
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Answer:

b) To convert a number from base b to base 10, we can use the formula:

anbn-1 + an-1bn-2 + ... + a1b0

where ai is the ith digit of the number in base b, and n is the number of digits.

Using this formula for 23₁0:

2×b^2 + 3×b^1 + 0×b^0 = 25

2b^2 + 3b = 25

This equation is not solvable because there are two unknowns (b^2 and b) and only one equation.

c) Using the same formula as above for b2,=2b1:

b^2×2 + b×1 = 2b + 1

b^2×2 - b×2 + b×3 - 2b - 1 = 0

2b^2 + b - 1 = 0

Using the quadratic formula, we find:

b = (-1 ± sqrt(1 + 8×2))/4

b = (-1 ± 3)/4

b = -1/2 or b = 1

However, b must be a positive integer, so the only solution is b = 1.

a) Adding numbers in binary:

111

+101

+110

1010

So 111, +101,+ 110, in binary, equals 1010 in binary, which is equal to 10 in base 10.

User Andre Rocha
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