Radix Value Calculation: Finding 'R' in Number Bases

The problem asks us to find the value of the radix 'R' given the equation \((1012)_3 = (112)_R\). To solve this, we need to convert both numbers to their equivalent base 10 (decimal) form and then solve the resulting algebraic equation for 'R'.

Number Conversion to Base 10

Let's convert the number \((1012)_3\) from base 3 to base 10. To do this, we multiply each digit by the base raised to the power of its position, starting from 0 for the rightmost digit.

• For \((1012)_3\):
• The digits are 1, 0, 1, 2.
• The positions (powers of the base) are 3, 2, 1, 0 respectively, from left to right.
• So, \((1012)_3 = 1 \times 3^3 + 0 \times 3^2 + 1 \times 3^1 + 2 \times 3^0\)
• Calculating the values:
  • \(1 \times 3^3 = 1 \times 27 = 27\)
  • \(0 \times 3^2 = 0 \times 9 = 0\)
  • \(1 \times 3^1 = 1 \times 3 = 3\)
  • \(2 \times 3^0 = 2 \times 1 = 2\)
• Adding these values: \(27 + 0 + 3 + 2 = 32\).
• Therefore, \((1012)_3 = 32_{10}\).

Next, let's convert the number \((112)_R\) from base R to base 10. We apply the same method:

• For \((112)_R\):
• The digits are 1, 1, 2.
• The positions (powers of the base) are 2, 1, 0 respectively, from left to right.
• So, \((112)_R = 1 \times R^2 + 1 \times R^1 + 2 \times R^0\)
• Calculating the values:
  • \(1 \times R^2 = R^2\)
  • \(1 \times R^1 = R\)
  • \(2 \times R^0 = 2 \times 1 = 2\)
• Adding these values: \(R^2 + R + 2\).
• Therefore, \((112)_R = (R^2 + R + 2)_{10}\).

Solving for Radix 'R'

Now that both numbers are in base 10, we can set their decimal equivalents equal to each other, as given in the original equation:

\((1012)_3 = (112)_R\)

Substituting their base 10 forms:

\(32 = R^2 + R + 2\)

To find 'R', we need to rearrange this into a standard quadratic equation form \(aR^2 + bR + c = 0\):

\(R^2 + R + 2 - 32 = 0\)

\(R^2 + R - 30 = 0\)

We can solve this quadratic equation by factoring. We need two numbers that multiply to \(-30\) and add up to \(1\) (the coefficient of R). These numbers are \(6\) and \(-5\).

So, we can factor the equation as:

\((R + 6)(R - 5) = 0\)

This gives us two possible values for R:

• \(R + 6 = 0 \implies R = -6\)
• \(R - 5 = 0 \implies R = 5\)

In number systems, a radix (or base) must be a positive integer and must be greater than the largest digit used in the number for that base. In \((112)_R\), the largest digit is 2.

Considering the properties of a radix:

• The radix cannot be negative, so \(R = -6\) is not a valid solution.
• The radix must be greater than the largest digit used. Here, the largest digit in \((112)_R\) is 2. So, \(R\) must be greater than 2 (\(R > 2\)). Our solution \(R = 5\) satisfies this condition, as \(5 > 2\).

Therefore, the valid value for the radix 'R' is 5.