GATE CS Mastery
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GATE CS 2014 Set 1 · Digital Logic

Question 18

NATPrinted as CS Q.81 markNo negative markingEasy (estimate)

Question

The base (or radix) of the number system such that the following equation holds is____________.

31220=13.1\frac{312}{20} = 13.1
Official answer and solution
Official answer
5
From
Official GATE 2014 answer key

Solution

Express numbers in polynomial form with respect to radix bb and solve the resulting algebraic equation.

Given equation in base bb:

312b20b=13.1b\frac{312_b}{20_b} = 13.1_b

Converting each term into base 10 polynomial form in terms of bb:

  • 312b=3b2+1b1+2b0=3b2+b+2312_b = 3b^2 + 1b^1 + 2b^0 = 3b^2 + b + 2
  • 20b=2b1+0b0=2b20_b = 2b^1 + 0b^0 = 2b
  • 13.1b=1b1+3b0+1b−1=b+3+1b13.1_b = 1b^1 + 3b^0 + 1b^{-1} = b + 3 + \frac{1}{b}

Substitute into the equation:

3b2+b+22b=b+3+1b\frac{3b^2 + b + 2}{2b} = b + 3 + \frac{1}{b}

Multiply both sides by 2b2b:

3b2+b+2=2b(b+3+1b)3b^2 + b + 2 = 2b \left(b + 3 + \frac{1}{b}\right)
3b2+b+2=2b2+6b+23b^2 + b + 2 = 2b^2 + 6b + 2

Rearranging terms:

3b2−2b2+b−6b+2−2=03b^2 - 2b^2 + b - 6b + 2 - 2 = 0
b2−5b=0b^2 - 5b = 0
b(b−5)=0b(b - 5) = 0

Since the digits include 3, the base must satisfy b>3b > 3.
Therefore, b=5b = 5.

Final answer

5

Common trap

Forgetting that digits in base bb must be less than bb, ruling out b=0b=0 and ensuring b>3b > 3.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2014 Computer Science and Information Technology (CS) Session 1
Question
18 of 65, printed as CS Q.8
Section
Computer Science and Information Technology

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