JAMIA 2026 — Computer PYQ

1. Expand the Left-Hand Side (LHS) in terms of base $b$:

To convert a number from an arbitrary base $b$ to base 10, we expand it using positional weights (powers of $b$) starting from the rightmost digit (position 0).

$$(1011{)}_b=1\cdot b^3+0\cdot b^2+1\cdot b^1+1\cdot b^0$$

$$(1011{)}_b=b^3+b+1$$

2. Set up the equation:

We are given that this expression is equal to $(131{)}_{10}$:

$$b^3+b+1=131$$

Subtract 1 from both sides:

$$b^3+b=130$$

Factor out $b$ on the left side:

$$b(b^2+1)=130$$

3. Solve for $b$ (Using Substitution/Option Checking):

Since $b$ must be a positive integer base, we can verify the given options to see which one satisfies the polynomial equation $b^3+b=130$.

• Test option (a) $b=2$:

  $$2^3+2=8+2=10\ne 130$$

• Test option (b) $b=3$:

  $$3^3+3=27+3=30\ne 130$$

• Test option (c) $b=5$:

  $$5^3+5=125+5=130$$

  This matches perfectly! Let's double check with the factored form: $5(5^2+1)=5(25+1)=5(26)=130$.

Correct Answer

The correct option is (c) 5.