If (125)_x + (32)_x = (201)_x, where x represents the base of the numerals, what is the va… — JAMB UTME

If $(125)_x + (32)_x = (201)_x$ , where $x$ represents the base of the numerals, what is the value of $x$ ?

The base

x

is equal to

6

Expanding the base-

x

numbers into base 10 polynomials yields

(x^2 + 2x + 5) + (3x + 2) = 2x^2 + 1

. Simplifying this equation gives

x^2 - 5x - 6 = 0

, which factors as

(x - 6)(x + 1) = 0

. Since a number base must be positive and greater than the highest digit in the expression (which is 5), the only valid solution is 6.

Expand all numbers in base

x

into positional power notation (base 10 equivalent)

(125)_x = 1 \cdot x^2 + 2 \cdot x^1 + 5 \cdot x^0 = x^2 + 2x + 5

(32)_x = 3 \cdot x^1 + 2 \cdot x^0 = 3x + 2

, and

(201)_x = 2 \cdot x^2 + 0 \cdot x^1 + 1 \cdot x^0 = 2x^2 + 1

Converting all terms to base 10 allows setup of a algebraic equation in

x

Substitute the expanded terms back into the original equation and simplify

(x^2 + 2x + 5) + (3x + 2) = 2x^2 + 1 \implies x^2 + 5x + 7 = 2x^2 + 1

Combining like terms on the left-hand side prepares the expression for quadratic rearrangement.

Rearrange into standard quadratic form

ax^2 + bx + c = 0

2x^2 - x^2 - 5x + 1 - 7 = 0 \implies x^2 - 5x - 6 = 0

Subtracting

(x^2 + 5x + 7)

from both sides sets the quadratic equation to zero.

Solve the quadratic equation by factoring and evaluate valid base conditions

(x - 6)(x + 1) = 0 \implies x = 6

x = -1

. Valid base:

x = 6

A number base must be a positive integer greater than any single digit present in the given numbers (digits up to 5 appear, so

x > 5

Conversion of numbers in arbitrary base

x

to base 10 via place-value expansion to solve polynomial equations.

Question