Given that 245_x - 157_x = 66_x, find the value of the base x. — JAMB UTME

Given that $245_x - 157_x = 66_x$ , find the value of the base $x$ .

Answer: 8

The value of the base

x

is 8.

Expanding all terms in powers of

x

gives

2x^2 + 4x + 5 - (x^2 + 5x + 7) = 6x + 6

. Simplifying gives

x^2 - 7x - 8 = 0

, which factors into

(x - 8)(x + 1) = 0

. Since a number base must be a positive integer greater than any digit present in the equation (the maximum digit here is 7), the only valid base is

x = 8

Convert each positional number into its base 10 polynomial expansion.

245_x = 2x^2 + 4x + 5

157_x = x^2 + 5x + 7

, and

66_x = 6x + 6

A number

d_2 d_1 d_0

in base

x

represents

d_2 x^2 + d_1 x^1 + d_0 x^0

in base 10.

Substitute the expanded expressions into the given subtraction equation.

(2x^2 + 4x + 5) - (x^2 + 5x + 7) = 6x + 6

This translates the base

x

relationship into a standard base 10 equation.

Simplify and rearrange into standard quadratic form.

x^2 - 7x - 8 = 0

Expanding the subtraction yields

x^2 - x - 2 = 6x + 6

. Subtracting

(6x + 6)

from both sides produces a quadratic set to zero.

Factorize the quadratic expression.

(x - 8)(x + 1) = 0 \implies x = 8 \text{ or } x = -1

Finding the roots provides potential mathematical values for

x

Apply number base constraints to select the valid root.

x = 8

A valid base must be a positive integer strictly greater than any individual digit in the expression. Since 7 appears in

157_x

x > 7

, ruling out

-1

and confirming

x = 8

Converting numbers from an unknown base

x

to base 10 polynomials to solve algebraic equations.

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