A binary operation \Delta defined on the set of integers modulo 11 is given by a \Delta b … — JAMB UTME

A binary operation $\Delta$ defined on the set of integers modulo $11$ is given by $a \Delta b \equiv (2a^2 - 3ab + b^2) \pmod{11}$ . Find the smallest non-negative integer $x$ that satisfies the equation $4 \Delta x \equiv 5 \pmod{11}$ .

Answer: 3

Answer

The smallest non-negative integer

x

that satisfies

4 \Delta x \equiv 5 \pmod{11}

is 3.

Evaluating

4 \Delta x

gives

32 - 12x + x^2 \equiv x^2 - x + 10 \pmod{11}

. Setting this congruent to

5 \pmod{11}

yields

x^2 - x + 5 \equiv 0 \pmod{11}

, which converts to

x^2 - x - 6 \equiv 0 \pmod{11}

. Factoring gives

(x - 3)(x + 2) \equiv 0 \pmod{11}

, which yields solutions

x \equiv 3 \pmod{11}

and

x \equiv 9 \pmod{11}

. The smallest non-negative integer among these solutions is

3

Step-by-Step Solution

Substitute the given value

a = 4

into the operation definition.

4 \Delta x = 2(4)^2 - 3(4)x + x^2 = 32 - 12x + x^2

This establishes the explicit algebraic polynomial in terms of

x

Reduce coefficients modulo

11

32 \equiv 10 \pmod{11}

and

-12x \equiv -x \pmod{11}

, giving

x^2 - x + 10 \pmod{11}

Simplifying coefficients reduces computational complexity during equation solving.

Form the modular quadratic equation and set it to zero.

x^2 - x + 10 \equiv 5 \pmod{11} \implies x^2 - x + 5 \equiv 0 \pmod{11} \implies x^2 - x - 6 \equiv 0 \pmod{11}

Expressing

5 \equiv -6 \pmod{11}

allows standard integer factorisation.

Factor the quadratic polynomial and solve for

x

(x - 3)(x + 2) \equiv 0 \pmod{11} \implies x \equiv 3 \text{ or } x \equiv -2 \equiv 9 \pmod{11}

Since

11

is prime, a product congruent to

0 \pmod{11}

implies at least one factor is congruent to

0 \pmod{11}

Select the smallest non-negative integer from the valid solution set

\{3, 9\}

x = 3

3

is non-negative and strictly smaller than

9

Key Concept

Modular Arithmetic Binary Operations and Quadratic Congruences

Estimated Time:3m 0s

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