If 5x \equiv 2 \pmod{11}, what is the value of (x^3 - 4x) \pmod{11} expressed in standard … — JAMB UTME

If $5x \equiv 2 \pmod{11}$ , what is the value of $(x^3 - 4x) \pmod{11}$ expressed in standard non-negative remainder form?

Solving the linear congruence

5x \equiv 2 \pmod{11}

gives

x \equiv 7 \pmod{11}

. Substituting

x = 7

into the expression

(x^3 - 4x)

yields

7^3 - 4(7) = 343 - 28 = 315

. Dividing 315 by 11 gives a quotient of 28 and a remainder of 7. Alternatively, working entirely modulo 11 gives

2 - 6 = -4 \equiv 7 \pmod{11}

. Thus, the canonical non-negative remainder is 7.

Find the modular inverse of 5 modulo 11

The inverse is 9, since

5 \times 9 = 45 \equiv 1 \pmod{11}

To isolate

x

in the linear congruence

5x \equiv 2 \pmod{11}

, multiply both sides by the modular inverse of 5.

Solve for x modulo 11

x \equiv 2 \times 9 = 18 \equiv 7 \pmod{11}

Simplifying 18 modulo 11 gives the canonical value of

x

Evaluate

x^3 - 4x

modulo 11

x^3 \equiv 7^3 = 343 \equiv 2 \pmod{11}

and

4x \equiv 4(7) = 28 \equiv 6 \pmod{11}

, so

x^3 - 4x \equiv 2 - 6 = -4 \pmod{11}

Substitute

x \equiv 7

into the polynomial expression and reduce each term modulo 11.

Convert negative remainder to canonical non-negative form

-4 + 11 = 7 \pmod{11}

Modular remainders must be expressed within the standard range

[0, 10]

Linear modular congruences and negative remainder canonical reduction

Estimated Time:2m 0s

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