If x is the smallest non-negative integer satisfying the modular congruence 7x + 4 \equiv … — JAMB UTME

If $x$ is the smallest non-negative integer satisfying the modular congruence $7x + 4 \equiv 2 \pmod{13}$ , find the value of $(x^3 + 2x) \pmod{13}$ expressed as a canonical non-negative remainder.

Answer: 6

Answer

The canonical non-negative remainder is 6.

Subtracting 4 from both sides of

7x + 4 \equiv 2 \pmod{13}

gives

7x \equiv -2 \equiv 11 \pmod{13}

. Multiplying by the modular inverse of 7 (which is 2, since

7 \times 2 = 14 \equiv 1 \pmod{13}

) yields

x \equiv 22 \equiv 9 \pmod{13}

. Evaluating

(9^3 + 2 \times 9) \pmod{13}

gives

(729 + 18) = 747

. Dividing 747 by 13 gives a quotient of 57 and a remainder of 6.

Step-by-Step Solution

Isolate the linear term in the congruence

7x \equiv -2 \equiv 11 \pmod{13}

Subtracting 4 from both sides simplifies the equation, and

-2 + 13 = 11

converts the negative remainder to positive form.

Solve for

x

by multiplying by the multiplicative inverse of 7 modulo 13

x \equiv 9 \pmod{13}

, so

x = 9

Since

7 \times 2 = 14 \equiv 1 \pmod{13}

, multiplying

7x \equiv 11 \pmod{13}

by 2 yields

x \equiv 22 \equiv 9 \pmod{13}

Evaluate

(x^3 + 2x) \pmod{13}

using modular reduction

6

9^3 = 729 = 56 \times 13 + 1 \equiv 1 \pmod{13}

and

2 \times 9 = 18 = 1 \times 13 + 5 \equiv 5 \pmod{13}

. Adding these gives

1 + 5 = 6

Key Concept

Solving linear modular congruences and modular polynomial evaluation

Estimated Time:1m 30s

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