Question

Difficulty: MediumModular Arithmetic

What is the smallest non-negative integer xx that satisfies the modular congruence 5x+7≡2(mod9)5x + 7 \equiv 2 \pmod{9}?

  1. 88Answer
  2. B
    44
  3. C
    11
  4. D
    77

Answer

The smallest non-negative integer xx is 8.
Subtracting 77 from both sides yields 5x≡−5(mod9)5x \equiv -5 \pmod{9}, which reduces to 5x≡4(mod9)5x \equiv 4 \pmod{9}. Multiplying both sides by the modular multiplicative inverse of 55 modulo 99 (which is 22, since 5×2=10≡1(mod9)5 \times 2 = 10 \equiv 1 \pmod{9}) gives x≡4×2=8(mod9)x \equiv 4 \times 2 = 8 \pmod{9}. Thus, the smallest non-negative integer value is 88.

Step-by-Step Solution

1
Isolate the term containing xx by subtracting 7 from both sides of the congruence.
5x≡2−7≡−5(mod9)5x \equiv 2 - 7 \equiv -5 \pmod{9}
Standard algebraic balance operations apply under modular arithmetic.
2
Convert the negative remainder −5-5 to its canonical non-negative representative in modulo 9.
−5+9=4  ⟹  5x≡4(mod9)-5 + 9 = 4 \implies 5x \equiv 4 \pmod{9}
Modular numbers are conventionally expressed in the non-negative range [0,n−1][0, n-1].
3
Find the modular multiplicative inverse of 5 modulo 9 and multiply both sides by it.
Since 5×2=10≡1(mod9)5 \times 2 = 10 \equiv 1 \pmod{9}, the inverse is 2. Multiplying gives x≡4×2=8(mod9)x \equiv 4 \times 2 = 8 \pmod{9}.
Multiplying by the modular inverse solves for xx when gcd⁡(5,9)=1\gcd(5, 9) = 1.

Key Concept

Linear Modular Congruences and Modular Inverses
Estimated Time:1m 30s
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