Find the least value of x which satisfies the equation 4x = 7(mod 9)

Analyze the question: We need the smallest positive x such that 4x leaves remainder 7 when divided by 9. In other words, 4x ≡ 7 (mod 9).

Concept: x satisfies 4x ≡ 7 (mod 9) if 4x − 7 is a multiple of 9. That is, 4x = 7 + 9k for some integer k. We need a value of k that makes 7 + 9k divisible by 4.

Solution:

• k = 0: 7 is not divisible by 4.
• k = 1: 16 is divisible by 4, so x = 16 ÷ 4 = 4.

Check: 4(4) = 16, and 16 = 9 + 7, so 16 ≡ 7 (mod 9). ✔

Now check the smaller options to be sure 4 is the least:

• x = 1: 4 → no
• x = 2: 8 → no
• x = 3: 12 ≡ 3 → no
• x = 4: 16 ≡ 7 → yes

So the least value is x = 4.

Note on the options: I also checked the others. x = 5 gives 20 ≡ 2, x = 6 gives 24 ≡ 6, and x = 7 gives 28 ≡ 1. None of these work, so only D is correct.

Tip: For modular equations like ax ≡ b (mod n), just keep adding n to b (7, 16, 25, ...) until the number is divisible by a. You can also test the options directly, as that is quick in an exam.