Find an integer $x$ such that $x \equiv 3$ (mod $4$) and $x \equiv 5$ (mod $9$)

Question

For this question I basically used the Chinese Remainder Theorem from my book. I just want to make sure my reasoning and logic was correct:

Notice that $4$ and $9$ are relatively prime so the criterion for the CRT is met. Also note that the answer for $x$ will be unique up to modulo $4 \cdot 9 = 36$

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Let $x_{1} = 1 - 4s = 9t$. So that,

• $x_{1} \equiv 1$ (mod $4$)
• $x_{1} \equiv 0$ (mod $9$)

Let $x_{2} = 1 - 9t = 4s$. So that,

• $x_{2} \equiv 1$ (mod $9$)
• $x_{2} \equiv 0$ (mod $4$)

We know from the CRT that:

• $x = 3 \cdot x_{1} + 5 \cdot x_{2}$
  • Now, here I tried to find values of $x_{1}$ and $x_{2}$ that would satisfy both congruences at once and found that $x_{1} = 9$ and $x_{2} = 28$ so that,
  • $x = 3 \cdot 9 + 5 \cdot 28 = 167$
  • Remembering that $x$ will be unique up to congruence modulo $36$ I checked to see that $167 \equiv 23$ mod($36$)
  • Thus $x = 23$

Now, I wanted to check another way and the CRT tells us that:

• $x \equiv 3$ (mod $4$) $= 3, 7, 11, 15, 19, 23, 27, 31, 34$
• $x \equiv 5$ (mod $9$) $= 5, 14, 23, ...$
• Noticed that the only number that coincides modulo $36$ is $23$ and thus $x = 23$

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1. Is this the correct answer?
2. Are both methods equally viable in determining the correct answer?