To find the smallest integer greater than $23$ with certain modular residues

Question

The question asks to find the smallest integer greater than $23$ that leaves the remainders $2,3,2$ when divided by $3,5,7$ respectively. The answer is given as $128$.

I used C.R.T. to solve this. $$x\equiv2\pmod 3\\ x\equiv3\pmod 5\\ x\equiv2\pmod 7 $$ where $M=m_1·m_2·m_3=3×5×7=105$.

From here I get solutions, $$x\equiv2\pmod 3\\ x\equiv1\pmod 5\\ x\equiv1\pmod 7$$
After that I find\begin{align*} x &\equiv 2×2×5×7+1×3×3×7+1×2×3×5\\ &\equiv 140+63+30\\ &\equiv 233 \pmod{105}, \end{align*} where I again find that $x=23$.

But one thing I have noticed that $233=105×1+128$. I cannot continue from here. I need help and any help is highly appreciated.

Answer 1

Applying directly the following proof of CRT, for example:

$$\begin{cases}y_1=\frac{105}3=35\;\implies y_1=2\pmod3\implies y_1^{-1}=2\pmod3\\{}\\y_2=\frac{105}5=21\implies y_2=1\pmod5\implies y_2^{-1}=1\pmod5\;\\{}\\y_3=\frac{105}7=15\implies y_3=1\pmod7\implies y_3^{-1}=1\pmod7\end{cases}$$

so that

$$2\cdot35\cdot2+3\cdot21\cdot1+2\cdot15\cdot1=233\;\;\text{ is a solution, and}$$

$$233=128\pmod{105}\;\;\text{is another solution, and}$$

$$128=23\pmod105\;\;\text{ is the unique solution modulo}\;105$$

Observe that unless you consider things modulo $\;3\cdot5\cdot7=105\;$ at the end , there is not "the" solution, but only one out of infinitely many

Answer 2

Let $x\equiv y \equiv 2 \mod 3,$ and $x\equiv y \equiv 3 \mod 5,$ and $x\equiv y \equiv 2 \mod 7.$ Then $x-y \equiv 0$ modulo $3,5,$ and $7.$

So $x-y$ is divisible by $3,5,$ and $7,$ so $x-y$ is divisible by $(3)(5)(7)=105.$ So if $y=23$ and $x>y$ then $x=23+105n$ for some $n\in \Bbb N.$

And if $x=23+105n$ for some $n\in \Bbb N$ then $x\equiv 23$ modulo $3,5, $ and $7.$