Solving using Chinese Remainder Theorem

Question

I now know my answer's incorrect but can someone explain why my approach is wrong? I tried to follow the logic from this YouTube tutorial https://www.youtube.com/watch?v=ru7mWZJlRQg:

x=1(mod3)
x=2(mod5)
x=3(mod7)

I got 222 by doing the following:

    mod 3      mod 5     mod 7
x= 5*7        +3*7      +3*5
x= 35         +21       +15

35(mod 3) = 2(mod3) and we want 1(mod3) so multiply 2(mod3)*2 = 4(mod3) = 1(mod3). So multiply 35*2 in the above chart.

Next, 21(mod5) = 1(mod5) and we want 2(mod5) so multiply 1(mod5)*7 = 7(mod5) = 2(mod5). So multiply 21*7 in the chart.

Last, 15(mod7) = 1(mod7), so multiply 3*1(mod7) = 3(mod7) and multiply 3*15 in the chart so get:

    mod 3      mod 5     mod 7
x= 35*2       +21*7      +15*3   =   222

Answer 1

The congruence

$$x \equiv 1 \mod 3$$ tells us that $x = 1 + 3k$ for some $k$. Plug this into the second equation to get

$$1 + 3k \equiv 2 \mod 5$$

So $3k \equiv 1$ and $k\equiv 2 \mod 5$. This means $k = 2 + 5j$ for some $j$, which in turn means $x = 1 + 3(2 + 5j) = 7 + 15j$. Now plug this into the third equation

$$7+15j \equiv 3 \mod 7$$

So $15j \equiv -4 \equiv 3 \mod 7$, whence $j \equiv 3 \mod 7$. This means $j = 3 + 7m$ for some $m$. Therefore $x = 7+15j = 7 + 15(3 + 7m) = 52 + 105m$.

Therefore $x \equiv 52 \mod 105$. Specifically, $52$ is a solution.

Answer 2

You have an error in your final sum, it is not $\,222\,$ but rather $\,262\equiv \color{#c00}{52}\pmod{\!105}$

Simpler $\,3,5,7\mid 2x\!+\!1\,\Rightarrow {\rm lcm}(3,5,7)=105\mid 2x\!+\!1\,$ so $\,{\rm mod}\ 105\!:\ 2x\equiv -1\equiv 104\,\Rightarrow\, x\equiv \color{#c00}{52}$

Remark $\ $ It is usually easier to solve the congruences successively in pairs. See this answer for an example and further discussion