Find a positive integer $x$ less than $105$ satisfying the following simultaneous congruence equations. 
$$x=2 mod 3$$
  $$x=3 mod 5$$
  $$x=4 mod 7$$

I have only learnt modulo for 2 weeks so far... really basic theorems. 

My attempt using definitions of modulo

From Equation 1, $3a=x-2 \rightarrow 15a=5x-10$
From Equation 2, $5b=x-3 \rightarrow 15b=3x-9$
Adding them together, $15(a+b)=8x-19$ which implies, $8x=19mod15$ or $8x=4mod15$
and hence i am stuck....
please help, i can't seem to understand this at all.

First, $x=3a+2=5b+3=7c+4$. From the second equality, you get that $3a+2=5b+3\Rightarrow 3a=5b+1$. If $b=3d$ or $b=3d+2$, you reach a contradiction, so $$b=3d+1\Rightarrow x=5b+3=5(3d+1)+3=15d+8.$$ From the last equation, $15d+8=7c+4\Rightarrow 15d+4=7c$. By contradiction you can check that $d=7e+3$, so $$x=15d+8=15(7e+3)+8=105e+53.$$ So, the number you ask for is $53$.
