When M divided by 4,the remainder is 3,so \(\left(M+1\right)⋮4\)

When M divided by 9,the remainder is 8,so \(\left(M+1\right)⋮9\)

\(\Rightarrow M+1⋮LCM\left(4;9\right)\) or \(M+1⋮36\)

=> When M divided by 36,the remainder is : 36 - 1 = 35

Write \(M=36k+r\left(k\in Z,0< r< 36\right)\) is remainder when M diveded by 36.
Deduced : r divided by 4 the remainder is 3 and r divided by 9  the remainder  is 8.
So: \(r-8\) is mutiple for 9 and r - 3 is mutiple for 4.
Thus \(r-8\in\left\{0;9;18;27,36\right\}\)
     \(\Rightarrow r\in\left\{8;17;26;35\right\}\) because r < 36
\(\Rightarrow r-3\in\left\{5;14;23;32\right\}\).
Because \(r-3⋮4\) \(\Rightarrow r-3=32\)\(\Leftrightarrow r=35\).
So when M divided by 35, the reaminder is 35.