A Division has different results (2)

By using Diophantine Equation , let $x$ and $y$ are different quotients:

$\begin{array}{l} 31+32x=37+42y\\ \rightarrow16x-21y=3\\ \rightarrow \begin {cases} x=12+21t\\y=9+16t \end{cases} \end{array}$

Then insert into original equation:

$\begin{array}{l}31+32(12+21t)\\ \rightarrow N=415+672t\\ \rightarrow N≡415 \mod{672}\\ \rightarrow (a,b)=(415,672) \end{array}$

Therefore, $b-a=\boxed{257}$ .

@Vincent Huang 's solution is great - I'm just adding this to show a method that can easily be extended to more complicated questions.

Since one of the quotients is a power of two, we can solve this just using parity. Any time we find a number $k$ that must be even, we'll replace it by $2k'$ ; any that must be odd we'll replace with $2k'+1$ . I am not good at mental arithmetic, so I'll do everything possible to keep numbers small; starting by rewriting the first congruence as $N=32x-1$ and the second as $N=42y-5$ for some $x,y$ : $\begin{aligned} 32x-1 &= 42y-5 \\ 32x+4 &= 42y \\ 16x+2 &= 21y \text{, so }y \text{ is even} \\ 16x+2 &= 42y' \\ 8x+1 &= 21y' \text{, so }y' \text{ is odd} \\ 8x+1 &= 42y''+21 \\ 4x &=21y''+10 \text{, so }y'' \text{ is even} \\ 4x &= 42y'''+10 \\ 2x &=21y'''+5 \text{, so }y''' \text{ is odd} \\ 2x &= 42y''''+26 \\ x &= 21y''''+13 \end{aligned}$

or, letting $u=y''''$ , we have $x=21u+13$ .

Hence $N=32(21u+13)-1=672u+415$ , ie $N \equiv 415 \pmod{672}$ and $b-a=\boxed{257}$ .