Schrodinger's Lamps!

For $N$ , we will have to make $k$ steps switching in each one the state of one of the $2n$ lamps. Thus, we will be able to make ${ (2n) }^{ k }$ steps. How many of those final states are those we want? It's easy to say ${ 2 }^{ 2n }$ , but only half of the states are possible, since after $m$ steps, the sum of the states ${ S }_{ S }$ (taking 0 as a lamp switch off and 1 as a switched, or viceversa) will satisfy ${ S }_{ S } \equiv m\quad (mod\quad 2)$ . So, the possible final solutions for $N$ are $N=\frac { { (2n) }^{ k } }{ \frac { { 2 }^{ 2n } }{ 2 } } =\frac { { (2n) }^{ k } }{ { 2 }^{ 2n-1 } }$ .

Same for M. The only difference in n is that we only switch the first $n$ lamps instead of the $2n$ lamps. So, the possible final solutions for $M$ will be $M=\frac { { n }^{ k } }{ \frac { { 2 }^{ n } }{ 2 } } =\frac { { (n) }^{ k } }{ { 2 }^{ n-1 } }$ .

Then $\frac { N }{ M } =\frac { \frac { { (2n) }^{ k } }{ { 2 }^{ 2n-1 } } }{ \frac { n^{ k } }{ { 2 }^{ n-1 } } } ={ \left( \frac { 2n }{ n } \right) }^{ k }·\frac { { 2 }^{ n-1 } }{ { 2 }^{ 2n-1 } } ={ 2 }^{ k }·{ 2 }^{ n-1-2n+1 }={ 2 }^{ k-n }$

Taking $k=n+a$ (yes, $a=10$ ): ${ 2 }^{ k-n }={ 2 }^{ n+a-n }={ 2 }^{ a }$ .

As $a=10$ , the solution will be ${ 2 }^{ a }={ 2 }^{ 10 }=\boxed { 1024 }$