Hi. Let a,b,c,d be the number of 150,100,75,60 wat bulbs used respsectively.
We have:
150a+100b+75c+60d=1000
And now we will keep reducing this equation:
30a+20b+15c+12d=200
From here on we see that c must be even and d must be a multiple of 5, hence let c=2k, d=5x:
30a+20b+30k+60x=200
3a+2b+3k+6x=20
Now we can see that b cannot be a multiple of 3, hence b is of the form b=3m+1 or b=3m+2 (mutually exclusive cases). By plugging in the second case we get:
3a+2(3m+2)+3k+6x=20
3a+6m+3k+6x=16
3(a+2m+k+2x)=16 which implies 16 has 3 as a factor, a contradiction. Hence only the first case is possible:
3a+2(3m+1)+3k+6x=20
3a+6m+3k+6x=18
a+2n+k+2x=6
And now find how many different solutions there are to that equation. Which is done manually or by computer or maybe generating functions (dunno what is quickest so far).
Now to finish it off:
(a+k)+2(n+x)=6
We have four cases:
a+k=0, 2(n+x)=6. 1 way to get a+k=0, 4 ways to get n+x=3
a+k=2, 2(n+x)=4. 3 ways to get a+k=2, 3 ways to get n+x=2
a+k=4, 2(n+x)=2. 5 ways to get a+k=4, 2 ways to get n+x=1
a+k=6, 2(n+x)=0. 7 ways to get a+k=6, 1 way to get n+x=0
Hence the final answer
