1Let, a=3ok+q, k ε {0,N}, 0≤q≤29, [a/2]+[a/3]+[a/5]≤31a/30
or, a≤31a/30
or,a≥0
Then [a/2]+[a/3]+[a/5]=31k + [q/2]+[q/3]+[q/5]
hence, 30k+q=31k + [q/2]+[q/3]+[q/5]
or, q-k= [q/2]+[q/3]+[q/5]..............(i)
R.h.s≥0, hence, q≥k
if we put, q= 0,.....,29 in (i)
we will get a k, for this 30 values of q,so, a unique pair(q,k) will be determined
hence, we shall get 30 solutions of a for each(q,k) pair
Correct ans- a0 30
