I simply used the subtract and conquer method:
if any recurrence relation in the form of ==== >» T(n)=aT(n-b)+f
if f=O(n^k) where k>=0
Then T(n)=O(n^k) if(a>1)
T(n)=O(n^(k+1)) if(a==1)
T(n)=O(n^k(a^(n/b))) if(a<1)
According to the problem :
The recurrence relation will be as
T(n)=T(n-4)+1
1===(O(n^0))
so, k=0;
applying the above theorem we get T(n)=O(n^(0+1)) => T(n)=O(n) =O(n^a(logn^b)
then a=1 and b=0
Then a+10b=1.
The answer is 1.