We may write, 7n=(10−3)n or 7n=10K+(−3)n (using expansion) ∴7n+3n=10K+(−3)n+3n ={
10k
n= odd
10k+2⋅3n
n= even .
Let n= even =2t,t∈N Then, 3n=32t=9t=(10−1)t =10p+(−1)t =10p±1 If n= even, then 7n+3n will never be multiple of 10. This implies n=0 dd n=11,13,15,...99 (since, n is two digit) ⇒10<n<100 Total possible 'n' are 45.