We may write, 7n=(10−3)n or 7n=10k+(−3)n (using expansion)
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= odd n‌‌=11,13,15,...99‌‌‌ (since, ‌n‌ is two digit) ‌ 10‌<n<100 Total possible ' n ' are 45 .