To get as close as possible to 7^120, I temporary state that n^80=7^120
This makes:
n^80/7^120=1
n^80/(7^(3/2))^80=1
n/7^(3/2)=1
7^3 is easy to calculate: 343.
Hence: n<SQU(343).
I know that 20^2=400 and 15^2=225. Thus, my answer is directly B) : 18.
This was a good question. Thank you Dtweah !
