The largest positive integer n such that n^80 < 7^120 is
(a) 12
(b) 18
(c) 21
(d) 24
(e) 30
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Largest Possible Integer
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ousek
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My strategy was :
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 !
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 !
