\(f(n)=1154∗1156=(1155-1)*(1155+1)=1155^2-1=(3^2*5^2*7^2*11^2)-1\)
We understand that the units digit of \(f(n)=1155^2-1\) must be 4.
(B) f(n)=5n−2 --> units digit of the result is either 3 or 8
(C) f(n)=5n+3 --> units digit of the result is either 3 or 8
Thus, we confidently eliminate choices (B) and (C), since both choices NEVER yield results with units digit of 4
A. f(n)=3n−2
\(f(n)=3n−2=(3^2*5^2*7^2*11^2)-1\)
\(3n=(3^2*5^2*7^2*11^2)+1\)
\(n=(3*5^2*7^2*11^2)+1/3\)
--> \(n\) is
NOT an integer, so
we confidently eliminate choice (A), since \(n\) has to be some integer
D. f(n)=7n−2
\(f(n)=7n−2=(3^2*5^2*7^2*11^2)-1\)
\(7n=(3^2*5^2*7^2*11^2)+1\)
\(n=(3^2*5^2*7*11^2)+1/7\)
--> \(n\) is
NOT an integer, so
we confidently eliminate choice (D), since \(n\) has to be some integer
E. 11n+10 \(f(n)=11n+10=(3^2*5^2*7^2*11^2)-1\)
\(11n=(3^2*5^2*7^2*11^2)-11\)
\(n=(3^2*5^2*7^2*11)-1\)
--> \(n\) is
an integer, so
we are confident that choice (E) is the CORRECT ANSWERFinal answer is (E)