321kumarsushant wrote:
the ans will be 3.
for this type of question which has small no as 2, 3 ,4.
you can calculate it by multiplying only the unit digit if the nos are countable on fingers,
else you should use the above method explained by Gmat Aspirants.
OA for this question is D (7).What is the last digit of \(3^{3^3}\)?(A) 1
(B) 3
(C) 6
(D) 7
(E) 9
If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).
So:
\((a^m)^n=a^{mn}\);
\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).
According to above:\(3^{3^3}=3^{(3^3)}=3^{27}\)
Cyclicity of 3 in positive integer power is four (the last digit of 3 in positive integer power repeats in the following patter {3-9-7-1}-{3-9-7-1}-...) --> the units digit of \(3^{27}\) is the same as for 3^3 (27=4*6+3) --> 7.
Answer: D (7).
ohh ! i got that question wrong.
the ans will be 7.