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PS I'll propose another solution: powers of 3 have following digits in recurring order: 3, 9, 7, 1, 3, 9, 7, 1 . . . So every 4th power of 3 ends with "1". It means that the last digit of \(3^{24}\) is one, and the last digit of \(3^{27}\) is 7.
I'm completely confused! Some people say that we should multiply exponents, others say that we should raise to the power... Can anybody give a certain rule for this part? Is the expression \(3^3^3\) is the same as \((3^3)^3\)?
Last edited by Igor010 on 10 Feb 2010, 06:09, edited 2 times in total.
RULE: 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 \((a^m)^n\).
So, \(3^{3^3}=3^{(3^3)}=3^{27}\)
Cyclicity of 3 in power is four. The units digit of \(3^{27}\) is the same as for \(3^3\) (27=4*6+3) --> \(7\). _________________
RULE: 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 \((a^m)^n\).
So, \(3^{3^3}=3^{(3^3)}=3^{27}\)
Cyclicity of 3 in power is four. The units digit of \(3^{27}\) is the same as for \(3^3\) (27=4*6+3) --> \(7\).
thanks for the explanation. Wikipedia also explains this "If exponentiation is indicated by stacked symbols, the rule is to work from the top down".
the bad thing is that when trying it on some calculators, it turns out to be 19683. seems that calculators do not obey the order
Is this a GMAT test question? Because I like the way it is explained in the screenshot above...rearrange until you have "1" as digit and then multiply with the rest! (kudos to the "creator")
3^1=3 3^3=9 3^3=27 3^4=81 3^5=243 ...... Pattern repeats with the last digits of 3,9,7,1 or every four 3^3^3=3^27 and 27/4=6 plus reminder of 3 ==> the third digit is 7.
RULE: 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 \((a^m)^n\).
So, \(3^{3^3}=3^{(3^3)}=3^{27}\)
Cyclicity of 3 in power is four. The units digit of \(3^{27}\) is the same as for \(3^3\) (27=4*6+3) --> \(7\).