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The digit 3 has the cyclicity of 4. means 3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81 3^5 = 243

So, the given problem can be stated as (3^3)^3, which is 3^27 Now 27 divided by 4 leaves a remainder of 3. Then, counting 3 back in the cyclicity , the Unit digit is 7.

Actually, in my opinion 3^3^3 should be reduced to 3^9 as the formula goes: a^x^y = a^(xy). 3^9 = 19,683 --> the unit number is 3, not 7 as some of you explained before.

Or we can use the method as proposed by some guy here: 9 mod 4 = 1 --> the last digit will be 3^1 = 3.

OA for this question is D (7).

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).

Can someone please explain why the exponent is 27 and not 9 (3x3)? Thanks.

Re: What is the last digit 3^{3^3} ? A. 1 B. 3 C. 6 D. 7 E. 9 [#permalink]

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23 Jul 2013, 11:52

elmagnifico wrote:

What is the last digit \(3^{3^3}\) ?

A. 1 B. 3 C. 6 D. 7 E. 9

please give detailed explanation.

OA is D (7).

There is something I don´t understand with this kind of exercise. When I resolve this in excel or in the calculator, the result is: 7,6256E+12

Which means \(7,6256*10^12 = 7.625.600.000.000\)

Then, the last digit is not 7 but zero. I think there is something wrong in my logic / understanding

_________________

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There is something I don´t understand with this kind of exercise. When I resolve this in excel or in the calculator, the result is: 7,6256E+12

Which means \(7,6256*10^12 = 7.625.600.000.000\)

Then, the last digit is not 7 but zero. I think there is something wrong in my logic / understanding

How can the last digit of \(3^{3^3}\) be zero? Excel rounds big numbers. Exact result is 7,625,597,484,987.

Complete solutions. 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).

OPNE DISCUSSION OF THIS QUESTION IS HERE: m25-73474.html _________________