Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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]

Show Tags

23 Jul 2013, 12:52

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

_________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

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 _________________