Author 
Message 
Manager
Joined: 09 Jun 2011
Posts: 103

Re: last digit of a power [#permalink]
Show Tags
01 Sep 2011, 21:45
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.
So the Answer is D.



Director
Joined: 23 Apr 2010
Posts: 581

Re: last digit of a power [#permalink]
Show Tags
22 Sep 2011, 01:54
Bunuel wrote: toannguyen wrote: 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 {3971}{3971}...) > 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.



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: last digit of a power [#permalink]
Show Tags
22 Sep 2011, 05:13
nonameee wrote: Can someone please explain why the exponent is 27 and not 9 (3x3)? Thanks. I'll have to repeat what Bunuel already said: (a^b)^c: Here, (a^b) should be executed first AND then the (result)^c should be executed. If there are no brackets: a^b^c: Then the precedence is from top to bottom. b^c should be executed first and then a^(result) So, 3^3^3=3^(result) i.e. 3^(27) 4^5^6: Execute 5^6 first. then find 4^(result obtained before) (4^5)^6: Now, because of the bracket. 4^5 should be executed first and then result^6. I really don't know who created this rule or what's the proof. I just accept is as a rule. 2+3*42=2+122=12. (It is another rule of precedence that multiplication should be done before addition or subtraction) As I always say, "A table is a table".
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Director
Joined: 23 Apr 2010
Posts: 581

Re: last digit of a power [#permalink]
Show Tags
22 Sep 2011, 05:23
Fluke, I've read Bunuel's post thoroughly. What I don't understand is why it is like this: 3^3 = 27 , instead of 3*3 = 9
I do understand that we go from the top to bottom; however, I don't understand why we don't simply multiply 3*3 and instead apply power (3^3).
If you multiply you get 3^9. In the second case you get 3^27.
Thanks.



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: last digit of a power [#permalink]
Show Tags
22 Sep 2011, 05:32
nonameee wrote: Fluke, I've read Bunuel's post thoroughly. What I don't understand is why it is like this: 3^3 = 27 , instead of 3*3 = 9
I do understand that we go from the top to bottom; however, I don't understand why we don't simply multiply 3*3 and instead apply power (3^3).
If you multiply you get 3^9. In the second case you get 3^27.
Thanks. The question is: 3^ 3^ 3Using the topdown: 3^ 3=3*3*3=27(Result) Then, 3^(27)=Answer. If the question were: ( 3^ 3)^ 3 Note the bracket here Using the downtop, 9^ 3=Answer. It fits in.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Director
Joined: 23 Apr 2010
Posts: 581

Re: last digit of a power [#permalink]
Show Tags
22 Sep 2011, 05:36
Quote: Using the downtop, 9^3=Answer. You mean 3^9, not 9^3?



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: last digit of a power [#permalink]
Show Tags
22 Sep 2011, 05:41
nonameee wrote: Quote: Using the downtop, 9^3=Answer. You mean 3^9, not 9^3? Sorry, I meant: (27)^3. as 3^3=27 It is also equal to (3^3)^3=3^(3*3)=3^9 The point is: this case holds good only if you use brackets.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Director
Joined: 23 Apr 2010
Posts: 581

Re: last digit of a power [#permalink]
Show Tags
22 Sep 2011, 05:47
I think I got it. The rule is this (quoted from GMAT Math Bible):
\((a^m)^n=a^{mn}\)
\(a^m^n=a^{(m^n)}\)
and not \((a^m)^n\)
I think that's it.



Current Student
Joined: 02 Apr 2012
Posts: 77
Location: United States (VA)
Concentration: Entrepreneurship, Finance
WE: Consulting (Consulting)

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



Math Expert
Joined: 02 Sep 2009
Posts: 39589

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, 13:09
Maxirosario2012 wrote: 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 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 {3971}{3971}...) > 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: m2573474.html
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Re: What is the last digit 3^{3^3} ? A. 1 B. 3 C. 6 D. 7 E. 9
[#permalink]
23 Jul 2013, 13:09



Go to page
Previous
1 2 3 4
[ 70 posts ]




