Author 
Message 
Director
Joined: 18 Feb 2008
Posts: 790

5
This post received KUDOS
What is the last digit of \(3^{3^3}\) ? (A) 1 (B) 3 (C) 6 (D) 7 (E) 9 Source: GMAT Club Tests  hardest GMAT questions The answer is wrong Should be B
Attachments
m09_32.JPG [ 15.56 KiB  Viewed 9514 times ]



Intern
Joined: 19 Dec 2008
Posts: 12

Re: m09 32 [#permalink]
Show Tags
12 Jan 2009, 05:55
1
This post received KUDOS
It's correct...the explanation is correct too.



Director
Joined: 29 Aug 2005
Posts: 859

Re: m09 32 [#permalink]
Show Tags
20 Jan 2009, 02:59
1
This post received KUDOS
who cares? billyjeans already got 770 on gmat!



Senior Manager
Joined: 11 Dec 2008
Posts: 478
Location: United States
GPA: 3.9

Re: m09 32 [#permalink]
Show Tags
30 Jul 2009, 09:00
This answer is clearly wrong! Mods can you please fix it? Thanks.



Manager
Joined: 12 Aug 2008
Posts: 61

Re: m09 32 [#permalink]
Show Tags
26 Sep 2009, 07:46
bipolarbear wrote: This answer is clearly wrong! Mods can you please fix it? Thanks. I agree. A power raised to a power is multipled, not treated as an exponential. I think there is a similar GMAT prep question, in which a power raised to a power is multipled.



Manager
Joined: 14 Dec 2009
Posts: 76

Re: m09 32 [#permalink]
Show Tags
10 Feb 2010, 01:16
MUST BE CHANGED!!!



Tuck Thread Master
Joined: 20 Aug 2009
Posts: 306
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)

Re: m09 32 [#permalink]
Show Tags
10 Feb 2010, 01:54
4
This post received KUDOS
The answer is correct, I don't see the mistake Can anyone elaborate more on this question? 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.



Manager
Joined: 14 Dec 2009
Posts: 76

Re: m09 32 [#permalink]
Show Tags
10 Feb 2010, 05:22
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, 07:09, edited 2 times in total.



Math Expert
Joined: 02 Sep 2009
Posts: 39622

Re: m09 32 [#permalink]
Show Tags
10 Feb 2010, 06:23
7
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
Last edited by Bunuel on 02 Sep 2010, 05:52, edited 1 time in total.



Manager
Joined: 14 Dec 2009
Posts: 76

Re: m09 32 [#permalink]
Show Tags
10 Feb 2010, 06:35
Bunuel wrote: Shalva is absolutely correct here.
RULE: The order of operation for exponents: \(a^m^n=a^{(m^n)}\) and not \((a^m)^n\). The rule is to work from the top down.
\(3^{3^3}=3^{(3^3)}=3^{27}\)
Cyclicity of 3 in power is four. The units digit of \(3^{27}\) is the same as for \(3^3\) (27=4*6+3) > \(7\). Thank you for your explanation!



Intern
Joined: 11 Sep 2009
Posts: 18
Location: Tampa ,FL

Re: m09 32 [#permalink]
Show Tags
11 Feb 2010, 09:49
Last digit should be 7 So correct answer is d



Intern
Joined: 25 Aug 2010
Posts: 41

Re: m09 32 [#permalink]
Show Tags
02 Sep 2010, 05:58
1
This post received KUDOS
Bunuel wrote: Shalva is absolutely correct here.
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



Intern
Joined: 30 Aug 2009
Posts: 26

Re: m09 32 [#permalink]
Show Tags
02 Sep 2010, 07:13
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")



Intern
Joined: 30 Aug 2010
Posts: 6

Re: m09 32 [#permalink]
Show Tags
02 Sep 2010, 07:34
Thank you, this question is very interesting!



Intern
Joined: 27 Aug 2010
Posts: 23

Re: m09 32 [#permalink]
Show Tags
02 Sep 2010, 07:39
1
This post received KUDOS
Another way to do this task:
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.



Senior Manager
Joined: 17 May 2010
Posts: 290

Re: m09 32 [#permalink]
Show Tags
02 Sep 2010, 07:42
Answer is 7. Step 1: 3^3= 27. Step 2: Therefore 3^(3^3)=3^27. = 3*3*3*.......3 (27 3s). Step 3: We know 3*3*3 = 27 Step 4: Hence step 2 can be simplified as 27*27*...27 (9 27s) Step 5: We know 27*27*27 ends in 3 (multiply it out. No need to do the entire multiplication. Only the last digits matter). Step 6: Hence we end up abc3*abc3*abc3 = a number ending with 7 (again only the last digits matter) Step 7: Voila!
_________________
If you like my post, consider giving me KUDOS!



Intern
Joined: 03 Jun 2010
Posts: 29

Re: m09 32 [#permalink]
Show Tags
02 Sep 2010, 12:24
Bunuel wrote: Shalva is absolutely correct here.
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) > \(7\). Greate explanation Bunuel



Manager
Joined: 27 Jul 2010
Posts: 194
Location: Prague
Schools: University of Economics Prague

Re: m09 32 [#permalink]
Show Tags
02 Sep 2010, 14:23
Oh, I made the same stupid mistake and chose B
_________________
You want somethin', go get it. Period!



Manager
Joined: 20 Apr 2010
Posts: 210
Schools: ISB, HEC, Said

Re: m09 32 [#permalink]
Show Tags
03 Sep 2010, 04:01
Thanks for explaination Bunuel I also arrived at 3 as answer



Intern
Joined: 29 Aug 2010
Posts: 4

Re: m09 32 [#permalink]
Show Tags
03 Sep 2010, 05:03
Answer is B as already explained by many people here.







Go to page
1 2
Next
[ 31 posts ]




