m09 32 : Retired Discussions [Locked]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 16 Jan 2017, 15:03

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m09 32

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Director
Joined: 18 Feb 2008
Posts: 798
Followers: 21

Kudos [?]: 119 [5] , given: 25

### Show Tags

12 Jan 2009, 03:57
5
KUDOS
What is the last digit of $$3^{3^3}$$ ?

(A) 1
(B) 3
(C) 6
(D) 7
(E) 9

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

The answer is wrong

Should be B
Attachments

m09_32.JPG [ 15.56 KiB | Viewed 9274 times ]

Intern
Joined: 19 Dec 2008
Posts: 12
Followers: 0

Kudos [?]: 13 [1] , given: 0

Re: m09 32 [#permalink]

### Show Tags

12 Jan 2009, 04:55
1
KUDOS
It's correct...the explanation is correct too.
Director
Joined: 29 Aug 2005
Posts: 877
Followers: 9

Kudos [?]: 361 [1] , given: 7

Re: m09 32 [#permalink]

### Show Tags

20 Jan 2009, 01:59
1
KUDOS
who cares? billyjeans already got 770 on gmat!
Senior Manager
Joined: 10 Dec 2008
Posts: 482
Location: United States
GMAT 1: 760 Q49 V44
GPA: 3.9
Followers: 40

Kudos [?]: 197 [0], given: 12

Re: m09 32 [#permalink]

### Show Tags

30 Jul 2009, 08:00
This answer is clearly wrong! Mods can you please fix it? Thanks.
Manager
Joined: 12 Aug 2008
Posts: 62
Followers: 2

Kudos [?]: 3 [0], given: 2

Re: m09 32 [#permalink]

### Show Tags

26 Sep 2009, 06:46
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: 13 Dec 2009
Posts: 79
Followers: 1

Kudos [?]: 37 [0], given: 20

Re: m09 32 [#permalink]

### Show Tags

10 Feb 2010, 00:16
MUST BE CHANGED!!!
Joined: 20 Aug 2009
Posts: 311
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Followers: 18

Kudos [?]: 140 [4] , given: 69

Re: m09 32 [#permalink]

### Show Tags

10 Feb 2010, 00:54
4
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: 13 Dec 2009
Posts: 79
Followers: 1

Kudos [?]: 37 [0], given: 20

Re: m09 32 [#permalink]

### Show Tags

10 Feb 2010, 04:22
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, 06:09, edited 2 times in total.
Math Expert
Joined: 02 Sep 2009
Posts: 36520
Followers: 7066

Kudos [?]: 92912 [7] , given: 10528

Re: m09 32 [#permalink]

### Show Tags

10 Feb 2010, 05:23
7
KUDOS
Expert's post
2
This post was
BOOKMARKED
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) --> $$7$$.
_________________

Last edited by Bunuel on 02 Sep 2010, 04:52, edited 1 time in total.
Manager
Joined: 13 Dec 2009
Posts: 79
Followers: 1

Kudos [?]: 37 [0], given: 20

Re: m09 32 [#permalink]

### Show Tags

10 Feb 2010, 05:35
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
Followers: 1

Kudos [?]: 1 [0], given: 2

Re: m09 32 [#permalink]

### Show Tags

11 Feb 2010, 08:49
Last digit should be 7 So correct answer is d
Intern
Joined: 25 Aug 2010
Posts: 46
Followers: 1

Kudos [?]: 12 [1] , given: 3

Re: m09 32 [#permalink]

### Show Tags

02 Sep 2010, 04:58
1
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
Followers: 0

Kudos [?]: 4 [0], given: 3

Re: m09 32 [#permalink]

### Show Tags

02 Sep 2010, 06:13
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
Followers: 0

Kudos [?]: 0 [0], given: 1

Re: m09 32 [#permalink]

### Show Tags

02 Sep 2010, 06:34
Thank you, this question is very interesting!
Intern
Joined: 26 Aug 2010
Posts: 23
Followers: 0

Kudos [?]: 18 [1] , given: 2

Re: m09 32 [#permalink]

### Show Tags

02 Sep 2010, 06:39
1
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: 299
GMAT 1: 710 Q47 V40
Followers: 4

Kudos [?]: 48 [0], given: 7

Re: m09 32 [#permalink]

### Show Tags

02 Sep 2010, 06:42

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: 02 Jun 2010
Posts: 29
Followers: 0

Kudos [?]: 6 [0], given: 4

Re: m09 32 [#permalink]

### Show Tags

02 Sep 2010, 11:24
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: 197
Location: Prague
Schools: University of Economics Prague
Followers: 1

Kudos [?]: 42 [0], given: 15

Re: m09 32 [#permalink]

### Show Tags

02 Sep 2010, 13:23
Oh, I made the same stupid mistake and chose B
_________________

You want somethin', go get it. Period!

Manager
Joined: 19 Apr 2010
Posts: 210
Schools: ISB, HEC, Said
Followers: 4

Kudos [?]: 77 [0], given: 28

Re: m09 32 [#permalink]

### Show Tags

03 Sep 2010, 03:01
Thanks for explaination Bunuel I also arrived at 3 as answer
Intern
Joined: 29 Aug 2010
Posts: 4
Followers: 0

Kudos [?]: 0 [0], given: 3

Re: m09 32 [#permalink]

### Show Tags

03 Sep 2010, 04:03
Answer is B as already explained by many people here.
Re: m09 32   [#permalink] 03 Sep 2010, 04:03

Go to page    1   2    Next  [ 31 posts ]

Similar topics Replies Last post
Similar
Topics:
1 m09 Q32 3 19 Nov 2010, 09:20
16 M09 Q11 18 19 Nov 2008, 06:03
m09 #34 6 10 Nov 2008, 03:10
19 M09 #20 19 10 Nov 2008, 02:24
10 m09 q22 41 07 Sep 2008, 12:18
Display posts from previous: Sort by

# m09 32

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.