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Director  Joined: 18 May 2008
Posts: 895

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1
11
What is the last digit of $$3^{3^3}$$ ?

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

Spoiler: :: OA
D

Source: GMAT Club Tests - hardest GMAT questions

What should be the approach in such types of questions?
VP  Joined: 17 Jun 2008
Posts: 1098

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4
1
3^x will have a pattern with unit digit as 3, 9, 7,1, 3,9,7,1,....

Thus, after every four values, the unit digit pattern will repeat.

Divide 27 by 4....3 is the remainder. Unit digit for the third value (from the pattern above) will be 7.
Director  Joined: 18 May 2008
Posts: 895

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wow! wht an easy approach! Thanks a lot!
scthakur wrote:
3^x will have a pattern with unit digit as 3, 9, 7,1, 3,9,7,1,....

Thus, after every four values, the unit digit pattern will repeat.

Divide 27 by 4....3 is the remainder. Unit digit for the third value (from the pattern above) will be 7.
Manager  Joined: 22 Jul 2009
Posts: 147

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25
4
To solve this kind of problems, it saves time to remember the following regarding the units digit:

.............................^1....^2.....^3.....^4.....^5......^6
0 has a cycle of 1.......0......0........0
1 has a cycle of 1.......1......1........1
2 has a cycle of 4.......2......4......8......16.......32.....64
3 has a cycle of 4.......3......9......27.....81......243....729
4 has a cycle of 2.......4......16......64....256
5 has a cycle of 1.......5......25......125
6 has a cycle of 1.......6......36......216
7 has a cycle of 4.......7......49......341...2401...16807..117649
8 has a cycle of 4.......8......64......512...4096..32768..262144
9 has a cycle of 2.......9......81......729....6561

This pattern of 1-1-4-4-2 goes on and on into infinity:
10 has a cycle of 1
11 has a cycle of 1
12 has a cycle of 4
13 has a cycle of 4
14 has a cycle of 2
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Intern  Joined: 08 Nov 2009
Posts: 40

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I got this question wrong because I interpreted 3^3^3 as (3^3)^3.

Is commonly accepted convention to start developing the exponents from the one on top to the one on the bottom (i.e. x^(a^(b^(c^d))))?
Math Expert V
Joined: 02 Sep 2009
Posts: 59236

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7
2
Marco83 wrote:
I got this question wrong because I interpreted 3^3^3 as (3^3)^3.

Is commonly accepted convention to start developing the exponents from the one on top to the one on the bottom (i.e. x^(a^(b^(c^d))))?

Yes: $$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$.

Check Number Theory chapter in Math Book (link below).
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Intern  Joined: 22 Dec 2009
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I didnt know you had to divide by 4 and to pick the number of the remainder..

How about the case for 3^12 where has zero remainder...?
Math Expert V
Joined: 02 Sep 2009
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gmatJP wrote:
I didnt know you had to divide by 4 and to pick the number of the remainder..

How about the case for 3^12 where has zero remainder...?

When remainder is zero you should take the base in the power of cyclisity. Last digit of 3^12 = last digit of 3^4, as the cyclisity of 3 if 4 and 12 divided by 4 leaves remainder zero.

For more see the Number Theory chapter in Math Book.
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Manager  Joined: 01 Nov 2010
Posts: 208
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
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the ans will be 3.
for this type of question which has small no as 2, 3 ,4.
you can calculate it by multiplying only the unit digit if the nos are countable on fingers,
else you should use the above method explained by Gmat Aspirants.
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Math Expert V
Joined: 02 Sep 2009
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321kumarsushant wrote:
the ans will be 3.
for this type of question which has small no as 2, 3 ,4.
you can calculate it by multiplying only the unit digit if the nos are countable on fingers,
else you should use the above method explained by Gmat Aspirants.

OA for this question is D (7).

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.

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Intern  Joined: 30 Nov 2010
Posts: 22
Location: Boston
Schools: Boston College, MIT, BU, IIM, UCLA, Babson, Brown

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wow! great question and useful tips.. thanks
Manager  Joined: 01 Nov 2010
Posts: 208
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)

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Bunuel wrote:
321kumarsushant wrote:
the ans will be 3.
for this type of question which has small no as 2, 3 ,4.
you can calculate it by multiplying only the unit digit if the nos are countable on fingers,
else you should use the above method explained by Gmat Aspirants.

OA for this question is D (7).

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.

ohh ! i got that question wrong.
the ans will be 7.
above explanation is appreciable.
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Intern  Status: Trying to get into the illustrious 700 club!
Joined: 18 Oct 2010
Posts: 41

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Bunuel wrote:
gmatJP wrote:
I didnt know you had to divide by 4 and to pick the number of the remainder..

How about the case for 3^12 where has zero remainder...?

When remainder is zero you should take the base in the power of cyclisity. Last digit of 3^12 = last digit of 3^4, as the cyclisity of 3 if 4 and 12 divided by 4 leaves remainder zero.

For more see the Number Theory chapter in Math Book.

Good point. I did everything like you guys except I chose the wrong starting point. I divided 27 by 4 and got 6 remainder 3 but thought 3^1 was the point of reference and chose 1 as the answer and not 7.

Thanks for clearing that up Bunuel!
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Intern  Joined: 23 Oct 2010
Posts: 48
Location: India

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D.

3^3^3 = 3^27
Now we need to find a pattern for last digits of different powers of 3 -->
1 2 3 4
3 9 7 1

after every 4 different unit digits the pattern repeates itself.
Hence for 4*6 = 24th power the units digit will be 1
27 - 4 = 3
the third unit's digit in the pattern above is 7
Manager  Joined: 10 Jan 2011
Posts: 112
Location: India
GMAT Date: 07-16-2012
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ritula wrote:
What is the last digit of $$3^{3^3}$$ ?

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

Spoiler: :: OA
D

Source: GMAT Club Tests - hardest GMAT questions

What should be the approach in such types of questions?

I solved the problem as below:

3^3^3 = 3^27= 3^9*3^9*3^9.

3^9 = 3^3*3^3*3^3 which has a unit digit of 3

unit digit of 3^27 = 3*3*3 = 27 hence 7

D
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-------Analyze why option A in SC wrong-------
Intern  Joined: 01 Aug 2011
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VERY SIMPLE
FIRST TWO POWER=9
THEN 3 RAISED TO POWER 9=(3.3).(3.3.).(3.3)(3.3)3
=(9.9.9.9.)3
=(81.81)3
=FIRST DIGIT AS1
Manager  Joined: 21 Nov 2010
Posts: 82

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D

Just take the last number when you square...and then look for order

3^{3^3} = 3^(27)

3^1 = 3
3^2 = 9
3^3 = 7 (remember only need the last number)
3^4 = 1 (same as above)
3^5 = 3 ...now the numbers are repeating

All you have to do now is see where 27 would land which is 3^3, so answer is 7.
Manager  Joined: 27 Oct 2011
Posts: 116
Location: United States
Concentration: Finance, Strategy
GPA: 3.7
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i broke it down to 3^27 => then 27 is divided by 3, 9 times. 3^3 = 27
then 7^9. break it down to pairs 7*7 = 49.
there will be four 9's and a remaining 7.
1*1*7 = 7
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Intern  Joined: 12 Dec 2012
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I had not find a topic for this question, so I have created this one.
The thing is - 3^3^2 = 3^27 in the official answer.
Is not it - 3^3^2 = 3^9 ?
Thank you for the attention.
Attachments d63d1fe678552ac0338e3bdaa5102421.png [ 261.88 KiB | Viewed 8203 times ]

Math Expert V
Joined: 02 Sep 2009
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KirikViktor wrote:
I had not find a topic for this question, so I have created this one.
The thing is - 3^3^2 = 3^27 in the official answer.
Is not it - 3^3^2 = 3^9 ?
Thank you for the attention.

Merging topics.

It's $$3^{3^3}=3^{(3^3)}=3^{27}$$, not $$3^{3^2}$$.

Hope it's clear.
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