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m25,#31

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m25,#31 [#permalink]

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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

What should be the approach in such types of questions?
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Re: m25,#31 [#permalink]

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New post 30 Aug 2009, 14:01
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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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Re: m25,#31 [#permalink]

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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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Re: m25,#31 [#permalink]

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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.

Answer: D (7).
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Re: m25,#31 [#permalink]

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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.
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M25-31 [#permalink]

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New post 18 Jul 2013, 12:19
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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.
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New post 18 Jul 2013, 12:27
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Re: m25,#31 [#permalink]

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New post 01 Dec 2008, 20:58
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.
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Re: m25,#31 [#permalink]

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New post 03 Jan 2010, 16:24
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))))?
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Re: m25,#31 [#permalink]

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New post 11 Jan 2010, 19:51
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...?
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Re: m25,#31 [#permalink]

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New post 11 Jan 2010, 22:47
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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Re: m25,#31 [#permalink]

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New post 01 Dec 2010, 08:43
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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Re: m25,#31 [#permalink]

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New post 01 Dec 2010, 08:59
wow! great question and useful tips.. thanks
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Re: m25,#31 [#permalink]

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New post 01 Dec 2010, 10:02
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.

Answer: D (7).



ohh ! i got that question wrong.
the ans will be 7.
above explanation is appreciable.
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Re: m25,#31 [#permalink]

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New post 01 Dec 2010, 11:04
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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Re: m25,#31 [#permalink]

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New post 02 Dec 2010, 07:46
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
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Re: m25,#31 [#permalink]

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New post 05 Dec 2011, 23:09
ritula wrote:
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

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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Re: m25,#31 [#permalink]

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New post 06 Dec 2011, 03:34
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
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Re: m25,#31 [#permalink]

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New post 07 Dec 2011, 11:51
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.
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Re: m25,#31 [#permalink]

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New post 04 Feb 2012, 22:24
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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Re: m25,#31   [#permalink] 04 Feb 2012, 22:24
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