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# M25-31

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Math Expert
Joined: 02 Sep 2009
Posts: 49320

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16 Sep 2014, 01:24
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Difficulty:

45% (medium)

Question Stats:

53% (00:23) correct 47% (00:28) wrong based on 331 sessions

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

A. 1
B. 3
C. 6
D. 7
E. 9

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Math Expert
Joined: 02 Sep 2009
Posts: 49320

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16 Sep 2014, 01:24
2
1
Official Solution:

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 the above:

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

Now, the units digit of 3 in positive integer power repeats in groups of four: {3-9-7-1} - {3-9-7-1} - ... Hence, the units digit of $$3^{27}$$ is the same as the units digit of $$3^3$$, which is 7.

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Joined: 17 Jun 2014
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04 Oct 2014, 22:00
why cant we start from 3^0 instead of 3^1 ?
I did in this way: 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27 then the cycle repeat with 3^4 = 81, 3^5 =243
the pattern here is (1,3,9,7) the answer should be e ?
Math Expert
Joined: 02 Sep 2009
Posts: 49320

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05 Oct 2014, 02:16
langtuprovn2007 wrote:
why cant we start from 3^0 instead of 3^1 ?
I did in this way: 3^0 = 1, 3^1 = 3, 3^2 = 9, 3^3 = 27 then the cycle repeat with 3^4 = 81, 3^5 =243
the pattern here is (1,3,9,7) the answer should be e ?

That's just wrong. If this were the case, then all numbers in power which is a multiple of cyclicity would have the last digit of 1. You should always start with power of 1.

For more check UNITS DIGITS, EXPONENTS, REMAINDERS PROBLEMS in our Special Questions Directory.

Hope it helps.
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Joined: 17 Jun 2014
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05 Oct 2014, 03:09
can you give me some examples plz !
Math Expert
Joined: 02 Sep 2009
Posts: 49320

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05 Oct 2014, 09:08
langtuprovn2007 wrote:
can you give me some examples plz !

There are a lot of examples in the link given above.
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Joined: 28 Jul 2014
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07 Nov 2015, 13:44
My 2 cents: It doesn't matter whether you start your powers at 0 or at 1; all that matters is recognizing the pattern.
let mn = 0,1,2,3,...; then let last digit = x = 1,3,9,7,....; Because x repeats in these 4 numbers; we can tell that x = 1 implies mn = 0,4,8,12....; x=3 implies mn = 1,5,9...; x=9 implies mn = 2,6,10,...; x=7, implies mn = 3,7,11,.... The question that becomes, which series gets you to 27? x=1 and x =3 gives evens, so rule out, leaving x=3 and x=7. From here you can manually add 4 to each series and see which one gets to 27. The answer is x=7. Or if you like algebra, find the series formula for each. For x=3; we have mn = 1,5,9...= 4a -3. There is no integer a such that 4a-3 = 27. On the other hand, for x=7, we have mn = 3,7,11,... or 4a-1. And 4a-1=27 yields an integer value of a, meaning that this series will reach 27
Intern
Joined: 10 Jun 2017
Posts: 23

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21 Apr 2018, 05:18
Hi guys,

when we divide 27 by 3, that yields a remainder of 3 (maximum possible remainder since R < Divisor)

Now when:

3^1 = 3 R=0
3^2 = 9 R=1
3^3 = 27 R=2
3^4 = 81 R=3

Now since the remainder was 3, I've associated to the last digit of 1.

Can anyone please tell me where I've went wrong in my thought proccess?

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 49320

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21 Apr 2018, 05:31
1
Miracles86 wrote:
Hi guys,

when we divide 27 by 3, that yields a remainder of 3 (maximum possible remainder since R < Divisor)

Now when:

3^1 = 3 R=0
3^2 = 9 R=1
3^3 = 27 R=2
3^4 = 81 R=3

Now since the remainder was 3, I've associated to the last digit of 1.

Can anyone please tell me where I've went wrong in my thought proccess?

Thanks

You should divide the power, which is 27, by cyclicity number, which is 4 to find the remainder: 27 divided by 4 gives the remainder of 3. Hence, the units digit of 3^27 is the same as the units digit of 3^3, which is 7.
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21 Apr 2018, 05:35
Got it! I was confusing this exercise with those when u need to find the digit of a non terminating repeating decimal. On those, the value of the remainder gets you straight to the answer...

Thanks a lot!
Intern
Joined: 27 Jan 2012
Posts: 2
Location: India
GMAT 1: 720 Q50 V38
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19 Jul 2018, 06:16
First you should solve for the index i.e. 3 raise to 3 = 27

den 3 raise to 27

so cycle of 4 is. 3 , 9, 7, 1.

27 comes 3rd in this cycle... so answer will be 7

Hope its clear now.
Re: M25-31 &nbs [#permalink] 19 Jul 2018, 06:16
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# M25-31

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