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# m09 Q32

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Manager
Joined: 16 Feb 2010
Posts: 225
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Kudos [?]: 316 [0], given: 16

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19 Nov 2010, 10:20
 What is the last digit of $$3^{3^3}$$ ?(C) 2008 GMAT Club - m09#32 * 1 * 3 * 6 * 7 * 9$$3^{3^3} = 3^{27} = ((3^4)^6) * (3^3) = 27 * (81^6)$$$$81^6$$ ends with 1, so $$27*(81^6)$$ ends with 7.The correct answer is D.

ps. used excel to verify !
I disagree as $$3^{3^3} = (3^3)(3^3)(3^3)= 27x27x27$$
therefore the answer should be 19683 thus B
Math Expert
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19 Nov 2010, 10:28
1
KUDOS
Expert's post
zisis wrote:
 What is the last digit of $$3^{3^3}$$ ?(C) 2008 GMAT Club - m09#32 * 1 * 3 * 6 * 7 * 9$$3^{3^3} = 3^{27} = ((3^4)^6) * (3^3) = 27 * (81^6)$$$$81^6$$ ends with 1, so $$27*(81^6)$$ ends with 7.The correct answer is D.

ps. used excel to verify !
I disagree as $$3^{3^3} = (3^3)(3^3)(3^3)= 27x27x27$$
therefore the answer should be 19683 thus B

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) --> $$7$$.

Discussed here: last-digit-of-a-power-70624.html and here: m09-74730.html

Hope it's clear.
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Manager
Joined: 16 Feb 2010
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20 Nov 2010, 01:16
gotta - makes perfect sense and kudos
Manager
Joined: 01 Nov 2010
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20 Nov 2010, 10:52
3^1=3
3^2=9
3^3=27
3^4=81
3^5=243

Means after every 4th power, the unit's digit will be repeated.

thus for 27 powers, it will be 27/4= 6 with reminder-3. where units digit is - 7..

Re: m09 Q32   [#permalink] 20 Nov 2010, 10:52
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# m09 Q32

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