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m09 32

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12 Jan 2009, 03:57
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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

The answer is wrong

Should be B
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12 Jan 2009, 04:55
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It's correct...the explanation is correct too.
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20 Jan 2009, 01:59
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who cares? billyjeans already got 770 on gmat!
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30 Jul 2009, 08:00
This answer is clearly wrong! Mods can you please fix it? Thanks.
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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.
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10 Feb 2010, 00:16
MUST BE CHANGED!!!
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Re: m09 32 [#permalink]

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10 Feb 2010, 00:54
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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.
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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.
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10 Feb 2010, 05:23
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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) --> $$7$$.
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Last edited by Bunuel on 02 Sep 2010, 04:52, edited 1 time in total.
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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!
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11 Feb 2010, 08:49
Last digit should be 7 So correct answer is d
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02 Sep 2010, 04:58
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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
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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")
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02 Sep 2010, 06:34
Thank you, this question is very interesting!
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02 Sep 2010, 06:39
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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.
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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!
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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
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02 Sep 2010, 13:23
Oh, I made the same stupid mistake and chose B
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03 Sep 2010, 03:01
Thanks for explaination Bunuel I also arrived at 3 as answer
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03 Sep 2010, 04:03
Answer is B as already explained by many people here.
Re: m09 32   [#permalink] 03 Sep 2010, 04:03

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