m09 32

billyjeans · **Posted:** Mon Jan 12, 2009 3:57 am

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

shalva · **Posted:** Wed Feb 10, 2010 12:54 am

The answer is correct, I don't see the mistake :shock:

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.

Bunuel · **Posted:** Wed Feb 10, 2010 5:23 am

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.

Mbadream4ever · **Posted:** Mon Jan 12, 2009 4:55 am

It's correct...the explanation is correct too.

seofah · **Posted:** Tue Jan 20, 2009 1:59 am

who cares? :lol:

billyjeans already got 770 on gmat!

bipolarbear · **Posted:** Thu Jul 30, 2009 8:00 am

This answer is clearly wrong! Mods can you please fix it? Thanks.

sid3699 · **Posted:** Sat Sep 26, 2009 6:46 am

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.

Igor010 · **Posted:** Wed Feb 10, 2010 12:16 am

MUST BE CHANGED!!! :shock:

Igor010 · **Posted:** Wed Feb 10, 2010 4:22 am

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?

Igor010 · **Posted:** Wed Feb 10, 2010 5:35 am

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!

Bhuvi · **Posted:** Thu Feb 11, 2010 8:49 am

Last digit should be 7 So correct answer is d

cnrnld · **Posted:** Thu Sep 02, 2010 4:58 am

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

thb · **Posted:** Thu Sep 02, 2010 6:13 am

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")

Nguyento · **Posted:** Thu Sep 02, 2010 6:34 am

Thank you, this question is very interesting!

Kronax · **Posted:** Thu Sep 02, 2010 6:39 am

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.

gijoedude · **Posted:** Thu Sep 02, 2010 6:42 am

Answer is 7.

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!

surjoy · **Posted:** Thu Sep 02, 2010 11:24 am

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

craky · **Posted:** Thu Sep 02, 2010 1:23 pm

Oh, I made the same stupid mistake and chose B

prashantbacchewar · **Posted:** Fri Sep 03, 2010 3:01 am

Thanks for explaination Bunuel I also arrived at 3 as answer

sravani · **Posted:** Fri Sep 03, 2010 4:03 am

Answer is B as already explained by many people here.

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