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What is the last digit 3^{3^3} ? A. 1 B. 3 C. 6 D. 7 E. 9

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Re: last digit of a power [#permalink] New post 01 Sep 2011, 21:45
The digit 3 has the cyclicity of 4.
means
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243

So, the given problem can be stated as (3^3)^3, which is 3^27
Now 27 divided by 4 leaves a remainder of 3.
Then, counting 3 back in the cyclicity , the Unit digit is 7.

So the Answer is D.
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Re: last digit of a power [#permalink] New post 22 Sep 2011, 01:54
Bunuel wrote:
toannguyen wrote:
Actually, in my opinion 3^3^3 should be reduced to 3^9 as the formula goes: a^x^y = a^(xy). 3^9 = 19,683 --> the unit number is 3, not 7 as some of you explained before.

Or we can use the method as proposed by some guy here: 9 mod 4 = 1 --> the last digit will be 3^1 = 3.


OA for this question is D (7).

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).


Can someone please explain why the exponent is 27 and not 9 (3x3)? Thanks.
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Re: last digit of a power [#permalink] New post 22 Sep 2011, 05:13
nonameee wrote:
Can someone please explain why the exponent is 27 and not 9 (3x3)? Thanks.


I'll have to repeat what Bunuel already said:

(a^b)^c: Here, (a^b) should be executed first AND then the (result)^c should be executed.

If there are no brackets:

a^b^c: Then the precedence is from top to bottom. b^c should be executed first and then a^(result)

So,

3^3^3=3^(result) i.e. 3^(27)

4^5^6: Execute 5^6 first. then find 4^(result obtained before)

(4^5)^6: Now, because of the bracket. 4^5 should be executed first and then result^6.

I really don't know who created this rule or what's the proof. I just accept is as a rule.

2+3*4-2=2+12-2=12. (It is another rule of precedence that multiplication should be done before addition or subtraction)

As I always say,
"A table is a table".
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Re: last digit of a power [#permalink] New post 22 Sep 2011, 05:23
Fluke, I've read Bunuel's post thoroughly. What I don't understand is why it is like this:
3^3 = 27 , instead of 3*3 = 9

I do understand that we go from the top to bottom; however, I don't understand why we don't simply multiply 3*3 and instead apply power (3^3).

If you multiply you get 3^9. In the second case you get 3^27.

Thanks.
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Re: last digit of a power [#permalink] New post 22 Sep 2011, 05:32
nonameee wrote:
Fluke, I've read Bunuel's post thoroughly. What I don't understand is why it is like this:
3^3 = 27 , instead of 3*3 = 9

I do understand that we go from the top to bottom; however, I don't understand why we don't simply multiply 3*3 and instead apply power (3^3).

If you multiply you get 3^9. In the second case you get 3^27.

Thanks.


The question is:

3^3^3

Using the top-down:
3^3=3*3*3=27(Result)

Then,
3^(27)=Answer.

If the question were:
(3^3)^3-- Note the bracket here

Using the down-top,
9^3=Answer.

It fits in.
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Re: last digit of a power [#permalink] New post 22 Sep 2011, 05:36
Quote:
Using the down-top,
9^3=Answer.


You mean 3^9, not 9^3?
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Re: last digit of a power [#permalink] New post 22 Sep 2011, 05:41
nonameee wrote:
Quote:
Using the down-top,
9^3=Answer.


You mean 3^9, not 9^3?


Sorry, I meant:

(27)^3. as 3^3=27

It is also equal to (3^3)^3=3^(3*3)=3^9

The point is: this case holds good only if you use brackets.
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Re: last digit of a power [#permalink] New post 22 Sep 2011, 05:47
I think I got it. The rule is this (quoted from GMAT Math Bible):

(a^m)^n=a^{mn}

a^m^n=a^{(m^n)}

and not (a^m)^n

I think that's it.
Re: last digit of a power   [#permalink] 22 Sep 2011, 05:47
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