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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]

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05 Aug 2009, 22:23
bipolarbear wrote:
Is 3^3^3 definitely taken as 3^(3^3)?

Is reading it as (3^3)^3 incorrect?

yes bracket makes a different case altogether
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Re: last digit of a power [#permalink]

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07 Aug 2009, 13:36
elmagnifico wrote:
What is the last digit $$3^{3^3}$$ ?

* 1
* 3
* 6
* 7
* 9

(a^m)^n = a^mn

(3^3)^3 = 3^(3*3)=3^9.

Since cyclicity of 3 is 4.units digit of 3^9 is 3.

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Re: last digit of a power [#permalink]

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11 Aug 2009, 03:19
3^27

and
3^1=3
3^2=9
3^3=7
3^4=1

and it repeats

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Re: last digit of a power [#permalink]

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11 Aug 2009, 04:10
elmagnifico wrote:
What is the last digit $$3^{3^3}$$ ?

* 1
* 3
* 6
* 7
* 9

SOL:
3^3^3 = 3^27

We know that for any integer the power cycle has a periodicity of 4.
=> LD (3^27) = LD (3^3) ................... (Since 27 MOD 4 = 3)
=> LD (3^3) = LD (27)
=> 7

ANS: 7
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Re: last digit of a power [#permalink]

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18 Aug 2009, 08:33
Somebody needs to back me up here...

First of all: 3^3^3 is neither 27^3 nor 3^27

I can prove it:

3^3^3 is equal to (3^3)*(3^3)*(3^3)

and we know that (a^n)*(a^m) = a^(n+m)

so...

(3^3)*(3^3)*(3^3) = 3^(3+3+3)

= 3^9

The actual property of powers is;
(a^n)^m = a^(n*m)

so for 3^3^3 = 3^(3*3)

or

3^9

With that being said, if you want to find the last number of any product, you need only to multiply the last digits:

3*3 = 9

so we can simplify 3*3*3*3*3*3*3*3*3 to 9*9*9*9*3

and because 9*9 = 81

we can simplify further to 1*1*3

ANSWER: the last digit is 3

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Re: last digit of a power [#permalink]

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18 Aug 2009, 09:38
h2polo wrote:
(3^3)*(3^3)*(3^3) = 3^(3+3+3)

= 3^9

This interpretation is incorrect! 3^3^3 cannot be equal to 3^9.....

A simple check would be to type the expression 3^3^3 on MS-Excel or google. The answer would be same as that of 3^27 which is surely different from 3^9.
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Re: last digit of a power [#permalink]

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18 Aug 2009, 11:05

in SciLab, 3^3^3=3^(3^3)=3^27

For this question, the concern would be in power computing, for instance X^Y^Z, which one has the higher priority, in other words, which part should be computed first?

another example would be 2^2^3 or 2^3^2? How to compute these two?
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Re: last digit of a power [#permalink]

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18 Aug 2009, 11:29
http://en.wikipedia.org/wiki/Exponentiation

according to this, the answer of "the last digit of 3^3^3" should be 7.

if you know the sequence of computing exponentiation, this question is a piece of cake, otherwise you would never get the answer right.
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Re: last digit of a power [#permalink]

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18 Aug 2009, 12:29
1
KUDOS
flyingbunny wrote:

in SciLab, 3^3^3=3^(3^3)=3^27

For this question, the concern would be in power computing, for instance X^Y^Z, which one has the higher priority, in other words, which part should be computed first?

another example would be 2^2^3 or 2^3^2? How to compute these two?

Well, I hadn't tried in Excel as I was sure it would give the correct answer! I wonder why it computes it wrongly.

Anyway we need to remember that in case of powers the computation need to be done from right to left and not from left to right. Thus 3^3^3 = 3^(3^3) = 3^27
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Re: last digit of a power [#permalink]

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26 Aug 2009, 03:22
Guys I have a querie:
3 has a cycle of four:
3^0=1
3^1=3
3^2=9
3^3=7
3^4=1
So the cycle is 1,3,9,7,,,,
Now we need 3^27.27 is 4*6+3 which is 6 full cycles of 4.Remainder is 3.So the 3rd. no. in the cycle is the answer which is 7.
HELP!!!!
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Re: last digit of a power [#permalink]

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26 Aug 2009, 13:43
tejal777 wrote:
Guys I have a querie:
3 has a cycle of four:
3^0=1
3^1=3
3^2=9
3^3=7
3^4=1
So the cycle is 1,3,9,7,,,,
Now we need 3^27.27 is 4*6+3 which is 6 full cycles of 4.Remainder is 3.So the 3rd. no. in the cycle is the answer which is 7.
HELP!!!!

Hi Tejal

Are you worried that you have got the rigth by using a perfect method!! Why do you need help? Yours is a perfect solution.
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Re: last digit of a power [#permalink]

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26 Aug 2009, 17:58
it can be written as 3^27..

Now,
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243

This implies that 3^4 has a unit digit of 1. Hence, 3^24 will also have a unit digit of 1.

=> now, we left with (unit digit 1 from 3^24) * 3^3 = (unit digit 1 from 3^24) * 27 = (unit digit 1 from 3^24) * (unit digit 7) = unit digit 7

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Re: last digit of a power [#permalink]

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26 Aug 2009, 18:44
onceatsea wrote:
Here's another way of looking at it !

Here the given number is $$(xyz)^n$$
z is the last digit of the base.
n is the index

To find out the last digit in $$(xyz)^n$$, the following steps are to be followed.
Divide the index (n) by 4, then

Case I
If remainder = 0
then check if z is odd (except 5), then last digit = 1
and if z is even then last digit = 6

Case II
If remainder = 1, then required last digit = last digit of the base (i.e. z)
If remainder = 2, then required last digit = last digit of the base $$(z)^2$$
If remainder = 3, then required last digit = last digit of the base $$(z)^3$$

Note : If z = 5, then the last digit in the product = 5

Example:
Find the last digit in (295073)^130

Solution: Dividing 130 by 4, the remainder = 2
Refering to Case II, the required last digit is the last digit of $$(z)^2$$, ie $$(3)^2$$ = 9 , (because z = 3)

Great explanation .. kudos
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Re: last digit of a power [#permalink]

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28 Aug 2009, 04:31
bhushan252 wrote:
3^3^3 to solve this we have to take top down approach...we cannot deduce 27^3......it should be 3^27....hence answer is 7

Hey buddy, how do you come from 27^3 to 3^27????
Something here I cant understand...

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Re: last digit of a power [#permalink]

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28 Aug 2009, 05:14
27^ 3 is (3^3) ^ 3 and not as 3^3^3 t
here is difference between both - (3^3) ^ 3 means (3*3*3) whole raise to 3 & 3^3^3 means 3^(3*3*3)

27^3 = 19683
3^ 27 = 7625597484987 (just for cross chk)

Ans 7
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Re: last digit of a power [#permalink]

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28 Aug 2009, 05:22
defoue wrote:
bhushan252 wrote:
3^3^3 to solve this we have to take top down approach...we cannot deduce 27^3......it should be 3^27....hence answer is 7

Hey buddy, how do you come from 27^3 to 3^27????
Something here I cant understand...

hi defoue....as i said plz follow top down apprach ...i mean if you have 2^3^4^2 this means 2^(3^(4^2))
hence 2^3^16......2^43046721.......now use our cyclicity table and hence answer is 2...sorry fortaking such weird number but couldn't think of anything else
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Re: last digit of a power [#permalink]

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07 Oct 2009, 10:35
Assuming we follow the order of operations, wouldn't we take the equation from left to right? that being (3^3)^3 which would actually yield 27^3 or even 3^9?

if this is true, the answer would be 3 as the units digit.
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Re: last digit of a power [#permalink]

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07 Oct 2009, 11:06
azule45 wrote:
Assuming we follow the order of operations, wouldn't we take the equation from left to right? that being (3^3)^3 which would actually yield 27^3 or even 3^9?

if this is true, the answer would be 3 as the units digit.

RULE: The order of operation for exponents: x^y^z=x^(y^z) and not (x^y)^z. The rule is to work from the top down.

3^3^3=3^(3^3)=3^27

Cycle 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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Re: last digit of a power [#permalink]

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02 Jan 2010, 22:18
thanks bubuel. For reminding me to read from the bottom up on these forums when there's a debate!

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Re: last digit of a power [#permalink]

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31 Jan 2010, 10:51
elmagnifico wrote:
What is the last digit $$3^{3^3}$$ ?

* 1
* 3
* 6
* 7
* 9

Using the cyclisity methodology... :
3^(3^3) = 3^27

Cyclisity of 3 = 4....

27 mod 4 = 3....

Therefore last digit would be 3^3 = 7......

Ans is D
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Re: last digit of a power   [#permalink] 31 Jan 2010, 10:51

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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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