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Re: last digit of a power [#permalink]
11 Aug 2009, 03:10

elmagnifico wrote:

What is the last digit \(3^{3^3}\) ?

* 1 * 3 * 6 * 7 * 9

please give detailed explanation.

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

Re: last digit of a power [#permalink]
18 Aug 2009, 08:38

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. _________________

Re: last digit of a power [#permalink]
18 Aug 2009, 10:05

in google, 3^3^3=3^(3^3)=3^27 in excel, 3^3^3=(3^3)^3=27^3, and also in google spreadsheet, 3^3^3=(3^3)^3=27^3;

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? _________________

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. _________________

Re: last digit of a power [#permalink]
18 Aug 2009, 11:29

1

This post received KUDOS

flyingbunny wrote:

in google, 3^3^3=3^(3^3)=3^27 in excel, 3^3^3=(3^3)^3=27^3, and also in google spreadsheet, 3^3^3=(3^3)^3=27^3;

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 _________________

Re: last digit of a power [#permalink]
26 Aug 2009, 02:22

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!!!! _________________

Re: last digit of a power [#permalink]
26 Aug 2009, 12:43

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. _________________

Re: last digit of a power [#permalink]
26 Aug 2009, 17:44

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 _________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

Re: last digit of a power [#permalink]
28 Aug 2009, 04:22

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 _________________

Bhushan S. If you like my post....Consider it for Kudos

Re: last digit of a power [#permalink]
07 Oct 2009, 09:35

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. _________________

Re: last digit of a power [#permalink]
07 Oct 2009, 10:06

Expert's post

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. _________________

Re: last digit of a power [#permalink]
31 Jan 2010, 09:51

elmagnifico wrote:

What is the last digit \(3^{3^3}\) ?

* 1 * 3 * 6 * 7 * 9

please give detailed explanation.

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 _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

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gmatclubot

Re: last digit of a power
[#permalink]
31 Jan 2010, 09:51