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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 _________________

Now the last digit of 122 is 2. We require only this number to determine the last digit of 122 raised to a positive power.

so the problem is essentially reduced to find the last digit of 2^94.

Now we know 2 has a cyclicity of 4. So we divide 94 by 4. The remainder for 94/4 is 2.

so last digit of 2^94 is same as that of 2^2 which is 4.

so last digit of 122^94 is 4

Remember: 1) Numbers 2,3,7 and 8 have a cyclicity of 4 2) Numbers 0,1,5 and 6 have a cyclicity of 1 (ie) all the powers will have the same unit digit. eg. 5^245 will have "5" as unit digit, 5^2000 will aslo have "5" as unit digit. Same holds for 0,1 and 6 3) If 4 is the number in the unit place of the base number then the unit digit will be "4" if the power is odd and it will be "6" if the power is even. eg. 4^123 will have unit digit of 4 since 123 is odd. 4) Similarly, for 9 the unit digit will be "9" for odd powers and "1" for even powers. eg 9^234 has unit digit as "1" since 234 is even.

There is something wrong with rule 3!

Imagine we have 4^124 --> following your rule you would say unit digit = 6.

But, 4^124 =2^(2*124) = 2^248 since 2 has a cyclicity of 4 --> 248/4 yields a remainder = 0. then the units digit is 2^0=1.

Am I correct?? Thank you! It was a great post amitdgr...Kudos!

I always transform a 4 into a 2^2 and use rule 1

Cyclicity of 4 is 2

4^1 = 4 4^2 =..6 4^3 = ..4 4^........

Therefore last digit of 4^124 would be... 124 mod 2 (cyclicity of 4) = 0.... Now if remainder is 0.. then last digit is 4^2(cyclicity) = 6

Your approach .... 4^124 = 2^248

Cyclicity of 2 is 4 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = ..6 2^5 = ..2

Therefore last digit of 2^248 = 248 mod 4(cyclicity of 2) = 0... Now if remainder is 0.. then last digit is 2^4(cyclicity) = 6...

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

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Re: What is the last digit 3^{3^3} ? A. 1 B. 3 C. 6 D. 7 E. 9 [#permalink]

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23 Jul 2013, 13:09

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This post received KUDOS

Expert's post

Maxirosario2012 wrote:

elmagnifico wrote:

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

A. 1 B. 3 C. 6 D. 7 E. 9

please give detailed explanation.

OA is D (7).

There is something I don´t understand with this kind of exercise. When I resolve this in excel or in the calculator, the result is: 7,6256E+12

Which means \(7,6256*10^12 = 7.625.600.000.000\)

Then, the last digit is not 7 but zero. I think there is something wrong in my logic / understanding

How can the last digit of \(3^{3^3}\) be zero? Excel rounds big numbers. Exact result is 7,625,597,484,987.

Complete solutions. What is the last digit of \(3^{3^3}\)? (A) 1 (B) 3 (C) 6 (D) 7 (E) 9

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).

OPNE DISCUSSION OF THIS QUESTION IS HERE: m25-73474.html _________________

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

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. _________________

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. _________________

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!!!! _________________

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. _________________

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

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

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. _________________

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. _________________