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Re: last digit of a power [#permalink]
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18 Aug 2009, 12:29
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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
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Re: last digit of a power [#permalink]
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28 Feb 2010, 13:39
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arturocb86 wrote: amitdgr wrote: Lets take another example Find the last digit of 122^94 A. 2 B. 4 C. 6 D. 8 E. 9 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... Hope this helps and clears it for u! More details refer this link of GMAT Math Book  mathnumbertheory88376.html?viewpost=666609#p666609. Check the Last digit section for explanation!
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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
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 {3971}{3971}...) > 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: m2573474.html
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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
please give detailed explanation. (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
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 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 MSExcel 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 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?
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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/Exponentiationaccording 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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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
So the answer is 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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