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Re: last digit of a power [#permalink]
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28 Feb 2010, 09:35
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



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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: last digit of a power [#permalink]
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02 Mar 2010, 10:13
amitdgr wrote: 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. Good points to remember.



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Re: last digit of a power [#permalink]
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05 May 2010, 16:05
please explain me what the term "cyclicity" means. I am not quite following the method used to find the unit digit.
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Re: last digit of a power [#permalink]
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06 May 2010, 08:16
valencia wrote: please explain me what the term "cyclicity" means. I am not quite following the method used to find the unit digit. Cyclicity in the context we are discussing means the frequency at which the last digit repeats itself for exponents. ie 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 2^6 = 64 2^7 = 128 2^8 = 256 2^ 9 = 512..... so on If you notice 2^1,2^5 and 2^ 9 have the same last digit(units digit) which is 2 so the last digit repeats itself after every 4 powers so cyclicity of 2 is 4. Another example 9^1 = 9 9^2 = 81 9^3 = 729 9^4 = 6561 9^5 = 59049 9^6 = 531441 In this case 9^1,9^3 and 9^5 have the same last digit so cyclicity of 9 is 2. Hope this clears it. Now to solve a question all you need to do is divide the power of the exponent by the cyclicty of the base and find the last digit. ie 9^898767 Now the power divided by 2 will give the remainder as 1 so last digit will be same as 9^1 = 9
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Re: last digit of a power [#permalink]
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29 Jun 2010, 04:49
Thanks, I understand what the term "cyclicity" means now. but can some one explain its application with GMAT question ?



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Re: last digit of a power [#permalink]
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02 Sep 2010, 21:29
i'm pretty sure this topic has been beaten to death, but the GMAT Club Math book talks about this under: Number Theory



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Re: last digit of a power [#permalink]
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18 Oct 2010, 12:13
Bunuel wrote: 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. Bunuel.. Can you pls tell me how much power does 2 carry in below mentioned expression and how.. ? 32^32^32 == 2^x ===> what is the value of X and how ? thanks in advance
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Re: last digit of a power [#permalink]
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19 Oct 2010, 15:37
hirendhanak wrote: Bunuel wrote: 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. Bunuel.. Can you pls tell me how much power does 2 carry in below mentioned expression and how.. ? 32^32^32 == 2^x ===> what is the value of X and how ? thanks in advance 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:\(2^x=32^{{32}^{32}}=32^{{(2^5)}^{32}}=32^{2^{160}}={(2^5)}^{2^{160}}=2^{(5*2^{160})}\) > \(x=5*2^{160}\). Hope it's clear.
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Re: last digit of a power [#permalink]
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12 Nov 2010, 21:10
chayanika 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. Wow !! Awesome I tried out this thing with a few numbers and matched the results with my scientific calculator. This method gives perfect answers. You deserve at least a dozen KUDOS for typing out all this patiently and sharing this knowledge with all of us. +1 from me. Guys pour in Kudos for this Chayanika 7 * zillion kudos from me. now wats the unit digit for this is something u will have to figure out , but the method is awesome



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Re: last digit of a power [#permalink]
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22 Nov 2010, 00:50
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.



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Re: last digit of a power [#permalink]
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22 Nov 2010, 02:45
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 {3971}{3971}...) > the units digit of \(3^{27}\) is the same as for 3^3 (27=4*6+3) > 7. Answer: D (7).
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Re: last digit of a power [#permalink]
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24 Nov 2010, 11:49
I absolutely agree with arturocb86. The rule number 3 seems to have something weird. Just take a very simple example: 4^10 has the last digit of 6 because power 10 is an even number. However, if we transfer 4^10 to 2^20 and use the formula 1, we'll have the last digit of 1 instead of 6 because power 20:4= 5 without remainder, so 2^0 = 1 < the last digit. If fact, 2^20 = 1,048,576 has 6 as the last digit. Anybody please help me out.



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Re: last digit of a power [#permalink]
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24 Nov 2010, 16:49
hugonhoc wrote: I absolutely agree with arturocb86. The rule number 3 seems to have something weird. Just take a very simple example: 4^10 has the last digit of 6 because power 10 is an even number. However, if we transfer 4^10 to 2^20 and use the formula 1, we'll have the last digit of 1 instead of 6 because power 20:4= 5 without remainder, so 2^0 = 1 < the last digit. If fact, 2^20 = 1,048,576 has 6 as the last digit. Anybody please help me out. I don't think he said to use 0 as the exponent when the remainder is 0. When you have no remainder you would use 4 as the exponent. so 2 ^ 20 would have the same last digit as 2 ^4 which is 6. In fact, 1 is never a unit digit of 2 to any power because 2 to any power will have unit digit of 2, 4, 8, or 6.



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Re: last digit of a power [#permalink]
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07 Feb 2011, 13:02
this table may be helpful for all.. 1 1 2 4 3 4 4 2 5 1 6 1 7 4 8 4 9 2 10 1
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Re: last digit of a power [#permalink]
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07 Feb 2011, 13:55
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) Thanks for this post
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Re: last digit of a power [#permalink]
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22 Apr 2011, 16:36
3^27 has a last digit of 7.
Answer is D.



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Re: last digit of a power [#permalink]
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20 May 2011, 12:56
bipolarbear wrote: Is 3^3^3 definitely taken as 3^(3^3)?
Is reading it as (3^3)^3 incorrect? yes you have to consider it top down.



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Re: last digit of a power [#permalink]
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20 May 2011, 21:40
3^4 last digit = 1. thus (3^3)^3 = 3^27 27/4 = gives remainder = 3. hence 3^24 * 3^3 = 1* 7 (last digits products) 7
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Re: last digit of a power [#permalink]
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29 Aug 2011, 07:37
3^3^3 which can also be written as 27^3 ie 27*27*27 to obtain the last digit of the expression we just cube the last digit ie 7*7*7 which is 49*7 but we just need the last digit so it will be 9*7= 63 so the last digit is 3 the answer is B




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