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Joined: 23 Apr 2010
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Cyclicity of 2
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20 Oct 2010, 05:45
I would like math gurus to explain how to calculate the cyclicity of 2.
When calculating the cyclicity of a number (1 to 9), I have followed the following steps (say we want to find the last digit of \(3^{35}\)):
1) 3^0 = 1 3^1 = 3 3^2 = 9 3^3 = 27 3^4 = 81
So we can conclude that the cyclicity of 3 is 4 and the last digits are: 1 (0th position), 3 (1st position), 9 (2nd position), 7 (3rd position).
2) 35 mod 4 = 3 => 3rd position
3) The last digit of \(3^{35}\) is therefore 7
Now when I want to find the last digit of \(2^{1024}\) , the method doesn't work:
2^0 = 1 2^1 = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32
1024 mod 4 = 0 but it can't be 1.
So, I thought that in case of multiples 2, 4, 8 we should distinguish between odd and even powers and start with 1.
2^1 = 2 2^3 = 8 2^5 = 32
2^2 = 4 2^4 = 16 2^6 = 64
Cyclicity is 2. 1024 is an even power. 1024 mod 2 = 0 which corresponds to 4.
Are my thoughts correct?
Thank you.



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Re: Cyclicity of 2
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20 Oct 2010, 06:05



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Re: Cyclicity of 2
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20 Oct 2010, 06:18
Bunuel, thank you for your reply. So are you saying that the last digit of \(2^{1024}\) is the same as the last digit of 2^4 which is 6? Am I right? Quote: So for the last digit of 2^1024 equals to the last digit of 2^4, as the cyclicity of the last digit of 2 in power is 4.



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Re: Cyclicity of 2
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20 Oct 2010, 06:43
Cyclicity appears in powers of digits when we start the power from 1. 2, 3, 7 and 8 have a cyclicity of 4 each 2^1 = 2 3^1 = 3 2^2 = 4 3^2 = 9 2^3 = 8 3^3 = _7 2^4 = _6 3^4 = _1 2^5 = _2 3^5 = __3 2^6 = _4 3^6 = __9 and so on... Cyclicity of 3 is 3, 9, 7, 1. Therefore, if I have (1543)^35, the last digit will be the last digit of 3^35. I divide 35 in blocks of 4 i.e. I divide 35 by 4 and I get 3 as remainder. Since I have 3 leftover, I go to 3, 9, 7 and stop there. Last digit of 3^35 is 7. In case it is 3^36, i.e. no remainder, it means I made complete groups of 3,9,7,1 and ended at 1. Hence last digit of 3^36 is 1. Cyclicity of 2 is 2, 4, 8, 6. Similar is the case when I have 2^36. I dont have any remainder when I divide 36 by 4 which implies that while making groups of 2, 4,8, 6, I ended on the 6 and hence the last digit of 2^36 will be 6. Cycility of 4 is 4, 6 and that of 9 is 9, 1 so these cases are extremely easy. If the power is odd, they will end with 4 and 9 respectively and if the power is even, they will end with 6 and 1 respectively.
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Re: Cyclicity of 2
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21 Oct 2010, 01:31
Bunuel and VeritasPrepKarishma, thank you. I got it.



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Re: Cyclicity of 2
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18 Apr 2011, 21:49
i want to know how to find the units digit of the no 73^74.. I have understood the process of how to find the cyclicity but cant understand how to apply it to problems like these.. Please give tin solution stepwise.. Thank you..
Posted from my mobile device



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Re: Cyclicity of 2
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19 Apr 2011, 05:08
s433369 wrote: i want to know how to find the units digit of the no 73^74.. I have understood the process of how to find the cyclicity but cant understand how to apply it to problems like these.. Please give tin solution stepwise.. Thank you..
Posted from my mobile device What will be the unit's digit of 3^4? It will be 1 (3, 9, 7, 1) What will be the unit's digit of 3^6? It will be 9 (3, 9, 7, 1, 3, 9) What will be the unit's digit of 3^8? It will be 1 (3, 9, 7, 1, 3, 9, 7, 1) So power that is a multiple of 4 will give a unit's digit of 1. What will be the unit's digit of 3^17? It will be 3. This is so because 3^16 will have a unit's digit of 1. So the cycle of 4 ends there. Then a new cycle begins and that will begin with 3. What will be the unit's digit of 3^74? The closest multiple of 4 to 74 is 72. So a cycle will end at 72. A new one begins at 73. So 3^74 will have the last digit of 9. Now, what will be the unit's digit of 73^74? It will be the same as the unit's digit of 3^74 because the ten's digit (7) has to contribution is the unit's digit of 73^some power. So just focus on the unit's digit of the number and the power. Of course, cyclicity is different for different numbers 2, 3, 7, 8 have a cyclicity of 4 4, 9 have a cyclicity of 2 5, 6 have a cyclicity of 1 i.e. any power of 5 or 6 ends with a 5 or 6 only. But you can just remember 4 and work from there for every number. Or just take a few seconds to figure out the cyclicity of the concerned number.
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Re: Cyclicity of 2
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23 Sep 2018, 01:48
0 => 0,0,0, so 21 = 1 1 => 1,1,1, so 21 = 1 2 => 2,4,8,6,2,4,8,6, so 51 = 4 3 => 3,9,7,1,3,9,7, so 51 =4 4 => 4,6,4,6,4, so 31 = 2 5 => 5,5,5,5, so 21 = 1 6 => 6,6,6,6, so 21 = 1 7 => 7,9,3,1,7, so 51 = 4 8 => 8, 4, 2, 6, 8, so 51 = 4 9 => 9, 1, 9, 1, 9, so 31 = 2 Based on this, we can conclude that 4 will work for all cases, so you always can replace n^x with n^((x1)%4+1). Or n^(x%4) but remember to use 4 instead of 0 if 4 divide x.



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Cyclicity of 2
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23 Sep 2018, 01:54
CYCLICITY OF UNIT DIGIT It is not necessary to remember cyclicity order of different digits, In fact, all digits follow a common cyclicity order, they repeat itself after 4k+1 power.Steps to solve such questions: 1) divide the power by 4 and find remainder. 2) Now find the unit digit by raising it to exponent of remainder.(if remainder is 0, raise it to exponent 4) this method works for every digit (PS: for some of digits , we have simpler pattern method. 1) 0  always 0 2) 4  odd power = 4, even power = 6 3) 5 always 5 4) 6 always 6 5) 9  odd power 9, even power = 1)
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Re: Cyclicity of 2
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27 Sep 2018, 11:59
s433369 wrote: i want to know how to find the units digit of the no 73^74.. I have understood the process of how to find the cyclicity but cant understand how to apply it to problems like these.. Please give tin solution stepwise.. Thank you..
Posted from my mobile device Ignore everything about the base number except for its units digit. To solve this problem, you only need to find the units digit of 3^74. (Make sure that you don't get mixed up and simplify the exponent instead of the base! That will give you a different answer.) In fact, you can always ignore everything but the units digit(s) of the base number(s) whenever you're doing a 'find the units digit' type problem. Want to find the units digit of 230473204732^4234 + 304451^234? You're really finding the units digit of 2^4234 + 1^234. To finish your problem, find the units digit of 3^74 using the process described earlier in this thread.
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Re: Cyclicity of 2 &nbs
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27 Sep 2018, 11:59






