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What is the remainder when 2^83 is divided by 9?
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05 Sep 2018, 06:10
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What is the remainder when 2^83 is divided by 9? (A) 4 (B) 0 (C) 4 (D) 5 (E) 8
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Re: What is the remainder when 2^83 is divided by 9?
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05 Sep 2018, 06:31
What is the remainder when 2^83 is divided by 9?
When a power of 2 is divided by 9, the remainder start repeating itsel after 2^6:
2^1 => remainder = 2 2^2 => remainder = 4 2^3 => remainder = 8 2^4 => remainder = 7 2^5 => remainder = 5 2^6 => remainder = 1
2^7 => remainder = 2
....
83/6 gives 5 as remainder => 2^83 and 2^5 have the same remainder when divided by 9..
Option D




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What is the remainder when 2^83 is divided by 9?
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05 Sep 2018, 06:28
2^83/9 =(2^2*2^81)/9 = (4*(2^3)^27)/9 = (4*8^27)/9 = (4*(91)^27)/9 Leaving us with a negative remainder of 4, since remainder can't be negative; the remainder is 94=5! For further explanation check out: https://www.veritasprep.com/blog/2011/0 ... ekinyou/Similar question with better explanation: https://gmatclub.com/forum/whatisthe ... 70273.htmlhttps://gmatclub.com/forum/whatisthe ... 07423.html
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Re: What is the remainder when 2^83 is divided by 9?
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05 Sep 2018, 06:38
T1101 wrote: What is the remainder when 2^83 is divided by 9?
(A) 4 (B) 0 (C) 4 (D) 5 (E) 8 \(2^6 = 64\) \(\frac{2^6}{9} = Remainder \ 1\) Now, \(2^{83} = 2^{6*13 + 5}\) \(2^{78}\) will give a remainder of \(1\) when divided by \(9\) and \(2^5 = 32\) when divided by \(9\) will give a remainder of \(5\), Thus Answer must be (D) 5
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What is the remainder when 2^83 is divided by 9?
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05 Sep 2018, 10:18
T1101 wrote: What is the remainder when 2^83 is divided by 9?
(A) 4 (B) 0 (C) 4 (D) 5 (E) 8 OA : D \(2^{83}= 2^2*(2^3)^{27}=2^2*(91)^{27}\) Remainder of \((91)^{27}\), when divided by \(9\) would be \(8 \quad (91)\). So \((91)^{27}\) can be written as \(9m+8\), where \(m\) is a positive integer. \(2^2*(91)^{27}=2^2*(9m+8)=4*9m+32\). \(4*9m+32\) or \(32\) divided by \(9\) will leave \(5\) as remainder.



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Re: What is the remainder when 2^83 is divided by 9?
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15 Sep 2018, 09:03
Hi,
I tried to apply a formula which goes this way: 2*2*2 = 8. Remainder of 8/9 is 1. We can form 27 such groups of (2*2*2) and every group will give us the remainder as 1. After forming 27 groups we are left with 2*2 which when divided by 9 will give the remainder as 4.
I don't understand why this formula doesn't yield the right result. Can anybody help? Many thanks in advance!!



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Re: What is the remainder when 2^83 is divided by 9?
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19 Sep 2018, 00:33
Peddy wrote: Hi,
I tried to apply a formula which goes this way: 2*2*2 = 8. Remainder of 8/9 is 1. We can form 27 such groups of (2*2*2) and every group will give us the remainder as 1. After forming 27 groups we are left with 2*2 which when divided by 9 will give the remainder as 4.
I don't understand why this formula doesn't yield the right result. Can anybody help? Many thanks in advance!! Hi Peddy, hope i can help you! I have done this question with the same approach as described here: https://www.veritasprep.com/blog/2011/0 ... ekinyou/\(\frac{{2}^{83}}{9}\) =\(\frac{(2^2*{2}^{81})}{9}\) =\(\frac{4*({2}^{3})^{27}}{9}\) =\(\frac{4*({8})^{27}}{9}\) =\(\frac{4*({91})^{27}}{9}\) So \(4*9^{27}\) will yield a reminder of 0. And \(4*(1)^{27}\)= 4. But a reminder can not be negative! Therefore 4 is not the reminder... The reminder is arrived by 94=5 Does it help?
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What is the remainder when 2^83 is divided by 9?
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18 Oct 2018, 09:26
Hi ,T1101 Well, it seems strange for me when the remainder is arrived by 94=5 On the basis, I agree with your approach with 2^83 / 9 = [4*(91)^27] / 9 (91)^27 will leave (1)^27 = 1. When 1 / 9, the quotient is 1 and remainder is 8 Now this [4*(91)^27] / 9 = (4*8)/9 = 32 / 9 yield 5 as remainder I think this explanation is more precisely and easier to understand



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Re: What is the remainder when 2^83 is divided by 9?
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25 Jul 2019, 14:40
Can anybody please provide a further stepbystep explanation ? Thanks



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Re: What is the remainder when 2^83 is divided by 9?
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26 Jul 2019, 11:12
T1101 wrote: What is the remainder when 2^83 is divided by 9?
(A) 4 (B) 0 (C) 4 (D) 5 (E) 8 Notice that 2^6 = 64 and 64/9 = 7 R 1. Now, notice that 2^83 = (2^6)^13 x 2^5 Since 2^6 has a remainder 1 when it’s divided by 9, (2^6)^13 will have a remainder of 1^13 = 1 when it’s divided by 9. Because of this, the remainder when 2^83 is divided by 9 will be same as the remainder when 2^5 is divided by 9. Since 2^5 = 32 and 32/9 = 3 R 5, the remainder is 5. Answer: D
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What is the remainder when 2^83 is divided by 9?
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10 Feb 2020, 11:54
Can anybody tell me why cyclity method does not work there as applied in other problems? 2 has four cycles: 2, 4, 8, 6 Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9?? I completely understood the above solution. I just want to satisfy my curiosity. Expert, please shade some light here.



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What is the remainder when 2^83 is divided by 9?
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10 Feb 2020, 12:20
T1101 wrote: What is the remainder when 2^83 is divided by 9?
(A) 4 (B) 0 (C) 4 (D) 5 (E) 8 Method 1We know that, if \(N\) leaves remainder \(r\) when divided by \(D, N^k \)leaves remainder \(r^k\) when divided by \(D\) (provided \(r^k\) is smaller than \(D\)); else you need to divided \(r^k\) again by \(D\) to get the actual remainder We can observe that \(2^6 = 64\) when divided by 9 leaves remainder 1 Thus, we express \(2^{83}\) in terms of \(2^6\): \(2^{83} = (2^6)^{13} * 2^5\) Since \(2^6\) when divided by 9 leaves remainder 1, the remainder when \((2^6)^{13}\) is divided by 9 is \(1^{13} = 1\) Thus, required remainder is simply the one when the leftover \(2^5 = 32\) is divided by 9 => Remainder = 5 Answer DMethod 2Look for patterns in remainders for powers of 2 when divided by 9: Remainder when \(2^1\) is divided by 9 = 2 Remainder when \(2^2\) is divided by 9 = 4 Remainder when \(2^3\) is divided by 9 = 8 Remainder when \(2^4\) is divided by 9 = 7 Remainder when \(2^5\) is divided by 9 = 5 Remainder when \(2^6\) is divided by 9 = 1 Remainder when \(2^7\) is divided by 9 = 2 ... and the cycle repeats Thus, the cycle is for 6 places Thus, we express \(2^{83}\) in terms of \(2^6\): \(2^{83} = (2^6)^{13} * 2^5 = 2^{78} * 2^5\) The 13th time the cycle repeats, the remainder comes to 1. Now, the cycle will repeat but will stop at the 5th position since we have \(2^5\) leftover Thus, the remainder corresponding to the 5th position in the cycle is 5 Answer D
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Re: What is the remainder when 2^83 is divided by 9?
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10 Feb 2020, 12:34
sajon wrote: Can anybody tell me why cyclity method does not work there as applied in other problems? 2 has four cycles: 2, 4, 8, 6 Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9?? I completely understood the above solution. I just want to satisfy my curiosity. Expert, please shade some light here. The cycle of 2 that you are referring to: 2, 4, 8, 6 is basically the cycle for the units digit This is also the cycle for remainder when powers of 2 are DIVIDED BY 10 Check: Unit digit of 2^1 = 2 <=> Remainder when 2^1 is divided by 10 is 2 Unit digit of 2^2 = 4 <=> Remainder when 2^2 is divided by 10 is 4 Unit digit of 2^3 = 8 <=> Remainder when 2^3 is divided by 10 is 8 Unit digit of 2^4 = 6 <=> Remainder when 2^4 is divided by 10 is 6 Hope this is clear now
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Re: What is the remainder when 2^83 is divided by 9?
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10 Feb 2020, 13:28
sajon wrote: Can anybody tell me why cyclity method does not work there as applied in other problems? 2 has four cycles: 2, 4, 8, 6 Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9?? I completely understood the above solution. I just want to satisfy my curiosity. Expert, please shade some light here. Btw I am no expert. The unit digit of the a number has no significance to the remainder (unless the divisor is 10). I will explain it with a few examples . (1) if your number is 8, then 8/9 = remainder 8 (2) if your number is 18, then 18/9 = remainder 0 (3) if your number is 118, then 118/9 = remainder 1 The unit digit is 8, but then the remainders are all different, so extending this logic you can't really say what you are suggesting. I hope this "satisfies your curiosity"!



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Re: What is the remainder when 2^83 is divided by 9?
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10 Feb 2020, 21:40
AnirudhaS wrote: sajon wrote: Can anybody tell me why cyclity method does not work there as applied in other problems? 2 has four cycles: 2, 4, 8, 6 Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9?? I completely understood the above solution. I just want to satisfy my curiosity. Expert, please shade some light here. Btw I am no expert. The unit digit of the a number has no significance to the remainder (unless the divisor is 10). I will explain it with a few examples . (1) if your number is 8, then 8/9 = remainder 8 (2) if your number is 18, then 18/9 = remainder 0 (3) if your number is 118, then 118/9 = remainder 1 The unit digit is 8, but then the remainders are all different, so extending this logic you can't really say what you are suggesting. I hope this "satisfies your curiosity"! Wow, now i got this point. Thanks a lot. Actually i solved few question using 10 and found cyclity method effective for all number. But, it is the only one that works.



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Re: What is the remainder when 2^83 is divided by 9?
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10 Feb 2020, 21:47
sujoykrdatta wrote: sajon wrote: Can anybody tell me why cyclity method does not work there as applied in other problems? 2 has four cycles: 2, 4, 8, 6 Then 2^83 should have 8 as the unit digit. So, why does 8 have not been the answer when divided by 9?? I completely understood the above solution. I just want to satisfy my curiosity. Expert, please shade some light here. The cycle of 2 that you are referring to: 2, 4, 8, 6 is basically the cycle for the units digit This is also the cycle for remainder when powers of 2 are DIVIDED BY 10 Check: Unit digit of 2^1 = 2 <=> Remainder when 2^1 is divided by 10 is 2 Unit digit of 2^2 = 4 <=> Remainder when 2^2 is divided by 10 is 4 Unit digit of 2^3 = 8 <=> Remainder when 2^3 is divided by 10 is 8 Unit digit of 2^4 = 6 <=> Remainder when 2^4 is divided by 10 is 6 Hope this is clear now Thanks for your reply. now i got my point




Re: What is the remainder when 2^83 is divided by 9?
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