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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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24 Aug 2010, 02:35
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What is the remainder when (18^22)^10 is divided by 7 ? А. 1 B. 2 C. 3 D. 4 E. 5
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Last edited by Financier on 15 Sep 2010, 06:57, edited 1 time in total.



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Re: Remainder [#permalink]
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Financier wrote: What is the remainder when (18^22)^10 is divided by 7 ?
А 1 B 2 C 3 D 4 E 5 I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way: \((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7. \(4^{220}=2^{440}\). 2^1 divided by 7 yields remainder of 2; 2^2 divided by 7 yields remainder of 4; 2^3 divided by 7 yields remainder of 1; 2^4 divided by 7 yields remainder of 2; 2^5 divided by 7 yields remainder of 4; 2^6 divided by 7 yields remainder of 1; ... So the remainder repeats the pattern of 3: 241. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146* 3+ 2). \(2^2\) divided by 7 yields remainder of 4. Answer: D.
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Re: Remainder [#permalink]
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03 Feb 2014, 22:51
idinuv wrote: VeritasPrepKarishma wrote: cumulonimbus wrote: Can you point out the mistake here: R(2^440)/7:
2^440 = 2*2^339 = 2*8^113=2*(7+1)^113=2*(7*I+1^113), here I is an integer.
R(2*(7*I+1^113))/7 = R(2*7*I+2)/7 = R(2/7) = 2 There's your mistake. If you take a 2 separately, you are left with 439 2s and not 339 2s. \(2^{440} = 4*2^{438} = 4*8^{146} = 4*(7 + 1)^{146}\) When you divide it by 7, remainder is 4. How can you say that remainder is 4? Using Binomial, we can say that when we divide \((7 + 1)^{146}\) by 7, the remainder will be 1. If this is not clear, check: http://www.veritasprep.com/blog/2011/05 ... ekinyou/When we expand \((7 + 1)^{146}\), we get lots of terms such that \(7^{146}\) is the first term and 1 is the last term. If you multiply \((7 + 1)^{146}\) by 4, you get the same terms except each is multiplied by 4 so the first term is \(4*7^{146}\) and the last one is 4. Every term will still have a 7 in it except the last term. Since the last term is 4, the remainder will be 4.
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Re: Remainder [#permalink]
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09 Jun 2013, 23:24
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cumulonimbus wrote: Can you point out the mistake here: R(2^440)/7:
2^440 = 2*2^339 = 2*8^113=2*(7+1)^113=2*(7*I+1^113), here I is an integer.
R(2*(7*I+1^113))/7 = R(2*7*I+2)/7 = R(2/7) = 2 There's your mistake. If you take a 2 separately, you are left with 439 2s and not 339 2s. \(2^{440} = 4*2^{438} = 4*8^{146} = 4*(7 + 1)^{146}\) When you divide it by 7, remainder is 4.
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Re: Remainder [#permalink]
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15 Oct 2010, 09:17
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my ans is D.. 18^220 /7 =2^440 /7 = (2^3)^146 * 4 /7 = 4 rem.



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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10 Jun 2013, 04:50
Financier wrote: What is the remainder when (18^22)^10 is divided by 7 ?
А. 1 B. 2 C. 3 D. 4 E. 5 ^ 1.(18)^220 = (((18) ^4)^5)^11 2. When 18 is divided by 7 the remainder is 4 3. Now 4^4 is 256. when divided by 7, the remainder is 4 4. Since the remainder is again 4, compute 4^5 = 1024. when divided by 7, the remainder is 2 5. Since 2 is the remainder, now we compute 2^11 = 2048. When divided by 7, the remainder is 4. This is the answer. The answer is choice D.
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2014, 08:23
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Bunuel wrote: gaurav1418z wrote: Bunuel
(18^22)^10 = 18^220
I remember you quoted that for (xyz)^n, if we are asked to find the remainder, we can find remainder for z^n
so i found remainder for 8^220, and got answer as 1
Where am i going wrong? Where did I write that? I think that you mean the following part from Number Theory booksaying that the last digit of \((xyz)^n\) is the same as that of \(z^n\). But the last digit of a number does not determine its remainder upon dividing by 7. For example, 8 divided by 7 gives the remainder of 1 while 18 divided by 7 gives the remainder of 4. Another approach to solve such kind of questions is using cyclicity. Lemme try and explain this. We have (18^22)^10 = 18^220 Now we try to see what are the remainders for various powers of 18 when it is divided by 7 18^1 leaves 4 as remainder 18^2 leaves 2 as remainder 18^3 leaves 1 as remainder 18^4 leaves 4 as remainder Hence, we see after 3 set of powers, the remainder starts repeating. Now, this means if the power is multiple of 3, the remainder is 1, if the power leaves remainder 1 when divided by 3, the actual answer will be 4, and so forth. We see, when 220 is divided by 3, we get 1 as remainder, showcasing 18^1 case, hence the final remainder is 4. This is because 18^220 = 18^219 x 18 = (18^3)^73 x 18 Thus, we need to find the remainder when 18 is divided by 4 as 18^219 gives 1 as remainder.



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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16 Feb 2016, 22:58
KarishmaParmar wrote: Just remember that all terms in the expression (a+b)^n are divisible by a except for the last term i.e b^n. My solution uses just this much. I need someone to validate this approach or did I just go absolutely berserk Yes, that's your binomial theorem concept applied to remainders. In case you are interested in checking out the details of this approach, look at this post: http://www.veritasprep.com/blog/2011/05 ... ekinyou/
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Re: Remainder [#permalink]
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24 Aug 2010, 08:22
Bunuel what is wrong with the approach below 18^220 = 2^220 3^440 this product will have last digit 6... Can't we use cyclisity or some other approach to solve? On the face of it it looks like a GMAT type problem... Posted from my mobile device
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Re: Remainder [#permalink]
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17 Oct 2010, 19:32
I had done the way Bunnel did but used powers of 4. Got D.
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Re: Remainder [#permalink]
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02 Jun 2013, 03:10
sudhanshushankerjha wrote: my ans is D.. 18^220 /7 =2^440 /7 = (2^3)^146 * 4 /7 = 4 rem. Hi Bunnel, Is the above solution correct? I didnt get it.



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Re: Remainder [#permalink]
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02 Jun 2013, 03:40
cumulonimbus wrote: sudhanshushankerjha wrote: my ans is D.. 18^220 /7 =2^440 /7 = (2^3)^146 * 4 /7 = 4 rem. Hi Bunnel, Is the above solution correct? I didnt get it. 18^220 =2^220 (3^2)^220 ={(2^3)^73 * 2^1} {(3^3)^146 * 3^2} =2*9 =18 which leaves a remainder of 4 when divided by 7



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Re: Remainder [#permalink]
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Bunuel wrote: Financier wrote: What is the remainder when (18^22)^10 is divided by 7 ?
А 1 B 2 C 3 D 4 E 5 I think this question is beyond the GMAT scope. It can be solved with Fermat's little theorem, which is not tested on GMAT. Or another way: \((18^{22})^{10}=18^{220}=(14+4)^{220}\) now if we expand this all terms but the last one will have 14 as multiple and thus will be divisible by 7. The last term will be \(4^{220}\). So we should find the remainder when \(4^{220}\) is divided by 7. \(4^{220}=2^{440}\). 2^1 divided by 7 yields remainder of 2; 2^2 divided by 7 yields remainder of 4; 2^3 divided by 7 yields remainder of 1; 2^4 divided by 7 yields remainder of 2; 2^5 divided by 7 yields remainder of 4; 2^6 divided by 7 yields remainder of 1; ... So the remainder repeats the pattern of 3: 241. So the remainder of \(2^{440}\) divided by 7 would be the same as \(2^2\) divided by 7 (440=146* 3+ 2). \(2^2\) divided by 7 yields remainder of 4. Answer: D. Hi Karishma/Bunnel, Can you point out the mistake here: R(2^440)/7: 2^440 = 2*2^339 = 2*8^113=2*(7+1)^113=2*(7*I+1^113), here I is an integer. R(2*(7*I+1^113))/7 = R(2*7*I+2)/7 = R(2/7) = 2



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Re: Remainder [#permalink]
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03 Feb 2014, 22:40
VeritasPrepKarishma wrote: cumulonimbus wrote: Can you point out the mistake here: R(2^440)/7:
2^440 = 2*2^339 = 2*8^113=2*(7+1)^113=2*(7*I+1^113), here I is an integer.
R(2*(7*I+1^113))/7 = R(2*7*I+2)/7 = R(2/7) = 2 There's your mistake. If you take a 2 separately, you are left with 439 2s and not 339 2s. \(2^{440} = 4*2^{438} = 4*8^{146} = 4*(7 + 1)^{146}\) When you divide it by 7, remainder is 4. How can you say that remainder is 4?



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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22 Apr 2014, 19:07
(18^{22})^{10}=18^{220}=(7+11)^{220}. If we expand this equation all terms will be divisible by 7 except the last one.
The last one will be 11^{220}. So we should find the remainder when 11^{220} is divided by 7.
11^{220}=?
11^1 divided by 7 yields remainder of 4; 11^2 divided by 7 yields remainder of 2; 11^3 divided by 7 yields remainder of 1; 11^4 divided by 7 yields remainder of 4; Now we have a pattern 4,2,1,4,2,1
Conclusion: the remainder repeats the pattern of 3: 421. So the remainder of 11^{220} divided by 7 would be the same as 11*1 (that is because 220 is 73*3+1) Answer: D.



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2014, 06:57
Bunuel
(18^22)^10 = 18^220
I remember you quoted that for (xyz)^n, if we are asked to find the remainder, we can find remainder for z^n
so i found remainder for 8^220, and got answer as 1
Where am i going wrong?



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2014, 07:58



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2014, 10:27
You are quite right there Bunuel, my bad, apologies. I am clear now, thanks I will ask another doubt i asked on a different thread( when5125isdividedby13theremainderobtainedis13022020.html) , to make it convenient for you to reply How did you get 12 as remainder when 1 is divided by 13, i didnt quite understand how did you substitute quotient and the remainder in the formula below 1 = 13(1) + 12



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2014, 23:52
gaurav1418z wrote: You are quite right there Bunuel, my bad, apologies. I am clear now, thanks I will ask another doubt i asked on a different thread( http://gmatclub.com/forum/when5125is ... 2020.html) , to make it convenient for you to reply How did you get 12 as remainder when 1 is divided by 13, i didnt quite understand how did you substitute quotient and the remainder in the formula below 1 = 13(1) + 12 Hi there, Remember that remainder can never be negative so when you get remainder negative as in above case then you add divisor to it So we have remainder 1+13 =12 Posted from my mobile device
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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18 May 2014, 02:40
Thanks Wounded Tiger, can i take this as a ground rule?




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