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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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13 Dec 2015, 07:26
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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 Here, 18^22 can be written as 18^(3k+1),k being muliple of 7 Hence (3k+1)^10 div by 7 will yield rem 1 in any case so,18^1/7 Rem=4
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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15 Feb 2016, 01:47
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. 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. (18^22)^10 = (3*3*2)^22^10 = ((71)* 3 )^22^10 = [(71)^22 * 3^22]^10 (71)^10 divided by 7 will have Remainder = 1 > ANow for 3^22 =9^11= (7+2)^11 , the whole term will be divisible by 7 except 2^11 2^11= 2*2* 8^3 2*2* (7+1)^3 or, Remainder =4 > BFrom statement A and B , 18^22^10 = (Remainder 1*Remainder4)^10 4^10 = 4 * 64^3 = 4* (63+1)^3 4* remainder 1 Answer D I need someone to validate this approach or did I just go absolutely berserk
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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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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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16 Feb 2016, 23:44
VeritasPrepKarishma wrote: 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. Thanks! I have only started using the concept after reading your post.And now solve all questions using it. Very useful.
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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24 Sep 2016, 06:44
Just thought I'd share a solution I didn't see here yet:
18^22^10 is the same as (14+4)^22^10, so we can focus on the remainder 4^22^10 when divided by 7.
4^n provides remainders of 4, 2 and 1 when divided by 7 for values of n = 3k+1, 3k+2, and 3k+3 respectively, so we need to find the remainder of 22^10 when divided by 3.
22^10 is the same as (21+1)^10, so we can focus on the remainder of 1^10 when divided by 3, which is 1.
Thus, remainder of 4^22^10 is the same as the remainder of 4^(3k+1) which is 4.
Therefore our solution is D!



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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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04 Oct 2016, 12:25
Financier wrote: What is the remainder when (18^22)^10 is divided by 7 ?
А. 1 B. 2 C. 3 D. 4 E. 5 =we can express the given expression by ((14+4)^22)^10 and when 14 is divided by 7 remainder is zero so expression reduces to =(4^22)^10= (16^11)^10=((14+2)^11)^10 again after dividing with 7 it reduces to (2^11)^10 =2048^10= (2044+4)^10, when 2048 is divided by 7 remainder is 4, so remaining expression is 4^10. Now when 4 and its exponents are divided by 7 following pattern it shows. Power 1,4,7,10  remainder 4 Power 2,5,8  remainder 2 Power 3,6,9  remainder 1 hence answer is 4.



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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28 Nov 2016, 09:39
Financier wrote: What is the remainder when (18^22)^10 is divided by 7 ?
А. 1 B. 2 C. 3 D. 4 E. 5 \(\frac{(18^{22})^{10}}{7}\) \(18 = 4 (mod 7)\) \(\frac{(4^{22})^{10}}{7} = \frac{4^{220}}{7} = \frac{2^{440}}{7}\) \(2^3 = 1 (mod 7)\) \(\frac{2^{438}*2^2}{7} = \frac{(2^3)^{146}*2^2}{7} = \frac{1*4}{7}\) 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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28 Nov 2016, 20:42
Financier wrote: What is the remainder when (18^22)^10 is divided by 7 ?
А. 1 B. 2 C. 3 D. 4 E. 5 Quickly solve this question by using modular arithmetic \(\begin{split} 18 &\equiv 4 &\pmod{7} \\ 18^{22} &\equiv 4^{22} = 2^{44} &\pmod{7}\\ (18^{22})^{10} &\equiv (2^{44})^{10} =2^{440} &\pmod{7}\\ \end{split} \) we have \(\begin{split} 2^3=8 &\equiv 1 &\pmod{7} \\ (2^3)^{146} &\equiv 1 &\pmod{7} \\ 2^{438} &\equiv 1 &\pmod{7} \\ 2^{438} \times 2^2 &\equiv 1 \times 2^2 &\pmod{7} \\ 2^{440} &\equiv 4 &\pmod{7} \\ \implies (18^{22})^{10} &\equiv 4 &\pmod{7} \end{split} \) The answer is D
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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02 Feb 2018, 14:24
Hi, At this point 2^440/7. 2 has a cycle of 2486 and 440 is divisible by 4. thus the units digit = 6 6/7 = remainder = 6.. What am I missing here? Please explain 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.



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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03 Feb 2018, 05:16
cuhmoon wrote: Hi, At this point 2^440/7. 2 has a cycle of 2486 and 440 is divisible by 4. thus the units digit = 6 6/7 = remainder = 6.. What am I missing here? Please explain 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. You are confusing the concept of cyclicity (units digit) with the concept of remainders. When you divide any number which ends in 6 by 7, you do not get a remainder of 6. This works only when you divide a number that ends in 6 by 10. In that case the remainder will be 6 only. e.g. 16 / 7 gives remainder 2 Cyclicity and unit's digit can help you solve remainders questions only in very specific cases. Those are discussed here: https://www.veritasprep.com/blog/2015/1 ... questions/https://www.veritasprep.com/blog/2015/1 ... nspart2/
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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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03 Feb 2018, 08:20
Financier wrote: What is the remainder when (18^22)^10 is divided by 7 ?
А. 1 B. 2 C. 3 D. 4 E. 5 The equation can be written as \((18^2)^{110}... (18^2)\) will give us a remainder of 2... So, we have now.. \((2^3)^{36} * 4\) divided by 7... Remainder = 4
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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03 Feb 2018, 11:04
Great artciles and very clear explanations! Thanks a lot! VeritasPrepKarishma wrote: cuhmoon wrote: Hi, At this point 2^440/7. 2 has a cycle of 2486 and 440 is divisible by 4. thus the units digit = 6 6/7 = remainder = 6.. What am I missing here? Please explain You are confusing the concept of cyclicity (units digit) with the concept of remainders. When you divide any number which ends in 6 by 7, you do not get a remainder of 6. This works only when you divide a number that ends in 6 by 10. In that case the remainder will be 6 only. e.g. 16 / 7 gives remainder 2 Cyclicity and unit's digit can help you solve remainders questions only in very specific cases. Those are discussed here: https://www.veritasprep.com/blog/2015/1 ... questions/https://www.veritasprep.com/blog/2015/1 ... nspart2/



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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18 Mar 2018, 06:27
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Jansaida, I think this is the explanation you were discussing on chat. Also the question needs to have the right brackets... (18^22)^10 is very different from 18^22^10... I spent quite some time trying to solve the later. Best, Gladi



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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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11 Apr 2018, 05:29
Responding to a pm: Quote: Can you please let me know how the following would be wrong for this problem ?
We have, 18^220=3^220 * 6^220.
Now, 6^220 = (71)^220  when divided by 7 this yields a remainder of 1. Am I correct ?
Yes, correct. Quote: If so, then 18^220 = 3^220 * (71)^220  therefore, now we ONLY need to check what is the remainder when 3^220 is divided by 7 and per the rule Cyclicity for 3, we can deduce that remainder will be 1 in this case also.
So, FINAL ans : remainder is 1 when (18^22)^10 is divided by 7.
Where am I getting it wrong ? Can you please help ?
Not correct. Unit's digit cyclicity is applicable only in case the divisor is 2 or 5 or 10 (or a multiple of 10). Check this post for more: https://www.veritasprep.com/blog/2015/1 ... questions/You need to handle it using binomial. \(3^{220} = 3 * 3^{219} = 3 * 27^{73} = 3 * (28  1)^{73}\) Remainder from \((28  1)^{73}\) will be 1 so overall remainder will be 3. Since divisor is 7, that is the same as remainder of 4. If you are not sure about negative remainders, check: https://www.veritasprep.com/blog/2014/0 ... thegmat/
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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15 May 2018, 11:44
VeritasPrepKarishma wrote: Responding to a pm: Quote: Can you please let me know how the following would be wrong for this problem ?
We have, 18^220=3^220 * 6^220.
Now, 6^220 = (71)^220  when divided by 7 this yields a remainder of 1. Am I correct ?
Yes, correct. Quote: If so, then 18^220 = 3^220 * (71)^220  therefore, now we ONLY need to check what is the remainder when 3^220 is divided by 7 and per the rule Cyclicity for 3, we can deduce that remainder will be 1 in this case also.
So, FINAL ans : remainder is 1 when (18^22)^10 is divided by 7.
Where am I getting it wrong ? Can you please help ?
Not correct. Unit's digit cyclicity is applicable only in case the divisor is 2 or 5 or 10 (or a multiple of 10). Check this post for more: https://www.veritasprep.com/blog/2015/1 ... questions/You need to handle it using binomial. \(3^{220} = 3 * 3^{219} = 3 * 27^{73} = 3 * (28  1)^{73}\) Remainder from \((28  1)^{73}\) will be 1 so overall remainder will be 3. Since divisor is 7, that is the same as remainder of 4. If you are not sure about negative remainders, check: https://www.veritasprep.com/blog/2014/0 ... thegmat/Hi, Can you tell me why the approach works with 18^220 = (14+4)^220, but it doesn't work when I try to solve it as 18^220=(213)^220?, in first case the remainder = 4 in second it = 1. I assume that (3)^220 gives 1 and not 3. And how in general to pick numbers to avoid any mistake? Thanks.
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2018, 05:35
Hero8888 wrote: VeritasPrepKarishma wrote: Responding to a pm: Quote: Can you please let me know how the following would be wrong for this problem ?
We have, 18^220=3^220 * 6^220.
Now, 6^220 = (71)^220  when divided by 7 this yields a remainder of 1. Am I correct ?
Yes, correct. Quote: If so, then 18^220 = 3^220 * (71)^220  therefore, now we ONLY need to check what is the remainder when 3^220 is divided by 7 and per the rule Cyclicity for 3, we can deduce that remainder will be 1 in this case also.
So, FINAL ans : remainder is 1 when (18^22)^10 is divided by 7.
Where am I getting it wrong ? Can you please help ?
Not correct. Unit's digit cyclicity is applicable only in case the divisor is 2 or 5 or 10 (or a multiple of 10). Check this post for more: https://www.veritasprep.com/blog/2015/1 ... questions/You need to handle it using binomial. \(3^{220} = 3 * 3^{219} = 3 * 27^{73} = 3 * (28  1)^{73}\) Remainder from \((28  1)^{73}\) will be 1 so overall remainder will be 3. Since divisor is 7, that is the same as remainder of 4. If you are not sure about negative remainders, check: https://www.veritasprep.com/blog/2014/0 ... thegmat/Hi, Can you tell me why the approach works with 18^220 = (14+4)^220, but it doesn't work when I try to solve it as 18^220=(213)^220?, in first case the remainder = 4 in second it = 1. I assume that (3)^220 gives 1 and not 3. And how in general to pick numbers to avoid any mistake? Thanks. No, the answer would be the same \((21  3)^{220}\) The last term will be \((3)^{220}\) which is same as \(3^{220}\). \(3* 3^{3*73} = 3 * (27)^{73} = 3 * (28  1)^{73}\) The last term now will be \(3*(1)^{73}\) which is 3 (a negative remainder) So the actual remainder will be 7  3 = 4
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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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17 May 2018, 10:34
VeritasPrepKarishma wrote: No, the answer would be the same
\((21  3)^{220}\)
The last term will be \((3)^{220}\) which is same as \(3^{220}\).
\(3* 3^{3*73} = 3 * (27)^{73} = 3 * (28  1)^{73}\)
The last term now will be \(3*(1)^{73}\) which is 3 (a negative remainder)
So the actual remainder will be 7  3 = 4
Thank you! I have tried just to find the last digit of \(( 3)^{220}\), since 3 is already less than 7. I thought it's a last stop. Is the last stop the format (x+/ 1)/x or we can stop with any number that is less than devisor? Or the power should be odd? I'm confused. Thank you in advance.
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Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
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19 May 2018, 23:31
Hero8888 wrote: VeritasPrepKarishma wrote: No, the answer would be the same
\((21  3)^{220}\)
The last term will be \((3)^{220}\) which is same as \(3^{220}\).
\(3* 3^{3*73} = 3 * (27)^{73} = 3 * (28  1)^{73}\)
The last term now will be \(3*(1)^{73}\) which is 3 (a negative remainder)
So the actual remainder will be 7  3 = 4
Thank you! I have tried just to find the last digit of \(( 3)^{220}\), since 3 is already less than 7. I thought it's a last stop. Is the last stop the format (x+/ 1)/x or we can stop with any number that is less than devisor? Or the power should be odd? I'm confused. Thank you in advance. Yes, but 3^220 is much much larger than 7. What you need is to bring it to the form (7a + 1)^n or (7a  1)^n. The reason for this is explained here: https://www.veritasprep.com/blog/2011/0 ... ekinyou/
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