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What is the remainder when (18^22)^10 is divided by 7 ?

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What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post 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
[Reveal] Spoiler: OA

Last edited by Financier on 15 Sep 2010, 06:57, edited 1 time in total.
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Re: Remainder [#permalink] New post 24 Aug 2010, 07:31
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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: 2-4-1. 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] New post 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...

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Re: Remainder [#permalink] New post 15 Oct 2010, 09:17
my ans is D..
18^220 /7 =2^440 /7 = (2^3)^146 * 4 /7 = 4 rem.
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Re: Remainder [#permalink] New post 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] New post 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] New post 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] New post 08 Jun 2013, 10:23
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: 2-4-1. 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] New post 09 Jun 2013, 23:24
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] New post 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] 10 Jun 2013, 04:50
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