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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]

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New post Updated on: 15 Sep 2010, 06:57
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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

Originally posted by Financier on 24 Aug 2010, 02:35.
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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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]

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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]

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New post 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: Remainder  [#permalink]

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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]

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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]

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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]

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New post 08 Jun 2013, 10:23
2
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]

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

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New post 10 Jun 2013, 04:50
1
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: Remainder  [#permalink]

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New post 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: Remainder  [#permalink]

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New post 03 Feb 2014, 22:51
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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 ... ek-in-you/


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

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New post 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: 4-2-1. 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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New post 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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New post 17 May 2014, 07:58
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 book (math-number-theory-88376.html) saying 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.
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Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

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New post 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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New post 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(when-51-25-is-divided-by-13-the-remainder-obtained-is-130220-20.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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New post 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/when-51-25-is ... 20-20.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

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Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

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New post 18 May 2014, 02:40
Thanks Wounded Tiger, can i take this as a ground rule?
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What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

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New post 13 Sep 2015, 20:10
Here in some cases i am taking = instead of congurent

Lets take \((18^22)^10 = x (mod 7)\) = x (mod 7)
here \((18^22)^10 = 18^220\)
now \(18 = 4 (mod 7)\)
we have formula if \(a=b(mod m)\) then \(a^k=b^k (mod m)\)
so \(18^220 = 4^220 (mod 7)\)
\(4^220 = (4^3)^73 . 4\)
and \(4^3 = 1 (mod 7)\)
so \(18^220 = 1.4 (mod 7)\) since we got \(4^3 = 1 (mod 7)\) then \((4^3)^73 = 1^73 = 1\)
so \(18^220 = 4 (mod 7)\)
therefore x = 7
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What is the remainder when (18^22)^10 is divided by 7 ?   [#permalink] 13 Sep 2015, 20:10

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