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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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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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24 Aug 2010, 07:31
6
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

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.

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03 Feb 2014, 22:51
4
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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09 Jun 2013, 23:24
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3
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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15 Oct 2010, 09:17
1
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
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.

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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
1
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
1
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 ... ek-in-you/
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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
1
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,
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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
1
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 = (7-1)^220 -- when divided by 7 this yields a remainder of 1. Am I correct ?

Yes, correct.
Quote:
If so, then 18^220 = 3^220 * (7-1)^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.

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 ... -the-gmat/
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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
1
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 ... ek-in-you/
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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...

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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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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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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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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.

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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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: 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)
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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
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] 17 May 2014, 07:58

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