GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 18 Jan 2020, 07:42

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 18 Jun 2010
Posts: 238
Schools: Chicago Booth Class of 2013
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

Updated on: 15 Sep 2010, 06:57
4
54
00:00

Difficulty:

75% (hard)

Question Stats:

52% (01:54) correct 48% (02:02) wrong based on 1268 sessions

### HideShow timer Statistics

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.
Math Expert
Joined: 02 Sep 2009
Posts: 60466

### Show Tags

24 Aug 2010, 07:31
6
26
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.

_________________
##### General Discussion
Director
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 574
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas

### Show Tags

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
Intern
Joined: 04 Sep 2010
Posts: 29

### Show Tags

15 Oct 2010, 09:17
1
my ans is D..
18^220 /7 =2^440 /7 = (2^3)^146 * 4 /7 = 4 rem.
Manager
Joined: 25 Jul 2010
Posts: 120
WE 1: 4 years Software Product Development
WE 2: 3 years ERP Consulting

### Show Tags

17 Oct 2010, 19:32
I had done the way Bunnel did but used powers of 4. Got D.
Manager
Joined: 14 Nov 2011
Posts: 114
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)

### Show Tags

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.
Intern
Joined: 21 Apr 2013
Posts: 7

### Show Tags

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
Manager
Joined: 14 Nov 2011
Posts: 114
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)

### Show Tags

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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9983
Location: Pune, India

### Show Tags

09 Jun 2013, 23:24
3
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.
_________________
Karishma
Veritas Prep GMAT Instructor

Director
Joined: 17 Dec 2012
Posts: 622
Location: India
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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.

_________________
Srinivasan Vaidyaraman
Sravna Test Prep
http://www.sravnatestprep.com

Holistic and Systematic Approach
Intern
Joined: 21 Mar 2013
Posts: 38
GMAT Date: 03-20-2014

### Show Tags

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?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9983
Location: Pune, India

### Show Tags

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.
_________________
Karishma
Veritas Prep GMAT Instructor

Intern
Joined: 26 Oct 2013
Posts: 20
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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)
Intern
Joined: 25 Jan 2014
Posts: 44
GMAT 1: 600 Q44 V29
GMAT 2: 710 Q48 V38
GPA: 3.35
WE: Analyst (Computer Software)
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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?
Math Expert
Joined: 02 Sep 2009
Posts: 60466
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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.
_________________
Intern
Joined: 17 May 2014
Posts: 36
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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.
Intern
Joined: 25 Jan 2014
Posts: 44
GMAT 1: 600 Q44 V29
GMAT 2: 710 Q48 V38
GPA: 3.35
WE: Analyst (Computer Software)
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

17 May 2014, 10:27
You are quite right there Bunuel, my bad, apologies. I am clear now, thanks

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
Director
Joined: 25 Apr 2012
Posts: 649
Location: India
GPA: 3.21
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

17 May 2014, 23:52
gaurav1418z wrote:
You are quite right there Bunuel, my bad, apologies. I am clear now, thanks

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
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
Intern
Joined: 25 Jan 2014
Posts: 44
GMAT 1: 600 Q44 V29
GMAT 2: 710 Q48 V38
GPA: 3.35
WE: Analyst (Computer Software)
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

18 May 2014, 02:40
Thanks Wounded Tiger, can i take this as a ground rule?
Intern
Joined: 13 Sep 2015
Posts: 1
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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
What is the remainder when (18^22)^10 is divided by 7 ?   [#permalink] 13 Sep 2015, 20:10

Go to page    1   2    Next  [ 38 posts ]

Display posts from previous: Sort by