GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 16 Aug 2018, 06:46

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
B
Status: One more try
Joined: 01 Feb 2015
Posts: 49
Location: India
Concentration: General Management, Economics
WE: Corporate Finance (Commercial Banking)
GMAT ToolKit User
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 13 Dec 2015, 07:26
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


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
_________________

Believe you can and you are halfway there-Theodore Roosevelt

Manager
Manager
User avatar
Joined: 15 Feb 2016
Posts: 62
GMAT 1: 710 Q48 V40
Reviews Badge
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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: 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.


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 = ((7-1)* 3 )^22^10 = [(7-1)^22 * 3^22]^10

(7-1)^10 divided by 7 will have Remainder = 1 ---> A

Now 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 ---->B

From 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 :cry:
_________________

It is not who I am underneath but what I do that defines me.

Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8187
Location: Pune, India
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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 :cry:


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/
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Manager
Manager
User avatar
Joined: 15 Feb 2016
Posts: 62
GMAT 1: 710 Q48 V40
Reviews Badge
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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 :cry:


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. :-D
_________________

It is not who I am underneath but what I do that defines me.

Intern
Intern
avatar
Joined: 11 Aug 2016
Posts: 9
Location: United States (PA)
Concentration: Finance, General Management
GMAT 1: 710 Q48 V40
GPA: 2.42
WE: Other (Mutual Funds and Brokerage)
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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!
Intern
Intern
avatar
B
Joined: 01 Jun 2013
Posts: 10
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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.
Senior Manager
Senior Manager
avatar
B
Joined: 13 Oct 2016
Posts: 367
GPA: 3.98
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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 \(4\).
Senior CR Moderator
User avatar
V
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1394
Location: Viet Nam
GMAT ToolKit User Premium Member
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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
_________________

Actual LSAT CR bank by Broall

How to solve quadratic equations - Factor quadratic equations
Factor table with sign: The useful tool to solve polynomial inequalities
Applying AM-GM inequality into finding extreme/absolute value

New Error Log with Timer

Manager
Manager
avatar
B
Joined: 30 Jun 2015
Posts: 60
Location: Malaysia
Schools: Babson '20
GPA: 3.6
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 02 Feb 2018, 14:24
Hi,

At this point 2^440/7.

2 has a cycle of 2-4-8-6 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.
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8187
Location: Pune, India
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 03 Feb 2018, 05:16
cuhmoon wrote:
Hi,

At this point 2^440/7.

2 has a cycle of 2-4-8-6 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 ... ns-part-2/
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Senior Manager
Senior Manager
avatar
G
Joined: 31 Jul 2017
Posts: 402
Location: Malaysia
WE: Consulting (Energy and Utilities)
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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
_________________

If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!

Manager
Manager
avatar
B
Joined: 30 Jun 2015
Posts: 60
Location: Malaysia
Schools: Babson '20
GPA: 3.6
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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 2-4-8-6 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 ... ns-part-2/
Manager
Manager
User avatar
G
Joined: 16 Sep 2016
Posts: 235
WE: Analyst (Health Care)
GMAT ToolKit User Premium Member CAT Tests
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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,
Gladi
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8187
Location: Pune, India
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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.

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 ... -the-gmat/
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Senior Manager
Senior Manager
User avatar
G
Joined: 29 Dec 2017
Posts: 337
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
GPA: 3.25
WE: Marketing (Telecommunications)
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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 = (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.


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 ... -the-gmat/


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=(21-3)^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.
_________________

I'm looking for a study buddy in NY, who is aiming at 700+. PM me.

Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8187
Location: Pune, India
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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 = (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.


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 ... -the-gmat/


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=(21-3)^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
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Senior Manager
Senior Manager
User avatar
G
Joined: 29 Dec 2017
Posts: 337
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
GPA: 3.25
WE: Marketing (Telecommunications)
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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.
_________________

I'm looking for a study buddy in NY, who is aiming at 700+. PM me.

Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8187
Location: Pune, India
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

Show Tags

New post 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/
_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here >

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Re: What is the remainder when (18^22)^10 is divided by 7 ? &nbs [#permalink] 19 May 2018, 23:31

Go to page   Previous    1   2   [ 38 posts ] 

Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.