NEW HERE? LEARN MORE
closeclose
Last visit was: 24 Apr 2024, 06:15 It is currently 24 Apr 2024, 06:15
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

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

SORT BY:
Date
Kudos
 1   2   
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
Financier
Manager
Manager
Joined: 18 Jun 2010
Posts: 230
Own Kudos [?]: 656 [116]
Given Kudos: 194
Schools:Chicago Booth Class of 2013
Send PM
Reviews Badge
What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post   Updated on: 15 Sep 2010, 06:57
7
Kudos
109
Bookmarks
00:00
A
B
C
D
E

Difficulty:

65% (hard)

Question Stats:

55% (01:55) correct 45% (02:02) wrong based on 1541 sessions
Hide Show timer Statistics
What is the remainder when (18^22)^10 is divided by 7 ?

А. 1
B. 2
C. 3
D. 4
E. 5
ShowHide Answer
Official Answer
D

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.
Most Helpful Reply
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92901
Own Kudos [?]: 618731 [53]
Given Kudos: 81586
Send PM
CAT Tests Forum Quiz Mode
Re: Remainder [#permalink] New post   24 Aug 2010, 07:31
13
Kudos
40
Bookmarks
Expert Reply
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.
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64892 [10]
Given Kudos: 426
Location: Pune, India
Send PM
Re: Remainder [#permalink] New post   09 Jun 2013, 23:24
6
Kudos
4
Bookmarks
Expert Reply
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.
General Discussion
User avatar
cumulonimbus
Manager
Manager
Joined: 14 Nov 2011
Posts: 100
Own Kudos [?]: 56 [3]
Given Kudos: 103
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE:Consulting (Manufacturing)
Send PM
GMAT ToolKit User
Re: Remainder [#permalink] New post   08 Jun 2013, 10:23
1
Kudos
2
Bookmarks
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
User avatar
S
SVaidyaraman
Director
Director
Joined: 17 Dec 2012
Posts: 589
Own Kudos [?]: 1519 [2]
Given Kudos: 20
Location: India
Send PM
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post   10 Jun 2013, 04:50
2
Kudos
Expert Reply
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.
User avatar
idinuv
Intern
Intern
Joined: 21 Mar 2013
Posts: 31
Own Kudos [?]: 550 [0]
Given Kudos: 56
Schools: INSEAD Jan '15 HBS Stanford '17 Haas '16 Kellogg 1YR '15
GMAT Date: 03-20-2014
Send PM
GMAT ToolKit User
Re: Remainder [#permalink] 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?
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64892 [4]
Given Kudos: 426
Location: Pune, India
Send PM
What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post   Updated on: 03 Oct 2023, 09:12
4
Kudos
Expert Reply
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.


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.

Originally posted by KarishmaB on 03 Feb 2014, 22:51.
Last edited by KarishmaB on 03 Oct 2023, 09:12, edited 1 time in total.
User avatar
GDR29
Intern
Intern
Joined: 26 Oct 2013
Posts: 18
Own Kudos [?]: 43 [0]
Given Kudos: 4
Send PM
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] 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.
avatar
gaurav1418z
Intern
Intern
Joined: 25 Jan 2014
Posts: 38
Own Kudos [?]: 20 [0]
Given Kudos: 4
Concentration: Strategy, International Business
GMAT 1: 600 Q44 V29
GMAT 2: 710 Q48 V38
GPA: 3.35
WE:Analyst (Computer Software)
Send PM
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] 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?
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92901
Own Kudos [?]: 618731 [2]
Given Kudos: 81586
Send PM
CAT Tests Forum Quiz Mode
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post   17 May 2014, 07:58
1
Kudos
1
Bookmarks
Expert Reply
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.
avatar
S
Alphatango
Intern
Intern
Joined: 01 Feb 2015
Status:One more try
Posts: 24
Own Kudos [?]: 29 [2]
Given Kudos: 164
Location: India
Concentration: General Management, Economics
Schools: Duke '19 ISB '18 Tepper '19
WE:Corporate Finance (Commercial Banking)
Send PM
GMAT ToolKit User
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post   13 Dec 2015, 07:26
2
Bookmarks
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
User avatar
KarishmaParmar
Manager
Manager
Joined: 15 Feb 2016
Posts: 52
Own Kudos [?]: 35 [0]
Given Kudos: 28
GMAT 1: 710 Q48 V40
Send PM
Reviews Badge
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] 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:
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64892 [1]
Given Kudos: 426
Location: Pune, India
Send PM
What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post   Updated on: 03 Oct 2023, 09:13
1
Kudos
Expert Reply
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.

Originally posted by KarishmaB on 16 Feb 2016, 22:58.
Last edited by KarishmaB on 03 Oct 2023, 09:13, edited 1 time in total.
User avatar
B
vitaliyGMAT
Senior Manager
Senior Manager
Joined: 13 Oct 2016
Posts: 300
Own Kudos [?]: 768 [0]
Given Kudos: 40
GPA: 3.98
Send PM
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] 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 \(4\).
avatar
B
cuhmoon
Intern
Intern
Joined: 30 Jun 2015
Posts: 37
Own Kudos [?]: 49 [0]
Given Kudos: 90
Location: Malaysia
Schools: Babson '20
GPA: 3.6
Send PM
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] 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.
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64892 [0]
Given Kudos: 426
Location: Pune, India
Send PM
What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post   Updated on: 03 Oct 2023, 09:14
Expert Reply
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.

Originally posted by KarishmaB on 03 Feb 2018, 05:16.
Last edited by KarishmaB on 03 Oct 2023, 09:14, edited 1 time in total.
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64892 [1]
Given Kudos: 426
Location: Pune, India
Send PM
What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post   Updated on: 03 Oct 2023, 09:16
1
Kudos
Expert Reply
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).


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://anaprep.com/number-properties-t ... emainders/

Originally posted by KarishmaB on 11 Apr 2018, 05:29.
Last edited by KarishmaB on 03 Oct 2023, 09:16, edited 1 time in total.
User avatar
G
Hero8888
Senior Manager
Senior Manager
Joined: 29 Dec 2017
Posts: 302
Own Kudos [?]: 307 [0]
Given Kudos: 273
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
GMAT 3: 710 Q50 V37
GPA: 3.25
WE:Marketing (Telecommunications)
Send PM
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] 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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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.
User avatar
L
KarishmaB
Tutor
Joined: 16 Oct 2010
Posts: 14817
Own Kudos [?]: 64892 [0]
Given Kudos: 426
Location: Pune, India
Send PM
Re: What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] New post   17 May 2018, 05:35
Expert Reply
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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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
User avatar
G
Hero8888
Senior Manager
Senior Manager
Joined: 29 Dec 2017
Posts: 302
Own Kudos [?]: 307 [0]
Given Kudos: 273
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
GMAT 3: 710 Q50 V37
GPA: 3.25
WE:Marketing (Telecommunications)
Send PM
What is the remainder when (18^22)^10 is divided by 7 ? [#permalink] 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.
GMAT Club Bot
gmatclubot
What is the remainder when (18^22)^10 is divided by 7 ? [#permalink]
17 May 2018, 10:34
 1   2   
Moderators:
Bunuel
Math Expert
92901 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions