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

 It is currently 20 Jan 2020, 14:26

### 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

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
Intern
Status: One more try
Joined: 01 Feb 2015
Posts: 37
Location: India
Concentration: General Management, Economics
WE: Corporate Finance (Commercial Banking)
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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
Joined: 15 Feb 2016
Posts: 58
GMAT 1: 710 Q48 V40
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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.

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

I need someone to validate this approach or did I just go absolutely berserk
_________________
It is not who I am underneath but what I do that defines me.
Manager
Joined: 15 Feb 2016
Posts: 58
GMAT 1: 710 Q48 V40
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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

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

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
Joined: 01 Jun 2013
Posts: 8
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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
Senior Manager
Joined: 13 Oct 2016
Posts: 354
GPA: 3.98
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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 $$4$$.
Retired Moderator
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1313
Location: Viet Nam
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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
$$\begin{split} 2^3=8 &\equiv 1 &\pmod{7} \\ (2^3)^{146} &\equiv 1 &\pmod{7} \\ 2^{438} &\equiv 1 &\pmod{7} \\ 2^{438} \times 2^2 &\equiv 1 \times 2^2 &\pmod{7} \\ 2^{440} &\equiv 4 &\pmod{7} \\ \implies (18^{22})^{10} &\equiv 4 &\pmod{7} \end{split}$$

_________________
Manager
Joined: 30 Jun 2015
Posts: 50
Location: Malaysia
Schools: Babson '20
GPA: 3.6
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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
Joined: 16 Oct 2010
Posts: 9989
Location: Pune, India
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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

Director
Joined: 31 Jul 2017
Posts: 505
Location: Malaysia
GMAT 1: 700 Q50 V33
GPA: 3.95
WE: Consulting (Energy and Utilities)
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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
Manager
Joined: 30 Jun 2015
Posts: 50
Location: Malaysia
Schools: Babson '20
GPA: 3.6
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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/
Senior Manager
Joined: 29 Dec 2017
Posts: 370
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)
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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.

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.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9989
Location: Pune, India
Re: What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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.

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

Senior Manager
Joined: 29 Dec 2017
Posts: 370
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)
What is the remainder when (18^22)^10 is divided by 7 ?  [#permalink]

### Show Tags

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.
What is the remainder when (18^22)^10 is divided by 7 ?   [#permalink] 17 May 2018, 10:34

Go to page   Previous    1   2   [ 38 posts ]

Display posts from previous: Sort by