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# What is the remainder when 333^222 is divided by 7?

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Current Student
Joined: 07 Jan 2016
Posts: 1083
Location: India
GMAT 1: 710 Q49 V36
What is the remainder when 333^222 is divided by 7?  [#permalink]

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09 Feb 2018, 01:42
1
jonyg wrote:
What is the remainder when 333^222 is divided by 7?

A. 3
B. 2
C. 5
D. 7
E. 1

The Euler's of 7 = 7 ( 1-1/7) = 6

333 mod 7 = 4

222 mod 6 = 0

4^0 = 1

333^222 mod 7 = 1

E

PS - These type of questions are very popular in the Indian Management Exam for the IIM's i.e CAT and this is the simplest way to solve such sums
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Joined: 14 Dec 2016
Posts: 11
Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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14 Mar 2018, 09:29
Bunuel wrote:
What is the remainder when 333^222 is divided by 7?
A. 3
B. 2
C. 5
D. 7
E. 1

$$333^{222}=(329+4)^{222}=(7*47+4)^{222}$$. Now if we expand this, all terms but the last one will have 7*47 as a multiple and thus will be divisible by 7. The last term will be $$4^{222}=2^{444}$$. So we should find the remainder when $$2^{444}$$ is divided by 7.

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;
...

The remainder repeats in blocks of three: {2-4-1}. So, the remainder of $$2^{444}$$ divided by 7 would be the same as $$2^3$$ divided by 7 (444 is a multiple of 3). $$2^3$$ divided by 7 yields remainder of 1.

Similar question to practice:
http://gmatclub.com/forum/what-is-the-r ... 34778.html
http://gmatclub.com/forum/when-51-25-is ... 30220.html
http://gmatclub.com/forum/what-is-the-r ... 26493.html
http://gmatclub.com/forum/what-is-the-r ... 00316.html
http://gmatclub.com/forum/what-is-the-r ... 99724.html

Theory on remainders problems: http://gmatclub.com/forum/remainders-144665.html

DS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198
PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199

Hope it helps.

why aren't we checking the cyclicity of 4^222 and then divide that number by 7 .
we followed the same approach here http://gmatclub.com/forum/what-is-the-r ... 26493.html .. plz explain , is it just because the division with 5 is different than division with other numbers
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WE: Engineering (Other)
Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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22 May 2018, 01:51
1
1
jonyg wrote:
What is the remainder when 333^222 is divided by 7?

A. 3
B. 2
C. 5
D. 7
E. 1

Another approach for similar kinds of sums is using Fermet's theorem. Refer attached photo
Attachments

WhatsApp Image 2018-05-22 at 3.18.54 PM.jpeg [ 90.53 KiB | Viewed 982 times ]

Intern
Joined: 07 Jul 2018
Posts: 6
What is the remainder when 333^222 is divided by 7?  [#permalink]

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08 Jul 2018, 13:29
1
1
I've read all the previous posts and it seems one step is missing (certainly too obvious for most of you guys) but for the others, check it out (I've eluded the parenthesis to simplify):
333^222=3^222 x 111^222

111^222 / 7 is found by actually dividing 111 by 7;
=>111=77+41=77+42-1
111 =77+35+6=77+35+7-1
=>with the remainder -1 comes -1^222=1, because 222 is even

3^222 =9^111 gives the following remainders when divided by 7:
9=7-2 => 2^111
8=7+1=> 1^37 (111 is a multiple of 3 as the sum of its digits is a multiple of 3, you don't need to find the quotient 37, you only to have 111=3x, to get 2^3=8)
=> Thus the remainder of this part is 1

Finally, the remainder for the whole number is 1x1 =1

Hope it'll help
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Welcoming critics is my way to improvement. So do not hesitate, tell me how I can improve. Thx
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Joined: 03 Jun 2019
Posts: 3184
Location: India
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WE: Engineering (Transportation)
Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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29 Mar 2020, 01:06
jonyg wrote:
What is the remainder when 333^222 is divided by 7?

A. 3
B. 2
C. 5
D. 7
E. 1

Asked: What is the remainder when 333^222 is divided by 7?

333^222mod 7 = 4^222mod7 = 4^{3*74}mod7 = 1

IMO E
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com
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Posts: 61
Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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08 May 2020, 00:52
I think we can write 333 as 343 -10 --> 7^3 -10
Now applying binomial theorem we have the term (-10)^222/7. From here:
remainders list
1. 10/7 = 3
2 100/7 = 2
3.1000/7 = 6
4. 10^4/7 = 4
5.10^5/7 = 5
6. 10^6/7 = 1
and the cycle repeats. So [-(10^222/7)] becomes 222=37*6 and so 37 =6*6+1 which leads us to the remainder for -10/7.
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Posts: 61
Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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08 May 2020, 07:33
Hi
Here is another approach:

(333^222)/7 = (343-10)^222/7 ==> (7^3-10)^222/7

Now: use binomial theorem and isolate the term [10^222/7]
If we observe, the remainders of 7 follow a cycle: 1,4,2,8,5,7 for every 10^6. So 222/6 ==> 37 and so we get 1 as the remainder

Hope this helps!
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Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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08 May 2020, 13:07
jonyg wrote:
What is the remainder when 333^222 is divided by 7?

A. 3
B. 2
C. 5
D. 7
E. 1

The exponents of 3 are 3, 9, 81, 243, 729, 2187... When divided by 7 the remainder : 3, 2, 4, 5, 1, 3....(pattern of 5). The power is 222. When divided by 5, we get remainder 2. So given the pattern of 5, the second remainder is 2.
Re: What is the remainder when 333^222 is divided by 7?   [#permalink] 08 May 2020, 13:07

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