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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

what is the reminder when 333^222 is divided by 7? a.3 b.2 c.5 d.7 e.1

official answer=>e

source- random internet

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.

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

Show Tags

28 Jul 2013, 03:05

1

This post was BOOKMARKED

(333)^222

Rem(333/7) = 4

=> (4) ^222

=> (16) ^111

Rem(16/7) = 2

(2)^111 = 2^100 * 2^11

Now let us observe the Rem(2^10)/7

2^10 = 1024 => Rem(1024/7) = 2

REM(2^100 * 2^11 ) be 7 = REM((2^10)^10 * 2^11) by 7 => REM((2)^10 * 2^10 * 2) by 7

=> Rem( 2* 2 *2 ) by 7

=> 8/7

=> 1

(E)
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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

Show Tags

31 Jul 2013, 03:27

333^222/7

3^222 *(111)^222 /7 =>(111)^222/7=> in terms of remainder (6)^222/7 or (-1)^222/7 which leaves 1 now the other part (3^2)^111/7 => (2)^111/7 =>(8)^27/7=>1^27 and this part is also one .
_________________

--It's one thing to get defeated, but another to accept it.

Re: what is the reminder when 333^222 is divided by 7? [#permalink]

Show Tags

01 Aug 2013, 12:37

1

This post was BOOKMARKED

Bunuel wrote:

jonyg wrote:

what is the reminder when 333^222 is divided by 7? a.3 b.2 c.5 d.7 e.1

official answer=>e

source- random internet

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.

Answer: E.

Hey can you explain to me how you get a remainder of 2 when you divide 2^1/7?

If x and y are positive integers, there exist unique integers q and r, called the quotient and remainder, respectively, such that \(y =divisor*quotient+remainder\)= xq + r and \(0\leq{r}<x.\)

For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since 15 = 6*2 + 3.

Notice that \(0\leq{r}<x\) means that remainder is a non-negative integer and always less than divisor.

As for your query, we can write \(2 = 0*7+2\), where 7 is the divisor, and 2 is the remainder.

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

Show Tags

02 Aug 2013, 04:18

Hi, My solution is as follows when 333/7 gives reminder 4 thus we have to find out 4^222 now 7 is a prime no so according to fermants littile therom (4^6)/7=1 now we have to see if 222 is divisble by 6 thus 222=6*37 hence 4^6k/7 =1 hence answer is 1 ie e

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

Show Tags

29 Apr 2014, 13:36

Hi, I am confused between 2 approaches for these kinds of problems Approach 1: Binomial Theorem. Approach 2: Find the unit's digit of the exponent and then find the remainder.

Unit's digit of 333^222 = unit's digit of 3^222. Then divide that by 7. Cyclicity of 3 = 4 {3,9,7,1}. 222/3 has a remainder of 2. 3^2 has a unit's digit of 9. 9/7 has a remainder of 2

In Approach 2, i don't always get the same ans as by using Approach 1. Which approach is preferred for these kinds of problems?

Hi, I am confused between 2 approaches for these kinds of problems Approach 1: Binomial Theorem. Approach 2: Find the unit's digit of the exponent and then find the remainder.

Unit's digit of 333^222 = unit's digit of 3^222. Then divide that by 7. Cyclicity of 3 = 4 {3,9,7,1}. 222/3 has a remainder of 2. 3^2 has a unit's digit of 9. 9/7 has a remainder of 2

In Approach 2, i don't always get the same ans as by using Approach 1. Which approach is preferred for these kinds of problems?

Thanks.

The units digit does not determine the remainder when dividing by 7. For example, 9 divided by 7 gives the remainder of 2, 19 divided by 7 gives the remainder of 5, 29 divided by 7 gives the remainder of 1, ...
_________________

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

Show Tags

15 Jul 2015, 21:01

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

25 Jul 2015, 04:43

Hi VeritasPrepKarishma : Bunuel Can you please solve my doubt, From 4^222 , 222 Is basically 55m+2. Since 4 has a cyclicity of { 4, 6} , the unit's digit here will be 6. When you divide this by 7, the remainder will be 6. But answer says remainder will be 1.

Hi VeritasPrepKarishma : Bunuel Can you please solve my doubt, From 4^222 , 222 Is basically 55m+2. Since 4 has a cyclicity of { 4, 6} , the unit's digit here will be 6. When you divide this by 7, the remainder will be 6. But answer says remainder will be 1.

Can you please help.

Hi, the units digit cannot determine the remainder except in the case of 2,5,10 etc... 6 will have remainder 6 but 16 will have 2 and so on.. the right way would be 4^222=(4^3)^74... now 4^3=64 and the remainder will be 1 when divided by 7.. so ans will be1^74=1 1 is the remainder.. hope it helps
_________________

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

Show Tags

15 Feb 2016, 09:17

1

This post received KUDOS

Bunuel wrote:

jonyg wrote:

what is the reminder when 333^222 is divided by 7? a.3 b.2 c.5 d.7 e.1

official answer=>e

source- random internet

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.

Answer: E.

Hope it helps.

I have no idea why this is a 95 % difficulty level question.

Just know that- Each term in the expression (x+y)^n is divisible by x except for the last term which is y^n

333^222 / 7

Try and bring the dividend to a form ( multiple of divisor +_ something). And usually such Qs are formed like that.

(333)^222 = (336-3)^222 Each term in the above expression is divisible by 7 ( since 336 is divisible by 7) but the last term which is 3^222

So now essentially our task is to find the remainder when 3^222 is divided by 7

Again the same routine

3^222 = (3^3)^74 = 27^12 or, (28-1)^12 All terms are divisible but the last i.e (-1)^12 12 being a positive power this is equal to 1

Answer E

Am I missing something ?
_________________

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

what is the reminder when 333^222 is divided by 7? a.3 b.2 c.5 d.7 e.1

official answer=>e

source- random internet

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.

Answer: E.

Hope it helps.

I have no idea why this is a 95 % difficulty level question.

Just know that- Each term in the expression (x+y)^n is divisible by x except for the last term which is y^n

333^222 / 7

Try and bring the dividend to a form ( multiple of divisor +_ something). And usually such Qs are formed like that.

(333)^222 = (336-3)^222 Each term in the above expression is divisible by 7 ( since 336 is divisible by 7) but the last term which is 3^222

So now essentially our task is to find the remainder when 3^222 is divided by 7

Again the same routine

3^222 = (3^3)^74 = 27^12 or, (28-1)^12 All terms are divisible but the last i.e (-1)^12 12 being a positive power this is equal to 1

Answer E

Am I missing something ?

Knowing binomial theorem expansion is a great help. Additionally, for the last part you can use cyclicity to aid you in finding the remainder.

Rem of \(3^1/7\) = 3 Rem of \(3^2/7\) = 2 Rem of \(3^3/7\) = 6 Rem of \(3^4/7\) = 4 Rem of \(3^5/7\) = 5 Rem of \(3^6/7\) = 1 ... and repeat

thus the cyclicity of \(3^n\) when divided by 7 = 6. 222/6 = 37 (exactly). Thus the remainder will be = 1.

P.S.: If a question seems "easy" to you need not necessarily be the same for some or in the case of this question, for the majority. 95% difficulty is not a manually inputted value but is calculated on the basis of number of incorrect attempts at this question.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...