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Intern  Joined: 12 Mar 2013
Posts: 7
What is the remainder when 333^222 is divided by 7?  [#permalink]

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80 00:00

Difficulty:   95% (hard)

Question Stats: 44% (01:58) correct 56% (01:31) wrong based on 1001 sessions

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

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

Originally posted by jonyg on 21 Jul 2013, 01:16.
Last edited by Bunuel on 21 Jul 2013, 02:30, edited 2 times in total.
Renamed the topic, edited the question and the tags.
Math Expert V
Joined: 02 Sep 2009
Posts: 64988
What is the remainder when 333^222 is divided by 7?  [#permalink]

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12
33
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.
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Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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1
1
(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

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

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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 .
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Posts: 10
Re: what is the reminder when 333^222 is divided by 7?  [#permalink]

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2
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

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.

Hey can you explain to me how you get a remainder of 2 when you divide 2^1/7?
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Posts: 562
Re: what is the reminder when 333^222 is divided by 7?  [#permalink]

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1
iNumbv wrote:
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

source- random internet

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

This might help : remainders-144665.html

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.

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

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

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2
1
A simple one line solution to this problem can be this:
Rem(333^222)/7 = Rem(4^222)/7 = Rem(64^74)/7=Rem((63+1)^74)/7 = 1
Math Expert V
Joined: 02 Sep 2009
Posts: 64988
Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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Bumping for review and further discussion.

For more on this kind of questions check Units digits, exponents, remainders problems collection.
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Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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1
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.
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Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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1
Hi All,

I used the following approach.

(333^222)/7

(333/7) = Remainder is 4

4^222 can be written as 2^444 which can be written as (2^3)^148

now what we have to do find is

((2^3)^148)/7

we can write the above expression as

((7+1)^148)/7

now apply remainder theorem.

Hence Remainder is 1.

Option E is correct
Math Expert V
Joined: 02 Sep 2009
Posts: 64988
Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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2
2
gmatcracker2407 wrote:
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, ...
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Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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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.

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Posts: 8751
Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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Shree9975 wrote:
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,
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
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Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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333 = (3*111) ^222
111/7 = (-1)^222 =1

now only 3^222
(3^2)^111
(9/7)^111 = 2^111
(2^3)^37

(8/7)^111 =1

ans is 1
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Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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3
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

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.

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

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

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1
KarishmaParmar wrote:
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

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.

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

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.

Read this cyclicity-on-the-gmat-213019.html for more on remainders and cyclicity.

Hope this helps.

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

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1
jonyg wrote:
What is the remainder when 333^222 is divided by 7?

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

$$333 = 4 (mod_7)$$

$$\frac{4^{222}}{7} = \frac{2^{444}}{7}$$

$$2^3 = 1 (mod_7)$$

$$\frac{(4^3)^{148}}{7} = \frac{1^{148}}{7} = \frac{1}{7}$$

Remainder is $$1$$
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Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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Hi Bunuel,

why is the remainder not 2?

With the questions I've been going through, the method was always the same. When a question asks for the remainder when dividing a number^integer, it is basically asking for the unit digit of the number.

And so I did the unit digit of 3 ^222 which would be 9. but then 9/7 gives a remainder of 2.

(Example of a similar question: 2^243 divided by 5, what is remainder? then the remainder is 2 because of cyclisity it ends up being 7, which if 7/5 gives a remainder of 2.
Or 43^86 divided by 5, the remainder is 4 because 9/5 gives a remainder of 5)

Could you please explain why this method didn't work for this example?

Thanks in advance! Math Expert V
Joined: 02 Sep 2009
Posts: 64988
Re: What is the remainder when 333^222 is divided by 7?  [#permalink]

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lisa.cornelia wrote:
Hi Bunuel,

why is the remainder not 2?

With the questions I've been going through, the method was always the same. When a question asks for the remainder when dividing a number^integer, it is basically asking for the unit digit of the number.

And so I did the unit digit of 3 ^222 which would be 9. but then 9/7 gives a remainder of 2.

(Example of a similar question: 2^243 divided by 5, what is remainder? then the remainder is 2 because of cyclisity it ends up being 7, which if 7/5 gives a remainder of 2.
Or 43^86 divided by 5, the remainder is 4 because 9/5 gives a remainder of 5)

Could you please explain why this method didn't work for this example?

Thanks in advance! Have you checked this post: https://gmatclub.com/forum/what-is-the- ... l#p1248585 ?
_________________ Re: What is the remainder when 333^222 is divided by 7?   [#permalink] 09 Feb 2018, 01:36

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