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

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What is the remainder when 333^222 is divided by 7? [#permalink] New post 21 Jul 2013, 01:16
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Jul 2013, 02:30, edited 2 times in total.
Renamed the topic, edited the question and the tags.
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Re: what is the reminder when 333^222 is divided by 7? [#permalink] New post 21 Jul 2013, 02:27
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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.

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Re: What is the remainder when 333^222 is divided by 7? [#permalink] New post 28 Jul 2013, 03:05
(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)
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Re: What is the remainder when 333^222 is divided by 7? [#permalink] New post 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 .
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Re: what is the reminder when 333^222 is divided by 7? [#permalink] New post 01 Aug 2013, 12:37
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?
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Re: what is the reminder when 333^222 is divided by 7? [#permalink] New post 02 Aug 2013, 03:28
Expert's post
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


official answer=>e

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] New post 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
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Re: What is the remainder when 333^222 is divided by 7? [#permalink] New post 30 Aug 2013, 10:58
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
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Re: What is the remainder when 333^222 is divided by 7? [#permalink] New post 09 Mar 2014, 12:14
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Re: What is the remainder when 333^222 is divided by 7? [#permalink] New post 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?

Thanks.
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Re: What is the remainder when 333^222 is divided by 7? [#permalink] New post 29 Apr 2014, 19:39
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
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Re: What is the remainder when 333^222 is divided by 7? [#permalink] New post 30 Apr 2014, 06:44
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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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: What is the remainder when 333^222 is divided by 7?   [#permalink] 30 Apr 2014, 06:44
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