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Re: what is the reminder when 333^222 is divided by 7? [#permalink]
21 Jul 2013, 02:27

Expert's post

3

This post was BOOKMARKED

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.

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

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

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

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]
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]
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?

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

Expert's post

1

This post was BOOKMARKED

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.

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