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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.

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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