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

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21 Jul 2013, 03:27

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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.

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

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28 Jul 2013, 04:05

(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]

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31 Jul 2013, 04:27

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]

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01 Aug 2013, 13:37

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]

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02 Aug 2013, 05:18

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]

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29 Apr 2014, 14:36

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]

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30 Apr 2014, 07:44

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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, ... _________________

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

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15 Jul 2015, 22:01

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

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25 Jul 2015, 05:43

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.

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

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25 Jul 2015, 06:01

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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.

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]

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15 Feb 2016, 10:17

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

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15 Feb 2016, 10:30

Expert's post

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

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. _________________

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