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Re: Find the remainder of the division (2^69)/9. [#permalink]

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24 Oct 2010, 20:09

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These type of questions can easily be solved with the help of remainder theorem which states that..When f(x) ,a polynomial function in x is divided by (x-a),the remainder will be f(a)

In the division since the numerator is in terms of power of 2,the denominator 9 also should be expressed in terms of power of 2 i.e as (2^3 + 1). Now numerator 2^69 can be written as (2^3)^23.

Now when (2^3)^23 is divided by {2^3 -(-1)} ,according to remainder theorem the remainder should be f(-1)

and f(-1)= {(-1)^3}^23 = -1 From this we are getting the remainder as -1 but to make it positive we have to add divisor. i.e -1 + 9 = 8 .

Hence answer is D.

Consider giving KUDOS if u find it informative and good.Thanks

Re: Find the remainder of the division (2^69)/9. [#permalink]

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24 Oct 2010, 21:35

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The remainder theorem is probably out of scope of the GMAT. Alternatively observe the cyclicity of the powers of 2 modulo 9 : 2^1 is 2 2^2 is 4 2^3 is 8 2^4 is 7 2^5 is 5 2^6 is 1 2^7 is 2 Pattern repeats ... 69 = 6*11 + 3 so remainder for 2^69 is 8

Re: Find the remainder of the division (2^69)/9. [#permalink]

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30 Nov 2010, 20:49

Nice answer. I did mine the long way, pretty much finding the pattern with the powers.

2^1=2 remainder 2 2^2=4 remainder 4 .............So on and so forth.

ankitranjan wrote:

These type of questions can easily be solved with the help of remainder theorem which states that..When f(x) ,a polynomial function in x is divided by (x-a),the remainder will be f(a)

In the division since the numerator is in terms of power of 2,the denominator 9 also should be expressed in terms of power of 2 i.e as (2^3 + 1). Now numerator 2^69 can be written as (2^3)^23.

Now when (2^3)^23 is divided by {2^3 -(-1)} ,according to remainder theorem the remainder should be f(-1)

and f(-1)= {(-1)^3}^23 = -1 From this we are getting the remainder as -1 but to make it positive we have to add divisor. i.e -1 + 9 = 8 .

Hence answer is D.

Consider giving KUDOS if u find it informative and good.Thanks

Re: Find the remainder of the division (2^69)/9. [#permalink]

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30 Nov 2010, 21:19

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Find the pattern of the remainders after each power:

2^1 remainder 2 2^2 remainder 4 2^3 remainder 8 2^4 remainder 7 2^5 remainder 5 2^6 remainder 1 -->this is where the cycle ends 2^7 remainder 2 -->this is where the cycle begins again

The remainder theorem is probably out of scope of the GMAT. Alternatively observe the cyclicity of the powers of 2 modulo 9 : 2^1 is 2 2^2 is 4 2^3 is 8 2^4 is 7 2^5 is 5 2^6 is 1 2^7 is 2 Pattern repeats ... 69 = 6*11 + 3 so remainder for 2^69 is 8 Answer : (d)

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Can somebody pls explain me what exatcly is going here?

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Re: Find the remainder of the division (2^69)/9. [#permalink]

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30 Nov 2010, 23:12

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krishnasty wrote:

shrouded1 wrote:

The remainder theorem is probably out of scope of the GMAT. Alternatively observe the cyclicity of the powers of 2 modulo 9 : 2^1 is 2 2^2 is 4 2^3 is 8 2^4 is 7 2^5 is 5 2^6 is 1 2^7 is 2 Pattern repeats ... 69 = 6*11 + 3 so remainder for 2^69 is 8 Answer : (d)

Posted from my mobile device

Can somebody pls explain me what exatcly is going here?

Re: Find the remainder of the division (2^69)/9. [#permalink]

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20 Jan 2015, 01:05

Quote:

Now when (2^3)^23 is divided by {2^3 -(-1)} ,according to remainder theorem the remainder should be f(-1)

and f(-1)= {(-1)^3}^23 = -1

can someone explain this step pls? I did understand the other methods which were discussed (with the cycles). but I can't understand this method. thanks in advance

Re: Find the remainder of the division (2^69)/9. [#permalink]

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26 Sep 2017, 01:12

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