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# Find the remainder of the division (2^69)/9.

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Manager
Joined: 08 Sep 2010
Posts: 178
Location: India
WE 1: 6 Year, Telecom(GSM)
Find the remainder of the division (2^69)/9.  [#permalink]

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24 Oct 2010, 19:38
13
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Difficulty:

55% (hard)

Question Stats:

57% (01:28) correct 43% (01:53) wrong based on 295 sessions

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Find the remainder of the division (2^69)/9.

A. 1
B. 4
C. 5
D. 8
E. 7
Manager
Joined: 08 Sep 2010
Posts: 178
Location: India
WE 1: 6 Year, Telecom(GSM)
Re: Find the remainder of the division (2^69)/9.  [#permalink]

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24 Oct 2010, 20:09
2
4
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 .

Consider giving KUDOS if u find it informative and good.Thanks
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Retired Moderator
Joined: 02 Sep 2010
Posts: 769
Location: London
Re: Find the remainder of the division (2^69)/9.  [#permalink]

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24 Oct 2010, 21:35
3
1
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

Posted from my mobile device
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Concentration: General Management, Finance
GPA: 3.59
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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 .

Consider giving KUDOS if u find it informative and good.Thanks
Manager
Joined: 17 Sep 2010
Posts: 180
Concentration: General Management, Finance
GPA: 3.59
WE: Corporate Finance (Entertainment and Sports)
Re: Find the remainder of the division (2^69)/9.  [#permalink]

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30 Nov 2010, 21:19
1
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

2^66 remainder 1
2^67 remainder 2
2^68 remainder 4
2^69 remainder 8

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

Posted from my mobile device

Can somebody pls explain me what exatcly is going here?

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Retired Moderator
Joined: 02 Sep 2010
Posts: 769
Location: London
Re: Find the remainder of the division (2^69)/9.  [#permalink]

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30 Nov 2010, 23:12
1
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

Posted from my mobile device

Can somebody pls explain me what exatcly is going here?

------------------------------------------------------------------------------

The remainder of any such sequence of powers always has a cyclical pattern to it. I am just trying to figure out the pattern

Initially I have shown how the cyclicity is 6, i.e, the 1st the 7th th 13th and so on powers are the same

Now we need the 69th power

69 = 6 * 11 + 3

Hence the remainder of the 69th power will be the same as that of the 3rd power, hence 8
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Re: Find the remainder of the division (2^69)/9.  [#permalink]

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21 Nov 2014, 18:09
Can you explain why does the cycle not start at 2^0? Then you would have the following.

2^0 is 1
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

Staying w/ the same formula
69=6*11 + 3; The remainder would then be 4
Math Expert
Joined: 02 Sep 2009
Posts: 50711
Re: Find the remainder of the division (2^69)/9.  [#permalink]

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22 Nov 2014, 05:34
1
2
judges32 wrote:
Can you explain why does the cycle not start at 2^0? Then you would have the following.

2^0 is 1
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

Staying w/ the same formula
69=6*11 + 3; The remainder would then be 4

Take it as a rule to start with the power of 1.
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Manager
Joined: 17 Dec 2013
Posts: 57
GMAT Date: 01-08-2015
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.
Math Expert
Joined: 02 Aug 2009
Posts: 7041
Re: Find the remainder of the division (2^69)/9.  [#permalink]

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20 Jan 2015, 02:07
Bunuel wrote:
judges32 wrote:
Can you explain why does the cycle not start at 2^0? Then you would have the following.

2^0 is 1
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

Staying w/ the same formula
69=6*11 + 3; The remainder would then be 4

Take it as a rule to start with the power of 1.

If you want to count 2^0 also then 2^69 becomes 70th term and ans will still remain the same
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Re: Find the remainder of the division (2^69)/9.  [#permalink]

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26 Sep 2017, 02:35
1
ankitranjan wrote:
Find the remainder of the division (2^69)/9.

A. 1
B. 4
C. 5
D. 8
E. 7

Using Binomial,

$$\frac{2^{69}}{9} = \frac{2^{3*23}}{9} = \frac{8^{23}}{9} = \frac{(9 - 1)^{23}}{9}$$

The remainder will be -1 i.e. 8

For details of this method, check:
https://www.veritasprep.com/blog/2011/0 ... ek-in-you/
https://www.veritasprep.com/blog/2014/0 ... -the-gmat/
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Re: Find the remainder of the division (2^69)/9.  [#permalink]

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29 Sep 2017, 09:24
ankitranjan wrote:
Find the remainder of the division (2^69)/9.

A. 1
B. 4
C. 5
D. 8
E. 7

Let’s find a remainder pattern:

2^1/9 has a remainder of 2

2^2/9 has a remainder of 4

2^3/9 has a remainder of 8

2^4/9 = 16/9 has a remainder of 7

2^5/9 = 32/9 has a remainder of 5

2^6/9 = 64/9 has remainder of 1

2^7/9 = 128/9 has a remainder of 2

We see the pattern of remainders is 2-4-8-7-5-1, so it repeats every 6 exponents.

Thus, 2^66/9 has a remainder of 1, 2^67/9 has a remainder of 2, 2^68/9 has a remainder of 4, and 2^69/9 has a remainder of 8.

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Re: Find the remainder of the division (2^69)/9. &nbs [#permalink] 29 Sep 2017, 09:24
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