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

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ankitranjan
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Find the remainder of the division (2^69)/9. [#permalink] New post   24 Oct 2010, 20:38
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Find the remainder of the division (2^69)/9.

A. 1
B. 4
C. 5
D. 8
E. 7
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D
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ankitranjan
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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   24 Oct 2010, 21:09
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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.

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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   24 Oct 2010, 22:35
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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

Answer : (d)

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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   30 Nov 2010, 21:49
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
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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   30 Nov 2010, 22:19
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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

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

And that is your answer.


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?

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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   01 Dec 2010, 00:12
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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?

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


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] New post   21 Nov 2014, 19:09
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
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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   22 Nov 2014, 06:34
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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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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   20 Jan 2015, 02:05
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
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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   20 Jan 2015, 03:07
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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] New post   26 Sep 2017, 03:35
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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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... ek-in-you/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/
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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   29 Sep 2017, 10:24
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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.

Answer: D
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Re: Find the remainder of the division (2^69)/9. [#permalink] New post   21 Sep 2023, 11:11
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