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# What is the least possible value that can be subtracted from 2^586 so

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Math Expert
Joined: 02 Sep 2009
Posts: 47084
What is the least possible value that can be subtracted from 2^586 so [#permalink]

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27 Apr 2016, 00:11
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85% (hard)

Question Stats:

58% (01:23) correct 42% (01:46) wrong based on 170 sessions

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What is the least possible value that can be subtracted from 2^586 so that the result is a multiple of 7?

A. 2
B. 3
C. 5
D. 7
E. 11

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Math Expert
Joined: 02 Aug 2009
Posts: 6235
Re: What is the least possible value that can be subtracted from 2^586 so [#permalink]

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27 Apr 2016, 02:31
3
3
Bunuel wrote:
What is the least possible value that can be subtracted from 2^586 so that the result is a multiple of 7?

A. 2
B. 3
C. 5
D. 7
E. 11

The Q is nothing but finding remainder when 2^586 is divided by 7..
Ofcourse since we are looking for least value, 7 and 11 cannot be the answer as the least value in those cases would be 0 or 4 and not 7 and 11..

we have to get the figure 2^586 in terms of a binomial expression..
$$2^{586} = (2^3)^{195} *2 = 8^195 * 2 = (7+1)^{195} * 2$$
$$(7+1)^{195}$$ will leave a remainder of 1 when divided by 7.
so remainder = 1*2 =2
A
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Re: What is the least possible value that can be subtracted from 2^586 so [#permalink]

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27 Apr 2016, 03:18
1
Bunuel wrote:
What is the least possible value that can be subtracted from 2^586 so that the result is a multiple of 7?

A. 2
B. 3
C. 5
D. 7
E. 11

We need to subtract something to make 2^586 a multiple of 7.

So let us try to write 2^586 in a from from which we can convert it in terms of 7
We know that 2^3 = 8 and 8 = (7+1)

Hence we will write 2^586 in powers of 8
2^586 = 2*2^585 = 2*8^195 = 2*(7+1)^195

All of the terms of (7+1)^195 will be multiple of 7 apart from the last term - "1"
Hence 2*(7+1)^195 = 2(Multiples of 7 + 1) = 2*Multiples of 7 + 2
Therefore we need to subtract 2 to make 2^586 a multiple of 7.

Correct Option: A
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What is the least possible value that can be subtracted from 2^586 so [#permalink]

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Updated on: 28 Apr 2016, 08:28
1
3
Bunuel wrote:
What is the least possible value that can be subtracted from 2^586 so that the result is a multiple of 7?

A. 2
B. 3
C. 5
D. 7
E. 11

A different approach

(2^1) / 7 will have remainder 2
(2^2) / 7 will have remainder 4
(2^3) / 7 will have remainder 1

(2^4) / 7 will have remainder 2
(2^5) / 7 will have remainder 4
(2^6) / 7 will have remainder 1

Thus cyclicity of 2^n / 7 is 3

$$2^{586}$$ = $$2^{195*3}$$ * $$2^1$$

$$2^{195*3}$$ / 7 will have remainder 1
$$2^1$$ / 7 will have remainder 2

Finally we have 2*1/7 , remainder will be 2

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Originally posted by Abhishek009 on 27 Apr 2016, 08:48.
Last edited by Abhishek009 on 28 Apr 2016, 08:28, edited 1 time in total.
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What is the least possible value that can be subtracted from 2^586 so [#permalink]

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28 Apr 2016, 04:03
Hi chetan2u/Janielle/Abhishek009,

Thanks for all of you methods.

However, after solving this question, I have another question. what if the same number is divided by 5? what would the reminder? I tried different approaches that you provided in you posts but seems I miss something because i have different answers.

Method 1 :
$$2^{586} = (2^3)^{195} *2 = 8^195 * 2 = (5+3)^{195} * 2$$
$$(5+3)^{195}$$ will leave a remainder of 3 when divided by 5
so remainder = 3*2 =6

Method 2:
$$2^{586} = (2^2)^{293} =(5-1)^{195} * 2$$

I do not how to find reminder!!!!!

Method 3:
(2^1) / 5 will have remainder 2
(2^2) / 5 will have remainder 4
(2^3) / 5 will have remainder 3
(2^4) / 5 will have remainder 1
(2^5) / 5 will have remainder 2

Thus cyclicity of 2^n / 5 is 4

2^286 = 2^2 * 2^284

2^2/5 will have reminder 4

2^284/5 will have reminder 1

so reminder 4*1=4

I welcome your comments to help me figure where my problem is

Thanks
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What is the least possible value that can be subtracted from 2^586 so [#permalink]

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28 Apr 2016, 08:27
Mo2men wrote:

Method 3:
(2^1) / 5 will have remainder 2
(2^2) / 5 will have remainder 4
(2^3) / 5 will have remainder 3
(2^4) / 5 will have remainder 1
(2^5) / 5 will have remainder 2

I welcome your comments to help me figure where my problem is

Thanks

(2^1) / 5 will have remainder 2
(2^2) / 5 will have remainder 4
(2^3) / 5 will have remainder 3
(2^4) / 5 will have remainder 1

(2^5) / 5 will have remainder 2
(2^6) / 5 will have remainder 4
(2^7) / 5 will have remainder 3
(2^8) / 5 will have remainder 1

So, Cyclicity will be 4

$$2^{586}$$ = $$2^{4*146}$$ * $$2^2$$

The blue part will have remainder 1 and the red part will always have the remainder 4

So IMHO result will be 4

Perfect as you stated.....

Mo2men wrote:

Method 2:
$$2^{586} = (2^2)^{293} =(5-1)^{195} * 2$$

$$2^{586}$$ = $$2^{2*293}$$ =>4^{293}

$$4^{293}$$ = $$(5 - 1)^{293}$$

Now 5^{293} divided by 5 will have remainder as 5

-1^{293} divided by 5 will have remainder as -1 { Since 293 is an odd power -1^Odd power = -1 }

Now you have 5 - 1 = 4
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Thanks and Regards

Abhishek....

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Re: What is the least possible value that can be subtracted from 2^586 so [#permalink]

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10 Oct 2016, 07:53
Bunuel wrote:
What is the least possible value that can be subtracted from 2^586 so that the result is a multiple of 7?

A. 2
B. 3
C. 5
D. 7
E. 11

my approach...
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64

I see a pattern here..when the exponent is a multiple of 3, I need to deduct 1 to have a number which is multiple of 7.
when the exponent is a multiple of 3+1, then I need to deduct 2.

585 - is a multiple of 3. (sum of digits 5+8+5 is a multiple of 3)
it means that from 2^586, we will need to deduct 2, to get a number divisible by 7.
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Re: What is the least possible value that can be subtracted from 2^586 so [#permalink]

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01 Jan 2018, 13:21
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Re: What is the least possible value that can be subtracted from 2^586 so   [#permalink] 01 Jan 2018, 13:21
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