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M11-13

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M11-13  [#permalink]

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New post 16 Sep 2014, 00:44
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Re M11-13  [#permalink]

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New post 16 Sep 2014, 00:44
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Official Solution:


Notice that \(10^n+8\) is divisible by 18 for any positive value of \(n\). In this case \(10^n+8=\text{even}+\text{even}=\text{even}\) so it's divisible by 2. Also, in this case, the sum of the digits of \(10^n+8\) is 9 so its divisible by 9. Since \(10^n+8\) is divisible by both 2 and 9, then it's divisible by \(2*9=18\) too.

On the other hand, if \(n=0\) then \(10^n+8=1+8=9\), so in this case \(10^n+8\) is not divisible by 9.

(1) \(n\) is a prime number. Hence, \(n\) is a positive integer. Sufficient.

(2) \(n\) is even. \(n\) can be zero as well as any positive even number. Not sufficient.


Answer: A
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Re M11-13  [#permalink]

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New post 10 Aug 2015, 17:34
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I think this is a high-quality question and I don't agree with the explanation. should say - On the other hand, if n=0 then 10n+8=1+8=9, so in this case 10n+8 is not divisible by "18".
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New post 17 Aug 2015, 03:31
bfwell wrote:
I think this is a high-quality question and I don't agree with the explanation. should say - On the other hand, if n=0 then 10n+8=1+8=9, so in this case 10n+8 is not divisible by "18".


If it's not divisible by 9, then it's also not divisible by 18.
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Re M11-13  [#permalink]

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New post 18 Aug 2015, 06:01
I think this is a high-quality question and I agree with explanation. I do agree with explanation but there is a mistake in it (a typo..)

On the other hand, if n=0 then 10n+8=1+8=9, so in this case 10n+8 is not divisible by 9

It must be :
On the other hand, if n=0 then 10n+8=1+8=9, so in this case 10n+8 is not divisible by 2
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Re: M11-13  [#permalink]

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New post 19 Oct 2015, 10:39
Bunuel wrote:
If \(n\) is an non-negative integer, is \(10^n+8\) divisible by 18?


(1) \(n\) is a prime number.

(2) \(n\) is even.

Official Solution:


Notice that \(10^n+8\) is divisible by 18 for any positive value of \(n\). In this case \(10^n+8=\text{even}+\text{even}=\text{even}\) so it's divisible by 2. Also, in this case, the sum of the digits of \(10^n+8\) is 9 so its divisible by 9. Since \(10^n+8\) is divisible by both 2 and 9, then it's divisible by \(2*9=18\) too.

On the other hand, if \(n=0\) then \(10^n+8=1+8=9\), so in this case \(10^n+8\) is not divisible by 9.

(1) \(n\) is a prime number. Hence, \(n\) is a positive integer. Sufficient.

(2) \(n\) is even. \(n\) can be zero as well as any positive even number. Not sufficient.


Answer: A


How can we consider zero for statement when it is non negative and non positive integer.
we can take only 2,4,6,8.................etc
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Re: M11-13  [#permalink]

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New post 19 Oct 2015, 22:01
Mechmeera wrote:
Bunuel wrote:
If \(n\) is an non-negative integer, is \(10^n+8\) divisible by 18?


(1) \(n\) is a prime number.

(2) \(n\) is even.

Official Solution:


Notice that \(10^n+8\) is divisible by 18 for any positive value of \(n\). In this case \(10^n+8=\text{even}+\text{even}=\text{even}\) so it's divisible by 2. Also, in this case, the sum of the digits of \(10^n+8\) is 9 so its divisible by 9. Since \(10^n+8\) is divisible by both 2 and 9, then it's divisible by \(2*9=18\) too.

On the other hand, if \(n=0\) then \(10^n+8=1+8=9\), so in this case \(10^n+8\) is not divisible by 9.

(1) \(n\) is a prime number. Hence, \(n\) is a positive integer. Sufficient.

(2) \(n\) is even. \(n\) can be zero as well as any positive even number. Not sufficient.


Answer: A


How can we consider zero for statement when it is non negative and non positive integer.
we can take only 2,4,6,8.................etc


Non-negative integers are those which are not negative: 0, 1, 2, 3, 4, ...
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Re: M11-13  [#permalink]

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New post 27 May 2016, 06:21
Bunuel wrote:
Official Solution:


Notice that \(10^n+8\) is divisible by 18 for any positive value of \(n\). In this case \(10^n+8=\text{even}+\text{even}=\text{even}\) so it's divisible by 2. Also, in this case, the sum of the digits of \(10^n+8\) is 9 so its divisible by 9. Since \(10^n+8\) is divisible by both 2 and 9, then it's divisible by \(2*9=18\) too.

On the other hand, if \(n=0\) then \(10^n+8=1+8=9\), so in this case \(10^n+8\) is not divisible by 9.

(1) \(n\) is a prime number. Hence, \(n\) is a positive integer. Sufficient.

(2) \(n\) is even. \(n\) can be zero as well as any positive even number. Not sufficient.


Answer: A


Bunuel:

Since any positive value of \(n\) would result in a divisible (by 18) number, is the explaination for statement 2 supposed to say any negative even number, not positive?

Thanks!
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M11-13  [#permalink]

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New post 27 May 2016, 06:35
Sallyzodiac wrote:
Bunuel wrote:
Official Solution:


Notice that \(10^n+8\) is divisible by 18 for any positive value of \(n\). In this case \(10^n+8=\text{even}+\text{even}=\text{even}\) so it's divisible by 2. Also, in this case, the sum of the digits of \(10^n+8\) is 9 so its divisible by 9. Since \(10^n+8\) is divisible by both 2 and 9, then it's divisible by \(2*9=18\) too.

On the other hand, if \(n=0\) then \(10^n+8=1+8=9\), so in this case \(10^n+8\) is not divisible by 9.

(1) \(n\) is a prime number. Hence, \(n\) is a positive integer. Sufficient.

(2) \(n\) is even. \(n\) can be zero as well as any positive even number. Not sufficient.


Answer: A


Bunuel:

Since any positive value of \(n\) would result in a divisible (by 18) number, is the explaination for statement 2 supposed to say any negative even number, not positive?

Thanks!


Hi,

It is already given that n is a non-negative integer, so 'n' can take value of either 0 or any positive integer..

Statement II tells us that n is even.....
so n can be 0...... then 10^n + 8 = 1+8=9, which is not div by 18.... ans NO
n can be positive EVEN integer, here ans will always be YES

so II is insuff and the EXPLANATION given is correct....
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Re: M11-13  [#permalink]

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New post 27 May 2016, 06:37
chetan2u wrote:
Sallyzodiac wrote:
Bunuel wrote:
Official Solution:


Notice that \(10^n+8\) is divisible by 18 for any positive value of \(n\). In this case \(10^n+8=\text{even}+\text{even}=\text{even}\) so it's divisible by 2. Also, in this case, the sum of the digits of \(10^n+8\) is 9 so its divisible by 9. Since \(10^n+8\) is divisible by both 2 and 9, then it's divisible by \(2*9=18\) too.

On the other hand, if \(n=0\) then \(10^n+8=1+8=9\), so in this case \(10^n+8\) is not divisible by 9.

(1) \(n\) is a prime number. Hence, \(n\) is a positive integer. Sufficient.

(2) \(n\) is even. \(n\) can be zero as well as any positive even number. Not sufficient.


Answer: A


Bunuel:

Since any positive value of \(n\) would result in a divisible (by 18) number, is the explaination for statement 2 supposed to say any negative even number, not positive?

Thanks!


Hi,

It is already given that n is a non-negative integer, so 'n' can take value of either 0 or any positive integer..

Statement II tells us that n is even.....
so n can be 0...... then 10^n + 8 = 1+8=9, which is not div by 18.... ans NO
n can be positive EVEN integer, here ans will always be YES

so II is insuff and the EXPLANATION given is correct....


Gotcha, thanks!
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Re: M11-13  [#permalink]

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New post 27 Jun 2016, 02:03
This is how I solved it –
Non negative integer means zero, 1,2,3 …….
Now as (1) says that n is prime, so it cannot be zero. Also (2) says that n is even so it cannot be zero. So forget zero.
Take n = 1,2,3…in each case, we will get 100+8 or 10000+8 or 1000000+3 ……
All of these numbers will be divisible by 18 because each of these sums are ending with 8 which is divisible by 2 AND
Sum of all digits is 9 ..(because there are lot of zeros and 1 and 8 in every sum).
As every number is divisible by 9 and 2 …that means it is divisible by 18.
Hence (1) is sufficient. So strike out answers BCE.
Now take (2). N is even. In this case also, it is divisible by 18 due to above mentioned logic. Hence strike out A.
Answer is D.
HOLD on! I just read answer by Bunuel and then googled that ZERO is considered as EVEN. Now, I will never forget it.
So (2) is not sufficient because if we take n=0 then we get answer 9 and 9 is not divisible by 18. Hence, answer is A.
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Re: M11-13  [#permalink]

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New post 19 Aug 2016, 10:35
Typo in explanation for st 2, kindly rectify
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Re: M11-13  [#permalink]

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New post 21 Aug 2016, 03:20
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Re: M11-13  [#permalink]

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New post 21 Aug 2016, 03:27
Bunuel wrote:
sanghar wrote:
Typo in explanation for st 2, kindly rectify


Can you please tell what typo are you talking about? Thank you.


On the other hand, if n = 0, 10^n + 8 is divisible by 9, but not by 2. Hence, not divisible by 18.

Its a minor discrepancy but makes a difference.

Posted from my mobile device
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Re M11-13  [#permalink]

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New post 09 Jun 2017, 07:12
I think this is a high-quality question and I agree with explanation. Small typo - "On the other hand, if n=0n=0 then 10n+8=1+8=910n+8=1+8=9, so in this case 10n+810n+8 is not divisible by 9". It should be "not divisible by 18"
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Re: M11-13  [#permalink]

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New post 19 Jun 2017, 22:13
bfwell wrote:
I think this is a high-quality question and I don't agree with the explanation. should say - On the other hand, if n=0 then 10n+8=1+8=9, so in this case 10n+8 is not divisible by "18".


To rule out a given information, we have to get two answers, so the answer explanation is perfect, as it tells when can we get an answer as a "yes" and a "no" with the information given in B.
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Re: M11-13  [#permalink]

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New post 22 Jul 2018, 02:21
Missed the 0 while considering even# and marked D!!! :roll:
Re: M11-13 &nbs [#permalink] 22 Jul 2018, 02:21
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