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15 Sep 2014, 23:44



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10 Aug 2015, 16:34
I think this is a highquality 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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17 Aug 2015, 02:31



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18 Aug 2015, 05:01
I think this is a highquality 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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19 Oct 2015, 09:39
Bunuel wrote: If \(n\) is an nonnegative 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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19 Oct 2015, 21:01
Mechmeera wrote: Bunuel wrote: If \(n\) is an nonnegative 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 Nonnegative integers are those which are not negative: 0, 1, 2, 3, 4, ...
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27 May 2016, 05:21
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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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 nonnegative 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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27 May 2016, 05:37
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 nonnegative 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: M1113
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27 Jun 2016, 01:03
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: M1113
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19 Aug 2016, 09:35
Typo in explanation for st 2, kindly rectify



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21 Aug 2016, 02:20



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21 Aug 2016, 02:27
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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09 Jun 2017, 06:12
I think this is a highquality 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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19 Jun 2017, 21:13
bfwell wrote: I think this is a highquality 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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22 Jul 2018, 01:21
Missed the 0 while considering even# and marked D!!!










