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If n is an non-negative integer is 10^n+8 divisible by 18?

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If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink] New post 18 Dec 2013, 10:43
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If n is an non-negative integer is 10^n+8 divisible by 18?

(1) n is a prime number
(2) n is even

m11 q13

[Reveal] Spoiler:
will post my doubt and OA after some discussion.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 18 Dec 2013, 10:56, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink] New post 18 Dec 2013, 11:18
Will go with A. 2 gives diff answers for zero n non zero
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Re: If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink] New post 18 Dec 2013, 11:47
[quote="PUNEETSCHDV"]If n is an non-negative integer is 10^n+8 divisible by 18?

(1) n is a prime number
(2) n is even


IMO A

1) 10^n+8 will never divisible by 18 if n is prime no. define no means A is sufficient
2) n is even, if n=10, 10^n+8 is divisible by 18, but if n=2 it is not, two diff ans, means not sufficient
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Re: If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink] New post 19 Dec 2013, 00:23
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PUNEETSCHDV 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

m11 q13

[Reveal] Spoiler:
will post my doubt and OA after some discussion.


If n is an non-negative integer is 10^n+8 divisible by 18?

Notice that 10^n+8 is divisible by 18 for any positive value of n. In this case 10^n+8=even+even=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 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 --> 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: If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink] New post 19 Dec 2013, 21:32
please use brackets as 10^n+8 is quiet confusing between 10^(n+8) or (10^n)+8
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Re: If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink] New post 20 Dec 2013, 02:57
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Notice that 10^n + 8 will have sum of digits = 9. Since, no matter what value 'n' may take, we will have 1 + 8 + 0 (depending on n) = 9.
Now this no will be even since last digit is 8 => divisible by 2. Hence, no of the form 10^n + 8 is always divisible by 18 when n > 0.
Only for n =0, the no becomes 9 which is not divisible by 18.

Stmt1 -> sufficient based on above, we know n is not equal to 0.
Stmt2 -> insufficient since n =0 (even no) is not divisible by 18 while others will be.

Hence (A)
Re: If n is an non-negative integer is 10^n+8 divisible by 18?   [#permalink] 20 Dec 2013, 02:57
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