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Re: If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink]

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18 Dec 2013, 12:47

[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

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 (the LCM of 2 and 9) 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.

Re: If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink]

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20 Dec 2013, 03:57

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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.

Re: If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink]

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03 Feb 2015, 16:10

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Notice we are considering non-negative numbers. We have to test 0 as well.

every positive integer power of 10 yields a reminder of 10.

statement 1: a prime is always a positive integer, thus reminder when 10^n is divided by 18 is 10; reminder when 8 is divided by 18 is 8. You can perform operations with reminders as long as you correct the excess (in case R>=divisor) R10+R8 = R18. Adjusting the excess you have R0, thus 10^prime+8 is a multiple of 18.

Sufficient.

statement 2: n is even, since we are dealing with non-negative numbers, zero is even. Assume n=0 and R=9

Not sufficient.

Answer A
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If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink]

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07 Feb 2015, 04:31

wastedyouth wrote:

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

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

The above justification in red is wrong, right? The point is that in case 2 "n" could be zero, as zero is an even number. In this case, 10^0 = 1 and 1+8=9, which is not divisible by 18 (it gives 1/2). In every other case we get a "yes". So, this statement is not sufficient.

In case 1 it is always divisible by 18, because any positive power of 10 (except for 1 of course) would only add zeroes at the end of 10, and 8 would be added to that. Only by testing n=2, 100+8=108 and 108/18=6 we answer the question as "yes".

Re: If n is an non-negative integer is 10^n+8 divisible by 18? [#permalink]

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01 Jul 2017, 10:17

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