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For example, \(\frac{17}{9}\) = 1 remainder 8 \(\frac{26}{9}\) = 2 remainder 8 \(\frac{35}{9}\) = 3 remainder 8

Therefore, n can be even or odd is a member of the set {17, 26, 35, ... }.

Step 2

Substitute the possible answers (1, 4, 9, 10, 17) to find a sum that is divisible by 18.

For example, 17 + 1 = 18, which is divisible by 18. Therefore the answer is 1 17 + 4 = 21, which is not divisible by 18 17 + 9 = 26, which is not divisible by 18 17 + 10 = 27, which is not divisible by 18 17 + 17 = 34, which is not divisible by 18

Re: Which sum is divisible by 18? [#permalink]
25 Apr 2013, 14:33

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stormbind wrote:

Grrr... Where did I go wrong?

Read carefully: When the even integer n is divided by 9, the remainder is 8. Which of the following, when added to n, gives a number that is divisible by 18? 17 is not even

Solution:

I use real numbers here I think it's easier

\(n=8\) reminder when divided by 9 is 8, to make it divisible by 18 it's easy to see that 10 is necessary \(8+10=18\) divisible by \(18\) D _________________

It is beyond a doubt that all our knowledge that begins with experience.

For example, \(\frac{17}{9}\) = 1 remainder 8 \(\frac{26}{9}\) = 2 remainder 8 \(\frac{35}{9}\) = 3 remainder 8

Therefore, n can be even or odd is a member of the set {17, 26, 35, ... }.

Step 2

Substitute the possible answers (1, 4, 9, 10, 17) to find a sum that is divisible by 18.

For example, 17 + 1 = 18, which is divisible by 18. Therefore the answer is 1 17 + 4 = 21, which is not divisible by 18 17 + 9 = 26, which is not divisible by 18 17 + 10 = 27, which is not divisible by 18 17 + 17 = 34, which is not divisible by 18

Grrr... Where did I go wrong?

Given that \(n=9q+8\) and n is even --> n could be 8, 26, 44, ...

What number from answer choices when added to 8 (for example) yields a multiple of 18? 10!

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