remainder problem

N = 9a + 8
As N is even 9a has to be even or a has to be even. For a = 2, N = 26
Now check answer choices.

We know that and q is even so. q/2 = integer

n = q9 + (8+x) ;

n/18 = 9*q / 18 + (8+x)/18;
= q/2 + (8+x)/18;
= (integer) + Integer

when will (8+x) /18 be an integer i.e. for what value of x ?

Ans : D i.e. 10

fangtray wrote:When the even integer n is dividd by 9, the remainder is 8. Which of the following when added to n, gives a sum that is divisible by 18?

a. 1
b. 4
c. 5
d. 10
e. 17

Hi!

Picking numbers is a great approach to solving many number property questions.

Here, our constraints are that n/9 must have a remainder of 8 and n must be even. What's the smallest number that satisfies these constraints? n=8 (8/9 has a quotient of 0 and a remainder of 8).

Now let's find the answer that, when we add it to 8, gives us a multiple of 18.

a) 1... 9 is not a multiple of 18... out!
b) 4... 12 is not a multiple of 18... out!
c) 5... 13 is not a multiple of 18... out!
d) 10... 18 IS a multiple of 18... ding ding ding we have a winner!

Of course, we could have also just applied a bit of logic. Since the answers are all pretty small, we can just ask "what number when added to 8 gives us 18, the first multiple of 18?" - and of course (d) would have jumped right out as correct.

n=2k

2k=9x+8
if x=1 then n=9+8=17 (out, since n is even)
if x=2 then 9*2+8=26 (bingo! n is even)

now just pay attention that 26 is 10 away from 36, which is divisible by 18

Here is how I would tackle this one.

Here's a general approach to any problem like this.

Consider a few integers with a remainder of 8 when divided by 9:

9*1 + 8 = 17
9*2 + 8 = 26
9*3 + 8 = 35
9*4 + 8 = 44
9*5 + 8 = 53
9*6 + 8 = 62
etc.

Notice how all the even ones are of the form 18*(something) + 8?

That means n = 18k + 8, where k is some integer. (For instance, if n = 26, k = 1, and if n = 44, k = 2.)

For this to divide by 18, we need to get rid of the remainder (which is currently 8). So if we add any number with a remainder of 10 when divided by 8, we'll get ...

(18k + 8) + (18m + 10) = 18k + 18m + (10+8) = 18k + 18m + 18, which must divide by 18.

For instance, suppose we have 26 + 28. This gives us 18 + 8 + 18 + 10, or 18 + 18 + 18. Success!

Similarly, ANY answer of the form 18*something + 10 works. Since 10 = 18*0 + 10, the answer must be D.

fangtray wrote:When the even integer n is dividd by 9, the remainder is 8. Which of the following when added to n, gives a sum that is divisible by 18?

a. 1
b. 4
c. 5
d. 10
e. 17

When the even integer n is divided by 9, the remainder is 8.
In other words, n is a MULTIPLE OF 9 plus 8:
n = 9a + 8, where a≥0.

Options for n, if n can be even or odd:
n = 8, 17, 26, 35...

Since n must be even, only the values in red are viable:
n = 8, 26...

When the correct answer choice is added to 8, the result must be a multiple of 18.
Only D works:
10 + 8 = 18.

The correct answer is D.

fangtray wrote:When the even integer n is dividd by 9, the remainder is 8. Which of the following when added to n, gives a sum that is divisible by 18?

a. 1
b. 4
c. 5
d. 10
e. 17

Could you please comment on my solution?

n = 9m+8 => (n+1) is divisible by 9

=> (n+1 + 9k) is divisible by 9

n is divisible by 18 when (n+1+9k) is an even number (or divisible by 2)

n is even => n+1 is odd

so 9k must be an odd number (odd+odd=even)

=> k must be odd

We have a lot of k but within this question's options, we just need to care about 1 values of k:

k=1 we have 1+9k = 10.


So 10 is the answer.