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
remainder problem
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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
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
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Hi!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
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.
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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
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
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Here is how I would tackle this one.
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Given : When the even integer n is divided by 9, the remainder is 8fangtray wrote:When the even integer n is divided 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
i.e. n could be 9x2+8 or 9x4+8 and so on...
i.e. n could be 26 or 44 and so on...
Question : Which of the following when added to n, gives a sum that is divisible by 18
Result will be divisible by 18 only When it's an Even integer therefore n must be added with an even integer only ELIMINATING OPTIONS A, C, and E
Checking Option B
26+4 = 18 NOT Divisible by 18 Ruled out
Therefore Option D must be correct
Checking option D
26+10 = 36 Divisible by 18
44+10 = 54 Divisible by 18
Answer: Option D
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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.
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.
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When the even integer n is divided by 9, the remainder is 8.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
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.
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There's a nice rule that say, "If N divided by D equals Q with remainder R, then N = DQ + R"fangtray wrote:When the even integer n is divided 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
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
When the even integer n is divided by 9, the remainder is 8
In other words, when n is divided by 9 we get some integer (say k) with remainder 8
So, applying the above rule, we can write n = 9k + 8 (where k is an integer)
IMPORTANT: We're told that n is EVEN.
In other words, 9k + 8 is even
Since 8 is even, we can conclude that 9k is even, which means k is EVEN.
If k is even, then 9k = 9(2 times some integer) = (18)(some integer)
So, we can conclude that 9k is divisible by 18
n = 9k + 8
We can see that, if we add 10 to n, we get 9k + 8 + 10, which simplifies to 9k + 18
Since 9k is divisible by 18 AND 18 is divisible by 18, we know that 9k + 18 is divisible by 18
So, if we add 10 to n, the sum is divisible by 18
Answer: D
Cheers,
Brent
Could you please comment on my solution?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
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.