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When positive integer N is divided by 18, the remainder is x
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05 Jun 2018, 07:54
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When positive integer N is divided by 18, the remainder is x. When N is divided by 6, the remainder is y. Which of the following are possible values of x and y?
i) x = 9 and y = 3 ii) x = 16 and y = 2 iii) x = 13 and y = 7
A) i only B) i and ii only C) i and iii only D) ii and iii only E) i,ii and iii
Re: When positive integer N is divided by 18, the remainder is x
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07 Jun 2018, 06:47
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GreenlightTestPrep wrote:
When positive integer N is divided by 18, the remainder is x. When N is divided by 6, the remainder is y. Which of the following are possible values of x and y?
i) x = 9 and y = 3 ii) x = 16 and y = 2 iii) x = 13 and y = 7
A) i only B) i and ii only C) i and iii only D) ii and iii only E) i,ii and iii
Let's examine each statement separately...
i) x = 9 and y = 3 Let's come up with a value of N that satisfies this condition. How about N = 9? 9 divided by 18 = 0 with remainder 9 (i.e., x = 9) ...and 9 divided by 6 = 1 with remainder 3 (i.e., y = 3) Perfect, statement i is TRUE Check the answer choices.....ELIMINATE D
ii) x = 16 and y = 2 Can you come up with a value of N that satisfies this condition? How about N = 16? 16 divided by 18 = 0 with remainder 16 (i.e., x = 16, which WORKS) However, 16 divided by 6 = 2 with remainder 4 (i.e., y = 4. NO GOOD)
How about N = 34? 34 divided by 18 = 1 with remainder 16 (i.e., x = 16, which WORKS) However, 34 divided by 6 = 5 with remainder 4 (i.e., y = 4. NO GOOD)
We can keep testing N-values until we convince ourselves that there are no values of N that makes those values (x = 16 and y = 2) possible. So, statement ii is FALSE Check the answer choices.....ELIMINATE B and E
HOWEVER, if you need more convincing that statement ii is FALSE, we can make the following observations: When positive integer N is divided by 18, the remainder is x: so, we can say that N = 18k + x for some integer k When positive integer N is divided by 6, the remainder is y: so, we can say that N = 6j + y for some integer j We can combine the two equations to get: 18k + x = 6j + y Isolate x to get: x = y + 6j - 18k Factor right side to get: x = y + 6(j - 3k) Rewrite as: x = y + some multiple of 6 This means we'll never have the case where x = 16 and y = 2, because 16 CANNOT be written as 2 + some multiple of 6 So, statement ii is FALSE
iii) x = 13 and y = 7 We must be careful with this one. While it is true that 13 CAN be written as 7 + some multiple of 6, we must also consider the following property of remainders: When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0
Based on the above property, when we divide N by 6, the remainder can be 5, 4, 3, 2, 1, or 0 So, the remainder CANNOT be 7 In other words, y CANNOT equal 7 So, statement iii is FALSE ELIMINATE C
Re: When positive integer N is divided by 18, the remainder is x
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15 Sep 2019, 12:51
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I don't feel this approach is right, but I got the right answer. Brent, please let me know if this approach has any validity.
So, I said:
18Q + x = N = 6Q + y
18Q + x = 6Q + y
Then I plugged in numbers for x and y, for example:
18Q + 9 = 6Q + 3
Then I imagined Q was 1, so we get
27 = 9
It's plain that the two sides aren't equal, but they are multiples of one another.
I used the same thing to look at the other solutions, and saw they weren't multiples of one another.
I read your solution again, and I think I may have made a mistake in writing Q for both sides instead of Q sub 1 and Q sub 2. We are told N is the same, but we're not told Q is the same.
So, I guess I could say
18Q_1 + 9 = 6Q_2 + 3
Given Q_1 and Q_2 could be different values.
The other solutions won't be multiples of each other, and therefore can't be equal whatever Q_1 and Q_2 are. For example, if x = 13 and y = 17, we get
18Q_1 + 13 = 6Q_2 + 7
if both Qs are 1, we get
31 = 13, which are clearly not multiples of each other, and therefore no matter what the Q's are for both sides, the two sides will never be equal.
I found the same to be true for the other answer choices. None of them resulted in multiples.
Was on the right path? Or was getting the right answer just a coincidence?
PS Sorry, if my line of reasoning was confusing. It felt confusing to me writing it.
Re: When positive integer N is divided by 18, the remainder is x
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16 Sep 2019, 06:54
1
arc601 wrote:
I don't feel this approach is right, but I got the right answer. Brent, please let me know if this approach has any validity.
So, I said:
18Q + x = N = 6Q + y
18Q + x = 6Q + y
Then I plugged in numbers for x and y, for example:
18Q + 9 = 6Q + 3
Then I imagined Q was 1, so we get
27 = 9
It's plain that the two sides aren't equal, but they are multiples of one another.
I used the same thing to look at the other solutions, and saw they weren't multiples of one another.
I read your solution again, and I think I may have made a mistake in writing Q for both sides instead of Q sub 1 and Q sub 2. We are told N is the same, but we're not told Q is the same.
So, I guess I could say
18Q_1 + 9 = 6Q_2 + 3
Given Q_1 and Q_2 could be different values.
The other solutions won't be multiples of each other, and therefore can't be equal whatever Q_1 and Q_2 are. For example, if x = 13 and y = 17, we get
18Q_1 + 13 = 6Q_2 + 7
if both Qs are 1, we get
31 = 13, which are clearly not multiples of each other, and therefore no matter what the Q's are for both sides, the two sides will never be equal.
I found the same to be true for the other answer choices. None of them resulted in multiples.
Was on the right path? Or was getting the right answer just a coincidence?
PS Sorry, if my line of reasoning was confusing. It felt confusing to me writing it.
It's fine to write 18Q_1 + x = 6Q_2 + y But it might be less confusing to write: 18k + x = 6j + y (where j and k are integers)
Now check the answer choices.. i) x = 9 and y = 3 We get: 18k + 9 = 6j + 3 (it this possible given that k and j are positive integers?) Rearrange to get: 6j - 18k = 9 - 3 Simplify to get: 6(j - 3k) = 6 Divide both sides by 6 to get: j - 3k = 1 We can see that there are many ways this can be possible. For example, j = 4 and k = 1 is one solution.
Keep going with other statements....
_________________
Re: When positive integer N is divided by 18, the remainder is x
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24 Mar 2020, 04:09
1
Please tell me if this is right N = 18k+ x N = 6j + y y-x = 18k - 6j y-x = 6(3k-j) This means y-x is a multiple of 6 So A) x=9 y=3 y-x=-6 which is divisible by 6 B) x=16 y=2 y-x=-14 which is not divisible by 6 so eliminate it C) x=13 y=7 since y<6 eliminate it Only A is left.
Re: When positive integer N is divided by 18, the remainder is x
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07 Nov 2021, 03:40
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