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When positive integer x is divided by 5, the remainder is 2.

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Joined: 15 Jun 2014
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When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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Updated on: 15 Jun 2014, 04:40
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When positive integer x is divided by 5, the remainder is 2. When positive integer y is divided by 4, the remainder is 1. Which of the following cannot be the sum x + y?

A. 12
B. 13
C. 14
D. 16
E. 21

Originally posted by Critique on 15 Jun 2014, 04:20.
Last edited by Bunuel on 15 Jun 2014, 04:40, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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15 Jun 2014, 04:41
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Critique wrote:
When positive integer x is divided by 5, the remainder is 2. When positive integer y is divided by 4, the remainder is 1. Which of the following cannot be the sum x + y?

A. 12
B. 13
C. 14
D. 16
E. 21

When positive integer x is divided by 5, the remainder is 2: x could be 2, 7, 12, 17, ...
When positive integer y is divided by 4, the remainder is 1: y could be 1, 5, 9, 13, 17, ...

If x is 7 and y is 5, then x + y = 12. Eliminate A.
If x is 12 and y is 1, then x + y = 13. Eliminate B.

If x is 7 and y is 9, then x + y = 16. Eliminate D.
If x is 12 and y is 9, then x + y = 21. Eliminate E.

By process of elimination the correct answer must be C.

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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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16 Jun 2014, 23:13
2

Please refer screenshot below for possible combinations
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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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20 Jun 2016, 20:54
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One way to solve this problem:-
since x=5a+2, where a is an integer.
and y=4b+1 where b is an integer
x+y=5a+4b+3..

now test for integer solutions of a and b. We need to subtract 3 from answer choice and then test whether the resultant can be expressed 5a+4b.
choice a) 12. Subtract 3 we get 9 . 9 = 4a+5b. take a=1, b=1.
choice b) 13. Subtract 3 we get 10. 10=4a+5b. take a=0 and b=2.
choice c) 14. Subtract 3 we get 11. 11 = 4a+5b. if we take a=0, 1,2, can we have b which is positive integer. let us check
a=0 ;11=5b , a = 1 ; 7 = 5b, a = 2 ; 3= 5b, a = 3 , now in this case -1 = 5b.. now from here on we can get negative values on the L.H.S. So we cannot express 11 = 4a+5b where a and b are integers. This is our answer.
choice d) 16 . Subtract 3 we get 13 . 13= 4a+5b. take a =2 and b = 1.
choice e) 21. subtract 3 we get 18. 18= 4a+5b. take a=2, b=2. Done!
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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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09 Nov 2016, 15:23
y=5r+2 (2,7,12,17,....)

x=4r+1 (1,5,9,13,....)

C is the only number you cannot create using any combination of x and y
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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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06 Sep 2017, 09:13
Bunuel how do you know the possible values of X? How does 2/5 give a remainder of 2?
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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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06 Sep 2017, 09:23
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bkastan wrote:
Bunuel how do you know the possible values of X? How does 2/5 give a remainder of 2?

1. If $$x$$ and $$y$$ are positive integers, there exist unique integers $$q$$ and $$r$$, called the quotient and remainder, respectively, such that $$y =divisor*quotient+remainder= xq + r$$ and $$0\leq{r}<x$$.

So, positive integer x is divided by 5, the remainder is 2 means that x = 5q + 2, thus x could be 2, 7, 12, ...

2. Let me ask you a question: how many leftover apples would you have if you had 2 apples and wanted to distribute in 5 baskets evenly? Each basket would get 0 apples and 2 apples would be leftover (remainder).

When a divisor is more than dividend, then the remainder equals to the dividend, for example:
3 divided by 4 yields the reminder of 3: $$3=4*0+3$$;
9 divided by 14 yields the reminder of 9: $$9=14*0+9$$;
1 divided by 9 yields the reminder of 1: $$1=9*0+1$$.

For more on this check:

5. Divisibility/Multiples/Factors

6. Remainders

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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06 Sep 2017, 09:40
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Bunuel thank you!!!

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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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06 Sep 2017, 22:31
Critique wrote:
When positive integer x is divided by 5, the remainder is 2. When positive integer y is divided by 4, the remainder is 1. Which of the following cannot be the sum x + y?

A. 12
B. 13
C. 14
D. 16
E. 21

x=5q+2
y=4p+1
x+y=5q+4p+3
12: q=1;p=1
13: q=2;p=0
16: q=1;p=2
21: q=2;p=2
14: no
C
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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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07 Sep 2017, 03:40
Note that 1 + 2 = 3. Deduct 3 from every proposed answer and try to construct the resulting figure with 5s and 4s.

A. 12 -3 = 9 = 5+4
B. 13 -3 =10= 5+ 5
C. 14 -3 = 11= correct answer
D. 16 -3 = 13 = 5+4+4
E. 21 -3 = 18= 4+4 +5+5
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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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11 Sep 2017, 11:35
Critique wrote:
When positive integer x is divided by 5, the remainder is 2. When positive integer y is divided by 4, the remainder is 1. Which of the following cannot be the sum x + y?

A. 12
B. 13
C. 14
D. 16
E. 21

We can create the following equations:

x = 5Q + 2

Some values for x are 2, 7, 12, 17.

y = 4Z + 1

Some values for y are 1, 5, 9, 13, 17.

Let’s go through each answer choice:

A) 12

Since 7 + 5 = 12, answer choice A is not correct.

B) 13

Since 12 + 1 = 13, answer choice B is not correct.

C) 14

Since we cannot get x + y to sum to 14, answer choice C is correct.

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Re: When positive integer x is divided by 5, the remainder is 2.  [#permalink]

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11 Feb 2019, 20:25
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Doubt regarding 0 being a postivei number  [#permalink]

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07 May 2019, 10:13
While solving the Number Properties Manhattan book, I came across this question.

When positive integer x is divided by 5, the remainder is 2. When positive integer y is divided by 4, the remainder is 1. Which of the following values CANNOT be the sum x+y?
A 12
B 13
C 14
D 16
E 21

Now the question is fairly straightforward. My doubt is regarding how come we are eliminating answer choice B i.e. 13. It is a possible value of x + y + 3 (remainder) as it can be 5 x 2 + 4 x 0 + 3, but the question has defined x and y to be positive integers only i.e. whole numbers greater than 0.

Hope by doubt is clear - Can someone explain to me how they eliminated Answer Choice B.
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Joined: 07 Jan 2019
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Doubt regarding 0 being a postivei number  [#permalink]

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07 May 2019, 10:42
naturalthing wrote:
While solving the Number Properties Manhattan book, I came across this question.

When positive integer x is divided by 5, the remainder is 2. When positive integer y is divided by 4, the remainder is 1. Which of the following values CANNOT be the sum x+y?
A 12
B 13
C 14
D 16
E 21

Now the question is fairly straightforward. My doubt is regarding how come we are eliminating answer choice B i.e. 13. It is a possible value of x + y + 3 (remainder) as it can be 5 x 2 + 4 x 0 + 3, but the question has defined x and y to be positive integers only i.e. whole numbers greater than 0.

Hope by doubt is clear - Can someone explain to me how they eliminated Answer Choice B.

Hi,

When you consider possible cases for x you can assume values to be 2,7,12,17 etc and when you consider the same case for y you can assume values to be 1,5,9,13,17 etc. Now when you take x=12 and y=1 you get the value as 13.

Hope this helps
Doubt regarding 0 being a postivei number   [#permalink] 07 May 2019, 10:42
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