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# M26-27

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Math Expert
Joined: 02 Sep 2009
Posts: 43867

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16 Sep 2014, 00:26
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Difficulty:

65% (hard)

Question Stats:

63% (01:19) correct 37% (01:45) wrong based on 51 sessions

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$$x$$, $$y$$ and $$z$$ are positive integers such that when $$x$$ is divided by $$y$$ the remainder is 3, and when $$y$$ is divided by $$z$$, the remainder is 8. What is the smallest possible value of $$x+y+z$$?

A. 12
B. 20
C. 24
D. 29
E. 33
[Reveal] Spoiler: OA

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Math Expert
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16 Sep 2014, 00:26
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Official Solution:

$$x$$, $$y$$ and $$z$$ are positive integers such that when $$x$$ is divided by $$y$$ the remainder is 3, and when $$y$$ is divided by $$z$$, the remainder is 8. What is the smallest possible value of $$x+y+z$$?

A. 12
B. 20
C. 24
D. 29
E. 33

Given $$x=qy+3$$, where $$q$$ is a quotient, an integer $$\ge 0$$. Which means that the least value of $$x$$ is when $$q=0$$, in that case $$x=3$$. This basically means that $$x$$ is less than $$y$$. For example 3 divided by 4 yields the remainder of 3.

According to the above:

$$x$$ is divided by $$y$$ the remainder is 3 means that the minimum value of $$x$$ is 3;

$$y$$ is divided by $$z$$ the remainder is 8 means that the minimum value of $$y$$ is 8 and the minimum value of $$z$$ is one more than 8, so 9 (8 divided by 9 yields the remainder of 8);

So, the smallest possible value of $$x+y+z$$ is $$3+8+9=20$$.

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Joined: 02 Sep 2014
Posts: 89
Location: United States
Schools: Haas EWMBA '20
GMAT 1: 770 Q50 V44
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08 Jun 2015, 20:58
wow I guessed this one but its a brilliant problem! Beautiful!
Intern
Joined: 14 Jun 2014
Posts: 7

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05 Dec 2015, 01:18
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4
Intern
Joined: 23 Mar 2016
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16 Sep 2016, 00:29
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4

please can anyone explain me this problem in easy way
Math Expert
Joined: 02 Sep 2009
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16 Sep 2016, 01:04
deospaima wrote:
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4

Not sure I understand your point. 2 divided by 4 gives the remainder of 2.
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16 Sep 2016, 01:21
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Karanagrawal wrote:
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4

please can anyone explain me this problem in easy way

See x has to be at least 3 for a remainder of 3 when x is divided by y.

then x = 3.

y has to be greater than 3 for x/y to get a remainder of 3,

and y has to be at least 8 for y/z to get a remainder of 8.

then y = 8.

z has to be greater than 8 for y/z to have a remainder of 8. Lowest possible integer greater than 8 is 9.

So going with z = 9.

x + y + z = 3 + 8 + 9 = 20
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Manager
Joined: 30 Oct 2012
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16 Sep 2016, 01:43
1
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Karanagrawal wrote:
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4

please can anyone explain me this problem in easy way

Remember this rule:
If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
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Math Expert
Joined: 02 Aug 2009
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16 Sep 2016, 02:06
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Expert's post
Karanagrawal wrote:
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4

please can anyone explain me this problem in easy way

Hi,
X as 2 and y as 4 is clearly incorrect...

Now for this Q, my advice would be talk of larger number first...

1) If y is divided by z, remainder is 8..
This tells us that least value of y is 8 and the remainder will be 8, ONLY if z is greater than 8, so take the next least possible value..
And it is 8+1=9...

2) now take the second part..
If x is divided by y, remainder is 3..
Since least value of y is 8, x will be 3 or 11 or 19 and so on to leave remainder of 3..
But we have to take least value and it is 3..

So x can be 3, y can be 8 and z can be 9..
Ans x+y+z=3+8+9=20
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Manager
Joined: 10 Feb 2017
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GMAT 1: 680 Q50 V30
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23 Mar 2017, 10:18
deospaima wrote:
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4

as stated minimum value of x=3 and y>3 it means for minimum value of X y>x

see eg: x=3 and y=4 this means when x is divided by y remainder is 3 and y>remainder i.e. 3 but in this case y if >3 will also be greater than x ultimately.
Intern
Joined: 12 Dec 2016
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12 Aug 2017, 07:12
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 07 Aug 2016
Posts: 17
Location: Brazil
Concentration: Economics, Marketing
WE: Project Management (Energy and Utilities)

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18 Oct 2017, 15:31
Would this method be wrong?

x/y -> remainder = 3
therefore y = x+3

y/z -> remainder of 8
therefore z = x+ 8

x+ 8 + x + x+ 3 = 11 + 3x - the least multiple of 3 that results in any answer is 3

therefore 3*3 + 11 = 20
Math Expert
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18 Oct 2017, 20:13
zzarur wrote:
Would this method be wrong?

x/y -> remainder = 3
therefore y = x+3

y/z -> remainder of 8
therefore z = x+ 8

x+ 8 + x + x+ 3 = 11 + 3x - the least multiple of 3 that results in any answer is 3

therefore 3*3 + 11 = 20

How does x divided by y gives the remainder of 3 translates to y = x + 3?
How does y divided by z gives the remainder of 8 translates to z = x + 8?

Check solution here: https://gmatclub.com/forum/m26-184466.html#p1415594
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24 Dec 2017, 03:48
Given:
x=y(integer) + 3

y=z(integer) + 8

Therefore, x=z(integer) +8 + 3

x + y + z = [z(integer) +8 + 3] + [z(integer) + 8] + z = z(0) + 8 + 3+ z(0) + 8 + z = 19 + z

Minimum possible value of z = 1

thus, x + y + z = 19 + 1 = 20
Re: M26-27   [#permalink] 24 Dec 2017, 03:48
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# M26-27

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