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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
M26-27  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 61% (01:32) correct 39% (02:11) wrong based on 263 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

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Math Expert V
Joined: 02 Sep 2009
Posts: 55271
Re M26-27  [#permalink]

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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 $$3+8+9=20$$.

Answer: B
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Current Student B
Joined: 02 Sep 2014
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Re: M26-27  [#permalink]

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wow I guessed this one but its a brilliant problem! Beautiful!
Intern  Joined: 14 Jun 2014
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Re M26-27  [#permalink]

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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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Re: M26-27  [#permalink]

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please explain..
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 V
Joined: 02 Sep 2009
Posts: 55271
Re: M26-27  [#permalink]

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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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M26-27  [#permalink]

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Karanagrawal wrote:
please explain..
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  B
Joined: 30 Oct 2012
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Location: India
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Re: M26-27  [#permalink]

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Karanagrawal wrote:
please explain..
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 V
Joined: 02 Aug 2009
Posts: 7686
M26-27  [#permalink]

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Karanagrawal wrote:
please explain..
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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Re: M26-27  [#permalink]

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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  B
Joined: 12 Dec 2016
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Re M26-27  [#permalink]

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I think this is a high-quality question and I agree with explanation.
Intern  B
Joined: 07 Aug 2016
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Location: Brazil
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Re: M26-27  [#permalink]

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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 V
Joined: 02 Sep 2009
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Re: M26-27  [#permalink]

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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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Re: M26-27  [#permalink]

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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
Retired Moderator G
Joined: 23 May 2018
Posts: 487
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M26-27  [#permalink]

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Hey Bunuel

If x/y leaves a remainder of 3 and y/z leaves a remainder of 8, how can I say that y is 4?

I get that 3/4 leaves a remainder of 3 which complies with the first part of the question, but (y=4)/z can not leave a remainder of 8.

So, if y/z leaves a remainder of 8, then y=8 and z=9 and with x/y leaving a remainder of 3, shouldn't x=8+3=11?

But the biggest problem is, the options do not have 8+9+11=28 as an option.
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Retired Moderator G
Joined: 23 May 2018
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Re: M26-27  [#permalink]

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I kept wondering about this question and I figured it out finally. Even with the denominator of 8, 3 could be the remainder if the numerator is 3. Thus x + y +z = 3 + 8 + 9 = 20

Therefore B: 20
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M26-27  [#permalink]

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Bunuel Can you please explain me how do we derive the minimum possible value of z?
I was able to understand how we get the minimum value of Y. But I could not understand your explanation for the minimum value of z to be 9.
Math Expert V
Joined: 02 Sep 2009
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Re: M26-27  [#permalink]

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ridhi9jain wrote:
Bunuel Can you please explain me how do we derive the minimum possible value of z?
I was able to understand how we get the minimum value of Y. But I could not understand your explanation for the minimum value of z to be 9.

y is divided by z the remainder is 8:

$$y = qz + 8$$, where $$q$$ is a quotient, an integer $$\ge 0$$. Which means that the least value of $$y$$ is when $$q=0$$, in that case $$y=8$$. This basically means that $$y$$ is less than $$z$$. For example 8 divided by 9 yields the remainder of 8.
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Joined: 12 Sep 2016
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Re: M26-27  [#permalink]

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Bunuel wrote:
ridhi9jain wrote:
Bunuel Can you please explain me how do we derive the minimum possible value of z?
I was able to understand how we get the minimum value of Y. But I could not understand your explanation for the minimum value of z to be 9.

y is divided by z the remainder is 8:

$$y = qz + 8$$, where $$q$$ is a quotient, an integer $$\ge 0$$. Which means that the least value of $$y$$ is when $$q=0$$, in that case $$y=8$$. This basically means that $$y$$ is less than $$z$$. For example 8 divided by 9 yields the remainder of 8.

Bunuel
Thank you. I get it now. Re: M26-27   [#permalink] 14 Oct 2018, 22:41
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