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M26-27

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M26-27  [#permalink]

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New post 16 Sep 2014, 01:26
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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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Re M26-27  [#permalink]

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New post 16 Sep 2014, 01:26
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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 \(3+8+9=20\).


Answer: B
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Re: M26-27  [#permalink]

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New post 08 Jun 2015, 21:58
wow I guessed this one but its a brilliant problem! Beautiful!
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Re M26-27  [#permalink]

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New post 05 Dec 2015, 02:18
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4
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Re: M26-27  [#permalink]

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New post 16 Sep 2016, 01:29
please explain..
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4
:( :( :roll:
please can anyone explain me this problem in easy way :cry:
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Re: M26-27  [#permalink]

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New post 16 Sep 2016, 02:04
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M26-27  [#permalink]

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New post 16 Sep 2016, 02:21
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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
:( :( :roll:
please can anyone explain me this problem in easy way :cry:


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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Re: M26-27  [#permalink]

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New post 16 Sep 2016, 02:43
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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
:( :( :roll:
please can anyone explain me this problem in easy way :cry:


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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M26-27  [#permalink]

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New post 16 Sep 2016, 03:06
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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
:( :( :roll:
please can anyone explain me this problem in easy way :cry:


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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New post 23 Mar 2017, 11:18
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.
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Re M26-27  [#permalink]

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New post 12 Aug 2017, 08:12
I think this is a high-quality question and I agree with explanation.
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Re: M26-27  [#permalink]

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New post 18 Oct 2017, 16:31
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
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Re: M26-27  [#permalink]

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New post 18 Oct 2017, 21:13
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M26-27  [#permalink]

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New post 24 Dec 2017, 04:48
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
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M26-27  [#permalink]

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New post 27 Aug 2018, 09:55
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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New post 28 Aug 2018, 10:00
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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Re: M26-27 &nbs [#permalink] 28 Aug 2018, 10:00
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