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# M30-17

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Intern
Joined: 19 Apr 2018
Posts: 2

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26 Dec 2018, 03:07
Can we take this approach? Bunuel
The equations are
x=3y+z [ Constraints: y>z and z>=0] As Remainder is greater than Quotient.
z=ay+2 [ Constraints: y>2]
Thus,
x=3y+ay+2
x=y(a+3) + 2
x=3y+2 or x =4y+2 or x=5y+2 [a can take values 0,1,2,3...]

Looking at the options:
I. 5 Can only fit if y=1, which breaks constraint of y>2, so NO
II. 8 Can only fit if y=2, which breaks constraint of y>2, so NO
III. 32 Fits the equation x=3*10+2 or x=5*6+2 So YES
Intern
Joined: 24 Dec 2011
Posts: 28
Location: India
GPA: 4
WE: General Management (Health Care)

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18 Jan 2019, 09:48
very nice question..
thank you for posting this. decent explanation.

Got a li'l doubt. same line of thinking as i came to y>2 bt substituted z in the first equation and tried to solve.
I got the equation x=3y+z
substituting z=y+2
comes to x=4y+2. (and y>2)
I'm not getting 32 either
Math Expert
Joined: 02 Sep 2009
Posts: 53063

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19 Jan 2019, 00:19
1
venkivety wrote:
very nice question..
thank you for posting this. decent explanation.

Got a li'l doubt. same line of thinking as i came to y>2 bt substituted z in the first equation and tried to solve.
I got the equation x=3y+z
substituting z=y+2
comes to x=4y+2. (and y>2)
I'm not getting 32 either

When z is divided by y, the remainder is 2 does not mean that z=y+2. It means that z =yq + 2.

Check the solution:
When $$z$$ is divided by $$y$$, the remainder is 2: when divisor ($$y$$ in our case) is more than dividend ($$z$$ in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2 < y$$.

So, we have that $$x=3y+2$$ and $$2 < y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.
_________________
Intern
Joined: 24 Dec 2011
Posts: 28
Location: India
GPA: 4
WE: General Management (Health Care)

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19 Jan 2019, 10:20
Bunuel wrote:
venkivety wrote:
very nice question..
thank you for posting this. decent explanation.

Got a li'l doubt. same line of thinking as i came to y>2 bt substituted z in the first equation and tried to solve.
I got the equation x=3y+z
substituting z=y+2
comes to x=4y+2. (and y>2)
I'm not getting 32 either

When z is divided by y, the remainder is 2 does not mean that z=y+2. It means that z =yq + 2.

Check the solution:
When $$z$$ is divided by $$y$$, the remainder is 2: when divisor ($$y$$ in our case) is more than dividend ($$z$$ in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, $$z=2$$ and $$2 < y$$.

So, we have that $$x=3y+2$$ and $$2 < y$$. This implies that the least value of $$x$$ is $$x=3*3+2=11$$: $$x$$ cannot be 5 or 8.

Could $$x$$ be 32? Yes. If $$y=10$$, then $$x=3*10+2=32$$.

thanq very much 4 the reply..
cleared..
M30-17   [#permalink] 19 Jan 2019, 10:20

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# M30-17

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