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M05-28

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Re: M05-28  [#permalink]

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New post 17 Aug 2018, 01:17
bonion wrote:
Bunuel wrote:
Official Solution:


We have system of equations with three distinct equations and three unknowns, so we can solve it. As \(x\) is squared we'll get two values for it and also two values for \(y\) and \(z\): two triplets. Hence we'll get two values for \(x^3 + y^2 + z\), one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

\(x^2 = y + 5\);

\(z = 2x\);

\(y = z - 2\), or which is the same \(y=(2x)-2\)

Solve for \(x\): \(x^2=(2x-2) + 5\). Rearrange: \(x^2-2x-3=0\). Either \(x=3\) or \(x=-1\)

So, the first triplet would be: \(x=3\), \(y=4\), \(z=6\). Which gives: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7;

The second triplet would be: \(x=-1\), \(y=-4\), \(z=-2\). Which gives: \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.

(2) \(y = 4\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.


Answer: D

Hello Bunuel I am not sure what i missed, but in statement 2, y = 4, then x^2 = 9, then x = 3 or -3,
If x = 3 the equation results 49 which is divisible by 7
if x = -3 the equation results -5 which is not divisible by 7
So, i think statement 2 is not sufficient.
Can you please clarify my answer.


x = -3 is not a solution.

y = 4, then z = 6 (from y = z - 2) and x = 3 (from z = 2x).
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M05-28  [#permalink]

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New post 24 Oct 2018, 23:23
HI @Bunuel/anyone.


I interpreted the given equation x^3+y^2+z divisible by 7 as x^3+(x^2-5)^2+2x divisible by 7
x^2=y+5 so y=x^2-5 ,z=2x. i have substituted these values in the above equation.
so now according to stmnt 1 x>0,
if x=1 value =19 not divisible by 7
if x=2 value =13 not divisible by 7
but if x=3 value =49 divisible by 7

so i have ignored a and marked b as answer
can anyone explain where am i wrong.
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New post 24 Oct 2018, 23:44
MaheshDuggirala wrote:
HI @Bunuel/anyone.


I interpreted the given equation x^3+y^2+z divisible by 7 as x^3+(x^2-5)^2+2x divisible by 7
x^2=y+5 so y=x^2-5 ,z=2x. i have substituted these values in the above equation.
so now according to stmnt 1 x>0,
if x=1 value =19 not divisible by 7
if x=2 value =13 not divisible by 7
but if x=3 value =49 divisible by 7

so i have ignored a and marked b as answer
can anyone explain where am i wrong.


If you read the solution carefully, you'll see that the system of equation we have (\(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\)) has the following roots: \(x=3\), \(y=4\), \(z=6\) OR \(x=-1\), \(y=-4\), \(z=-2\). No other values of x, y and z satisfy these three equations (so x can be neither 1 nor 2). If x = 3, then \(x^3+y^2+z\) IS divisible by 7 but if x = -1 then it's NOT.

(1) says that x > 0, so x = 3 and \(x^3+y^2+z\) IS divisible by 7.
(2) says that y = 4, so x = 3 and \(x^3+y^2+z\) IS divisible by 7.

Hope it's clear.
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New post 25 Oct 2018, 21:22
Thanks Bunuel for the great explanation.

So u meant to say that for any given equation we should try to find the variable values and then interpret the equation.
If we don't do so we gonna miss the limitations on those variable values.
is my understanding correct?.
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M05-28  [#permalink]

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New post 16 Nov 2018, 03:05
bonion wrote:
Bunuel wrote:
Official Solution:


We have system of equations with three distinct equations and three unknowns, so we can solve it. As \(x\) is squared we'll get two values for it and also two values for \(y\) and \(z\): two triplets. Hence we'll get two values for \(x^3 + y^2 + z\), one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.

Given:

\(x^2 = y + 5\);

\(z = 2x\);

\(y = z - 2\), or which is the same \(y=(2x)-2\)

Solve for \(x\): \(x^2=(2x-2) + 5\). Rearrange: \(x^2-2x-3=0\). Either \(x=3\) or \(x=-1\)

So, the first triplet would be: \(x=3\), \(y=4\), \(z=6\). Which gives: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7;

The second triplet would be: \(x=-1\), \(y=-4\), \(z=-2\). Which gives: \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.

(2) \(y = 4\). We deal with the first triplet: \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. Sufficient.


Answer: D

Hello Bunuel I am not sure what i missed, but in statement 2, y = 4, then x^2 = 9, then x = 3 or -3,
If x = 3 the equation results 49 which is divisible by 7
if x = -3 the equation results -5 which is not divisible by 7
So, i think statement 2 is not sufficient.
Can you please clarify my answer.


Hey bonion

I am not sure if u still have the same doubt. But here is the simple logic.

If we take x = -3 then does it satisfy ALL the other equations?
That is IF \(x = -3\) ,
Plug it in here \(z = 2x\) ----- We get \(z = -6\)
Next plug it here \(y = z - 2\) ----- We get \(y = -8\)

Now most importantly check \(y = -8\) in the first equation
Plug it here \(x^2 = y + 5\)
\(x^2 = -8 + 5\)
\(x^2 = 3\)

Do u see now that we cannot take x = -3? Because it does not satisfy all the equations.

Hope it helps! :)
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Re: M05-28  [#permalink]

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New post 29 Jan 2019, 04:34
Bunuel

I tried to plug in some numbers as follows:
X=3, Y=4 and Z=6.....Solving the equation the sum equals 49 which is divisible by 7
X=5, Y=8 and Z=10.....Solving the equation the sum equals 199 which is not divisible by 7

How is the correct answer choice D in that case?
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Re: M05-28  [#permalink]

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New post 29 Jan 2019, 04:42
Megha1119 wrote:
Bunuel

I tried to plug in some numbers as follows:
X=3, Y=4 and Z=6.....Solving the equation the sum equals 49 which is divisible by 7
X=5, Y=8 and Z=10.....Solving the equation the sum equals 199 which is not divisible by 7

How is the correct answer choice D in that case?


First of all X=5, Y=8 and Z=10 does not satisfy \(x^2 = y + 5\).

Also, (2) says that \(y = 4\)
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New post 09 Feb 2019, 03:39
Hi,

I don't understand why statement A alone is sufficient. I simply tried different options for X>0 and found Y and Z by substitution. Why must I solve for X before and thus restrict the field to -1 and 3?

Many thanks in advance!
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Re: M05-28  [#permalink]

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New post 09 Feb 2019, 04:02
Bunuel wrote:
If \(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\), is \(x^3 + y^2 + z\) divisible by 7?


(1) \(x \gt 0\)

(2) \(y = 4\)


The whole point was to utilize the given

\(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\) when utilized, will give x^2 -2x-3=0

Which on solving will give 2 solutions for x i.e. -1 and 3

which when put back in y 2x -2 will give y = -4 or y = 4 and give z = -2 or z =6

from 1, x > 0

one case will work here
y = 4 & z = 6 , x = 3

\(x^3 + y^2 + z\) divisible by 7
Sufficient

from 2, y =4
one case will work here
y = 4 & z = 6 , x = 3

\(x^3 + y^2 + z\) divisible by 7
Sufficient

D
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New post 09 Feb 2019, 04:07
pero88 wrote:
Hi,

I don't understand why statement A alone is sufficient. I simply tried different options for X>0 and found Y and Z by substitution. Why must I solve for X before and thus restrict the field to -1 and 3?

Many thanks in advance!


Hey pero88

If you were able to solve without looking for any values for x in A, that is really great

But since x is given as x^2, we can directly say that it will have 2 solutions, which will satisfy that particular expression.

Now when you know this, its difficult to guess and find the solutions rather than just utilize what is given, and try utilizing it in our benefit

x^2 in terms of y+5, wont be used in any way, but if we go ahead in the question, we see that we can substitute for y and find an expression in only one unknown value.
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Re M05-28  [#permalink]

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New post 20 Feb 2019, 19:07
I think this is a high-quality question and I agree with explanation.
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Re: M05-28  [#permalink]

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New post 02 Jul 2019, 16:26
This is such a sneaky, but good question!

You can solve for all variables based on the stem, there are going to be two conditions: x>0 or x<0 : that would alter the solutions as follows:

x=3 or x=-1
if x =3
y = 4
z = 6

if x=-1
y=-4
z=-2


(1) X> 0 therefore we can derive all other values to learn that the numerator = 49 which is divis by 7
sufficient

(2) y= 4
this can only happen when x>0. Thus x=3 and z = 6 -->sufficient
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Re: M05-28   [#permalink] 02 Jul 2019, 16:26

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