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M05-28

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Re: M05-28  [#permalink]

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New post 17 Aug 2018, 00:17
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, 22:23
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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Re: M05-28  [#permalink]

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New post 24 Oct 2018, 22:44
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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Re: M05-28  [#permalink]

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New post 25 Oct 2018, 20:22
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, 02:05
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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M05-28 &nbs [#permalink] 16 Nov 2018, 02:05

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