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Re: Divisibility by 7 [#permalink]
13 Jun 2010, 08:33

7

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

study 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. \(y = 4\)

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: \(x^2 = y + 5\) \(y = z - 2\) --> \(y=2x-2\). \(z = 2x\)

\(x^2 =2x-2+ 5\) --> \(x^2-2x-3=0\) --> \(x=3\) or \(x=-1\)

\(x=3\), \(y=4\), \(z=6\) - first triplet --> \(x^3 + y^2 + z=27+16+6=49\), divisible by 7; \(x=-1\), \(y=-4\), \(z=-2\) - second triplet --> \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\) --> we deal with first triplet. \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. sufficient. (2) \(y = 4\) --> --> we deal with first triplet. \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. sufficient.

Rephrase the stem: x^2 = y + 5 x^2 - y + 5 = 0 x^2 -(z-2) + 5 = 0 x^2 - z - 3 = 0 x^2 - 2x - 3 = 0 x = 3, -1 Substituting gives z = 6, -2 and y = 4, -4

Notice that both statements ask for same information, that is 1) x>0 -> x=3 and corresponding y=4 and z=6. 2) y=4 means x=3 and z=6

Substituting these values in the stem gives us 51, which is not div by 7. So, both suff.

D _________________

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DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: If x^2 = y + 5 , y = z - 2 and z = 2x , is x^3 + y^2 + z [#permalink]
21 Sep 2013, 14:07

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Re: Divisibility by 7 [#permalink]
27 Apr 2014, 05:17

Bunuel wrote:

study 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. \(y = 4\)

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: \(x^2 = y + 5\) \(y = z - 2\) --> \(y=2x-2\). \(z = 2x\)

\(x^2 =2x-2+ 5\) --> \(x^2-2x-3=0\) --> \(x=3\) or \(x=-1\)

\(x=3\), \(y=4\), \(z=6\) - first triplet --> \(x^3 + y^2 + z=27+16+6=49\), divisible by 7; \(x=-1\), \(y=-4\), \(z=-2\) - second triplet --> \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\) --> we deal with first triplet. \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. sufficient. (2) \(y = 4\) --> --> we deal with first triplet. \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. sufficient.

Answer: D.

Hope it helps.

Hi bunnel,

In second statement y=4 if we replace we found x^2 = 4+5 = 9

so can we write x= +3 and -3. if we replace x= -3 then sum will be -24+16+6 = -1. then how it can be divisible by 7

Re: Divisibility by 7 [#permalink]
28 Apr 2014, 01:41

Expert's post

PathFinder007 wrote:

Bunuel wrote:

study 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. \(y = 4\)

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: \(x^2 = y + 5\) \(y = z - 2\) --> \(y=2x-2\). \(z = 2x\)

\(x^2 =2x-2+ 5\) --> \(x^2-2x-3=0\) --> \(x=3\) or \(x=-1\)

\(x=3\), \(y=4\), \(z=6\) - first triplet --> \(x^3 + y^2 + z=27+16+6=49\), divisible by 7; \(x=-1\), \(y=-4\), \(z=-2\) - second triplet --> \(x^3 + y^2 + z=-1+16-2=13\), not divisible by 7.

(1) \(x \gt 0\) --> we deal with first triplet. \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. sufficient. (2) \(y = 4\) --> --> we deal with first triplet. \(x^3 + y^2 + z=27+16+6=49\), divisible by 7. sufficient.

Answer: D.

Hope it helps.

Hi bunnel,

In second statement y=4 if we replace we found x^2 = 4+5 = 9

so can we write x= +3 and -3. if we replace x= -3 then sum will be -24+16+6 = -1. then how it can be divisible by 7

Please clarify

Thanks.

Given: \(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\).

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