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Re: Divisibility by 7 [#permalink]
13 Jun 2010, 08:33
This post received KUDOS
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.
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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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