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Permutations and Combinations question

2009 · Shift 2 · Q31
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Permutations and Combinations question

2009 · Shift 2 · Q31

JEE AdvancedMathematicsPermutations and CombinationsNumerical+3 / −1
Let (x, y, z)\left( {x,\,y,\,z} \right)(x,y,z) be points with integer coordinates satisfying the system of homogeneous equation: 3x−y−z=0−3x+z=0−3x+2y+z=0\begin{matrix} {3x - y - z = 0} \\ { - 3x + z = 0} \\ { - 3x + 2y + z = 0} \\ \end{matrix}3x−y−z=0−3x+z=0−3x+2y+z=0​ Then the number of such points for which x2+y2+z2≤100x^2 + {y^2} + {z^2} \le 100x2+y2+z2≤100 is
Numerical answer
View written solutionFree

Correct answer: 7

  1. We need integer-coordinate points (x,y,z)(x,y,z)(x,y,z) satisfying
{3x−y−z=0−3x+z=0−3x+2y+z=0\begin{cases} 3x-y-z=0 \\ -3x+z=0 \\ -3x+2y+z=0 \end{cases}⎩⎨⎧​3x−y−z=0−3x+z=0−3x+2y+z=0​

and also x2+y2+z2≤100.x^2+y^2+z^2\le 100.x2+y2+z2≤100.

  1. Solve the linear system.

From the second equation, −3x+z=0  ⟹  z=3x.-3x+z=0 \implies z=3x.−3x+z=0⟹z=3x.

Substitute into the first equation: 3x−y−z=0  ⟹  3x−y−3x=0  ⟹  −y=0  ⟹  y=0.3x-y-z=0 \implies 3x-y-3x=0 \implies -y=0 \implies y=0.3x−y−z=0⟹3x−y−3x=0⟹−y=0⟹y=0.

Now check the third equation: −3x+2y+z=−3x+2(0)+3x=0,-3x+2y+z=-3x+2(0)+3x=0,−3x+2y+z=−3x+2(0)+3x=0, which is automatically satisfied.

So all solutions are of the form

(x,y,z)=(x,0,3x),(x,y,z)=(x,0,3x),(x,y,z)=(x,0,3x),

where xxx is an integer.

  1. Apply the condition x2+y2+z2≤100x^2+y^2+z^2\le 100x2+y2+z2≤100.

Substitute y=0y=0y=0 and z=3xz=3xz=3x:

x2+02+(3x)2≤100x^2+0^2+(3x)^2\le 100x2+02+(3x)2≤100 x2+9x2≤100x^2+9x^2\le 100x2+9x2≤100 10x2≤10010x^2\le 10010x2≤100 x2≤10.x^2\le 10.x2≤10.

Since xxx is an integer, x∈{−3,−2,−1,0,1,2,3}.x\in\{-3,-2,-1,0,1,2,3\}.x∈{−3,−2,−1,0,1,2,3}.

That gives 777 possible values of xxx, hence 777 integer-coordinate points.

  1. Therefore, the required number of points is 7.\boxed{7}.7​.

  2. Comparison with stored answer: Stored correct answer = 777.

Our derived answer also equals 777, so they agree.

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