Latest self Aptitude Question SOLUTION: If a cubic number y is added to a perfect square x and resultant sum is divided by a cubic number z we get result equal to y. Find x,y & z considering they are distinct and numbers

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If a cubic number y is added to a perfect square x and resultant sum is divided by a cubic number z we get result equal to y. Find x,y & z considering they are distinct and numbers are within 1 to 100.

Devendra Marghade (11 years ago)

Solved Vignesh Read Solution (3)

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Ashu  (11 years ago)

Pls check if given statement is correct.
If a cubic number y is added to a perfect square x and resultant sum is divided by a cubic number z we get result equal to y. Find x,y & z considering they are distinct and numbers are within 1 to 100.

If the statement is read as
If a cubic number y is added to a perfect square x and resultant sum is divided by a 'perfect square number' z we get result equal to y. Find x,y & z considering they are distinct and numbers are within 1 to 100.

then (y+x)/z=y
(2^3+ 8^2)/(3^2) = (8+64)/9= 72/9 =8 = 2^3 =y

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Vignesh  (11 years ago)

By considering 0 is a perfect square, answer will be

So many possibles

x=0; y=8(cube of 2); z=1(cube of 1)--> (8+0)/1 = 8
x=0; y=27(cube of 3); z=1(cube of 1)--> (27+0)/1 = 27
x=0; y=64(cube of 4); z=1(cube of 1)--> (64+0)/1 = 64

Other than these, there is no possible i think.

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Vignesh  (11 years ago)

Question may wrong or typical mistake.

Otherwise,

If 0 is a perfect square, answer will be

Three possibles
x=0;y=8;z=1
x=0;y=27;z=1
x=0;y=64;z=1

i.e,
x=0; y=8(cube of 2); z=1(cube of 1)--> (8+0)/1 = 8
x=0; y=27(cube of 3); z=1(cube of 1)--> (27+0)/1 = 27
x=0; y=64(cube of 4); z=1(cube of 1)--> (64+0)/1 = 64

Other than these, there is no possible i think.

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