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Bunuel
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x, y and z are different positive integers. If 24x^3y^3z^3 is a square 28 Aug 2019, 02:30
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D
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65% (hard)

Question Stats:

55% (01:49) correct 45% (01:52) wrong based on 247 sessions

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x, y and z are different positive integers. If \(24x^3y^3z^3\) is a square of an integer, what is the lowest common multiple of x, y and z?

A. 1
B. 2
C. 3
D. 6
E. 12
 
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D
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x, y and z are different positive integers. If 24x^3y^3z^3 is a square 28 Aug 2019, 04:30
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\(24x^3y^3z^3\) = X
√X ; 4*x*y*x * √6*x*y*z ; x*y*z has to be 6 so x=2,y=3;z=1
LCM x,y,z; 6
IMO D

Bunuel
x, y and z are different positive integers. If \(24x^3y^3z^3\) is a square of an integer, what is the lowest common multiple of x, y and z?

A. 1
B. 2
C. 3
D. 6
E. 12
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x, y and z are different positive integers. If 24x^3y^3z^3 is a square 28 Aug 2019, 04:42
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6
By looking answer
1*24=24 not square root of any number
2*24=48 not Sr
3*24=72 not Sr
6*24=144 sr of 12 ✓
12*24=288 not sr
And:6(1,2,3)

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x, y and z are different positive integers. If 24x^3y^3z^3 is a square 28 Aug 2019, 05:36
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Solution



Given

    • x, y and z are different positive integers
    • \(24x^3y^3z^3\) is a square of an integer

To find

    • Lowest common multiple of x, y and z
Approach
• \(2^4x^3y^3z^3\)= Perfect Square = \(N^2\)
• \(2^3 × 3 × x^3y^3z^3\)= \(N^2\)
• \(2 × 3 × 2^2 × x^3y^3z^3\)= \(N^2\)
• \(2 × 3 × 2^2 × (xyz)^3\)= \(N^2\)
• \(2^2\) is already a perfect square. Hence, 6 × \((xyz)^3\) must be a perfect square.
Therefore, (xyz)^3 must be of the form 6 * some perfect square for 6 × \((xyz)^3\) to be a perfect square.

Therefore, option D is the correct answer.
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x, y and z are different positive integers. If 24x^3y^3z^3 is a square 31 May 2022, 13:46
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Bunuel
x, y and z are different positive integers. If \(24x^3y^3z^3\) is a square of an integer, what is the lowest common multiple of x, y and z?

A. 1
B. 2
C. 3
D. 6
E. 12
Show more
Bunuel, I think the question needs a bit of fixing. The question should ideally read which of the following could be a lowest common multiple of x, y, and z. Because x=10, y=15, and z=25 also satisfy the condition that \(24x^3y^3z^3\) is a square of an integer, but the LCM of (10, 15, 25) = 150 is not an option :) similarly there can be infinitely many combinations­
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x, y and z are different positive integers. If 24x^3y^3z^3 is a square 13 Feb 2024, 12:25
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