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x, y, and z are integers such that |x| < |y| < |z| < 40. Also, x + y +
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08 Mar 2022, 00:17
08 Mar 2022, 00:17
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x, y, and z are integers such that |x| < |y| < |z| < 40. Also, x + y + z = 20. What is the maximum possible value of xyz?
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3510
Re: x, y, and z are integers such that |x| < |y| < |z| < 40. Also, x + y +
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16 Mar 2022, 13:03
16 Mar 2022, 13:03
2
-10, -9 , +39
multiplying them together would get you 3510
multiplying them together would get you 3510
Re: x, y, and z are integers such that |x| < |y| < |z| < 40. Also, x + y +
[#permalink]
24 Jun 2025, 00:08
24 Jun 2025, 00:08
1
x,y,z can be +ve, 0, -ve. But since it's given |x| < |y| < |z| < 40,
|x| can take +ve/ 0 values -> 0 till 37
|y| can take values -> 1 till 38 (because min value |x| can take is 0 and max |x| is 37, |y| need to be greater than x )
|z| can take values -> 2 till 39
Maximum possible value of xyz is when xyz is positive and that can happen if all 3 integers are +ve / 2 integers -ve and 1 +ve integer.
Now if we check we can see if all 3 +ve, that can lessen the value of xyz, because the value of x,y,z < 20. But if we take -ve values, the value |x|,|y|,|z| can be greater than 20, so the product can be way greater. So we know 2 from x,y,z is -ve and 1 +ve.
In x + y + z = 20, +ve number value will be way greater than the sum of -ve numbers to get 20, out of x,y,z the highest value they can have if either of them is +ve is 39 for z. So z will be +ve with value 39, x and y -ve numbers in the range -> 0 till -37 & -1 till -38 respectively.
39 + x + y = 20, So x + y = -19.
Now we need to find out which 2 -ve numbers added can give the highest product when x and y is multiplied, that is when x and y value is max -> when they're equal to -10, -9.
Here since |x| < |y|, x = -9, y = -10, z = 39. x * y * z = 39 * -9 * -10 = 3510
|x| can take +ve/ 0 values -> 0 till 37
|y| can take values -> 1 till 38 (because min value |x| can take is 0 and max |x| is 37, |y| need to be greater than x )
|z| can take values -> 2 till 39
Maximum possible value of xyz is when xyz is positive and that can happen if all 3 integers are +ve / 2 integers -ve and 1 +ve integer.
Now if we check we can see if all 3 +ve, that can lessen the value of xyz, because the value of x,y,z < 20. But if we take -ve values, the value |x|,|y|,|z| can be greater than 20, so the product can be way greater. So we know 2 from x,y,z is -ve and 1 +ve.
In x + y + z = 20, +ve number value will be way greater than the sum of -ve numbers to get 20, out of x,y,z the highest value they can have if either of them is +ve is 39 for z. So z will be +ve with value 39, x and y -ve numbers in the range -> 0 till -37 & -1 till -38 respectively.
39 + x + y = 20, So x + y = -19.
Now we need to find out which 2 -ve numbers added can give the highest product when x and y is multiplied, that is when x and y value is max -> when they're equal to -10, -9.
Here since |x| < |y|, x = -9, y = -10, z = 39. x * y * z = 39 * -9 * -10 = 3510
