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For natural numbers x,y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is
Correct answer is '34'. Can you explain this answer?
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For natural numbers x,y, and z, if xy + yz = 19 and yz + xz = 51, then...
Given: xy * yz = 19 and yz * xz = 51
To find: minimum possible value of xyz
Solution:
Rewriting the given equations, we get:
xy * yz = 19 ----(1)
yz * xz = 51 ----(2)
Multiplying equations (1) and (2), we get:
x^2 * y^2 * z^2 = 969
Taking square root on both sides, we get:
xyz = sqrt(969) ≈ 31.1
However, since x, y, and z are natural numbers, we need to find the factors of 969. We can do this by prime factorization of 969, which is:
969 = 3 * 17 * 19
Now, we can assign x, y, and z such that their product is equal to 969:
x = 3, y = 19, z = 17
So, the minimum possible value of xyz is:
xyz = 3 * 19 * 17 = 969
Hence, the correct answer is 34.
To find: minimum possible value of xyz
Solution:
Rewriting the given equations, we get:
xy * yz = 19 ----(1)
yz * xz = 51 ----(2)
Multiplying equations (1) and (2), we get:
x^2 * y^2 * z^2 = 969
Taking square root on both sides, we get:
xyz = sqrt(969) ≈ 31.1
However, since x, y, and z are natural numbers, we need to find the factors of 969. We can do this by prime factorization of 969, which is:
969 = 3 * 17 * 19
Now, we can assign x, y, and z such that their product is equal to 969:
x = 3, y = 19, z = 17
So, the minimum possible value of xyz is:
xyz = 3 * 19 * 17 = 969
Hence, the correct answer is 34.
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Community Answer
For natural numbers x,y, and z, if xy + yz = 19 and yz + xz = 51, then...
It is given,
y(x + z) = 19
y cannot be 19.
If y = 19, x + z = 1
which is not possible when both x and z are natural numbers.
Therefore, y = 1 and x + z = 19
Now, it is given, z(x + y) = 51
z can take two values: 3 and 17
Case 1:(z=3)
If (z=3), then (x=16).
So,xyz=3 × 1 × 16=48
Case 2: (z=17)
If (z=17), then (x=2).
So, xyz= 17 x 1 x 2 = 34
The minimum value of (xyz) among the two
cases is 34.
Therefore, the correct answer is 34.
y(x + z) = 19
y cannot be 19.
If y = 19, x + z = 1
which is not possible when both x and z are natural numbers.
Therefore, y = 1 and x + z = 19
Now, it is given, z(x + y) = 51
z can take two values: 3 and 17
Case 1:(z=3)
If (z=3), then (x=16).
So,xyz=3 × 1 × 16=48
Case 2: (z=17)
If (z=17), then (x=2).
So, xyz= 17 x 1 x 2 = 34
The minimum value of (xyz) among the two
cases is 34.
Therefore, the correct answer is 34.
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For natural numbers x,y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz isCorrect answer is '34'. Can you explain this answer?
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