If x, y and z are positive integers such that xy = 432, yz = 96 and z < 9, then the smallest possible value of x + y + z is
Answer
Correct Answer : c ) 46
Explanation :In the given question,
it is given that:
x, y and z are integers and x > 0, y > 0, z > 0
xy = 432 ......(1)
yz = 96 ........(2)
z < 9
To find:
smallest value of x + y + z
Let us find the prime factorization of yz
yz = 96 = 2 x 2 x 2 x 2 x 2 x 3 = 25 x 31
it is given that z < 9
hence, the possible values of z can be 2, 3, 4, 6, 8 (as these numbers can be obtained from combinations from prime factorization of 96)
based on values of z as 2, 3, 4, 6, 8,
values of y respectively will be 48, 32, 24, 16, 12
Now based on values of y, we can find values of x from eq. 1
Values of x on the basis of values of y as 48, 32, 24, 16, 12, respectively will be:
9, 13.5, 18, 27, 36
As x is an integer so, x = 13.5 is not possible.
Now, we have four sets of values of x, y and z
Set 1:
x, y, z = 2, 48, 9
x + y + z = 59
Set 2:
x, y, z = 4, 24, 18
x + y + z = 46
Set 3:
x, y, z = 6, 16, 27
x + y + z = 49
Set 4:
x, y, z = 8, 12, 36
x + y + z = 56
The minimum value of x + y + z is 46
Hence, option (c) is the correct answer.
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