If x, y and z are positive integers such that xy = 432, yz = 96 and z

Question

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

Abhilash Singh · Accepted Answer

Answer: 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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