Suppose x, y, z are three positive integers such that x< y<>z and xyz = 7 2. Which one of the following values of S yields more than one solution to the equation x + y + z-S?

examrobotsa's picture
Q: 46 (CAPF/2018)
Suppose x, y, z are three positive integers such that x< y<>z and xyz = 7
2.
Which one of the following values of S yields more than one solution to the equation x + y + z-S?

question_subject: 

Science

question_exam: 

CAPF

stats: 

0,6,3,2,6,1,0

keywords: 

{'positive integers': [0, 0, 0, 1], 'solution': [2, 0, 7, 14], 'values': [0, 2, 1, 3]}

The question states that x, y, and z are positive integers such that x < y < z, and xyz = 72. We need to find the value of S that yields more than one solution to the equation x + y + z - S.

First, let`s find the prime factorization of 72: 72 = 2^3 * 3^2.

To find the number of solutions to the equation x + y + z - S, we can use the concept of partitions. Each positive integer can be expressed as a sum of three positive integers.

For any given value of S, the sum of x + y + z should be equal to 72. We need to find the different ways to partition 72 into three positive integers.

Let`s consider option 2: 14.

To find the number of solutions, we need to find the number of different ways to partition 72 - 14 = 58.

The possible partitions of 58 are:

1 + 1 + 56

1 + 2 + 55

1 + 3 + 54

...

27 + 14 + 17

28 + 13 + 17

...

56 + 1 + 1

As we can see, there