Question

How many solutions in positive integers are there to the equation \({x_1} + {x_2} + {x_3} = 20\) with \(2 < {x_1} < 6\), \(6 < {x_2} < 10\), and \(0 < {x_3} < 5\)?

Short answer

There no positive integer solutions to the equation \({x_1} + {x_2} + {x_3} = 20\) with \(2 < {x_1} < 6,6 < {x_2} < 10,0 < {x_3} < 5\).

Step-by-step solution

In the problem given

Equation is \({x_1} + {x_2} + {x_3} = 20\) with \(2 < {x_1} < 6,6 < {x_2} < 10,0 < {x_3} < 5\).

The definition and the formula for the given problem

Positive integers are literally a part of a larger group of numbers called integers. Integers are all the full numbers, both positive and negative. By whole numbers we mean numbers without fractions or decimals. you'll also call positive integers your 'counting numbers' because they're the identical. you do not count with fractions or decimals or negative numbers.

Determining the sum in expanded form

Since maximum integer values of \({x_1},{x_2}\) and \({x_3}\) are \(5,9\) and \(4\)respectively.

The sum of three values is \(18\), which is less than \(20\).

Hence, there no positive integer solutions to the equation

\({x_1} + {x_2} + {x_3} = 20\) with \(2 < {x_1} < 6,6 < {x_2} < 10,0 < {x_3} < 5\).