What is the least possible positive integer-value of n such that sqrt{18\cdot n \cdot 34}$ is an integer?

Given √(18 x n x 34), for the expression to be an integer, the product must be a perfect square. We are looking for n, the smallest number to multiply 18 x 34 to make it a perfect square. We have:

18 x 34 = 2 x 9 x 2 x 17 = 2 x 3 x 3 x 2 x 17 = 2² x 3² x 17.

From here, we see that we need to multiply an extra 17 to make it a perfect square, therefore n = 17. So we have:

√(18 x 17 x 34) = √(10404) = 102

MrRoyal MrRoyal · Answer 2 · 2021-12-23T13:27:10+01:00

The least possible positive integer-value of n such that [tex]\sqrt{18\cdot n \cdot 34}[/tex] is an integer is 17

The expression is given as:

[tex]\sqrt{18\cdot n \cdot 34}[/tex]

Expand the equation

[tex]\sqrt{18\cdot n \cdot 34} = \sqrt{2 \times 9 \times n \times 2 \times 17}[/tex]

Express 8 as 3^2

[tex]\sqrt{18\cdot n \cdot 34} = \sqrt{2 \times 3^2 \times n \times 2 \times 17}[/tex]

Rearrange the factors as follows:

[tex]\sqrt{18\cdot n \cdot 34} = \sqrt{2 \times 2\times 3^2 \times n \times 17}[/tex]

Express 2 * 2 as 2^2

[tex]\sqrt{18\cdot n \cdot 34} = \sqrt{2^2\times 3^2 \times n \times 17}[/tex]

Evaluate all exponents

[tex]\sqrt{18\cdot n \cdot 34} = 2\times 3 \times \sqrt{n \times 17}[/tex]

[tex]\sqrt{18\cdot n \cdot 34} = 6 \times \sqrt{n \times 17}[/tex]

For the expression to be an integer, then the radical must be a perfect square.

Given that the expression in the radical is [tex]{n \times 17[/tex], then the value of n must equal 17, at the very least

i.e.

[tex]n = 17[/tex]

Hence, the least possible positive integer-value of n is 17