Step-by-step explanation:
Let the three consecutive positive integers be x, (x + 1) & (x + 2)
Therefore, according to the given condition:
[tex]x(x + 1) = (x + 2) + 167 \\ \\ \therefore \: {x}^{2} + x = x + 169 \\ \\ \therefore \: {x}^{2} + x - x = 169\\ \\ \therefore \: {x}^{2} = 169 \\ \\ \therefore \: {x} = \pm \sqrt{169} \\ \\ \therefore \: {x} = \pm 13 \\ \\ \because \: x \: is \: positive \: \\ \\ \therefore \: x \: \neq \: - 13 \\ \\ \therefore \: x \: = \: 13 \\ \\\therefore \: x + 2\: = \: 13 + 2 = 15[/tex]
Thus, the largest integer is 15.
