# Two consecutive integers have a product of 240, what are the integers?

Jun 2, 2015

AS CORRECTLY POINTED OUT BY @George C. THIS WORKS FOR THE SUM NOT THE PRODUCT..SORRY!
Call your starting integer $n$:
$n + \left(n + 1\right) + \left(n + 2\right) = 240$
$3 n + 3 = 240$
$3 n = 237$
$n = 79$
$79 + 80 + 81 = 192$

Jun 2, 2015

$240 = n \left(n + 1\right) = {n}^{2} + n$

When $n > 0$

${n}^{2} < n \left(n + 1\right) < {\left(n + 1\right)}^{2}$

${15}^{2} = 225$ and ${16}^{2} = 256$

So try $15 \times 16 = 240$ - yes

The other solution is $- 16$ and $- 15$