# How do you find two consecutive integers whose product is 58?

Jan 13, 2017

No such pair of integers. Hence, no solution.

#### Explanation:

As the only factors of $58$ are $\left\{1 , 2 , 29 , 58\right\}$, there are no two consecutive integers whose product is $58$.

Let us check it algebraically. Assume one integer is $x$ and next is $x + 1$. Hence, we have

$x \left(x + 1\right) = 58$ or ${x}^{2} + x = 58$ or ${x}^{2} + x - 58 = 0$

Using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, we have

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \times 1 \times \left(- 58\right)}}{2}$

= $\frac{- 1 \pm \sqrt{1 + 232}}{2}$

= $\frac{- 1 \pm \sqrt{233}}{2}$

and as we do not have a whole number as square root of $233$

we do not have any such pair of integers. Hence, no solution.

Jan 13, 2017

There are no such factors. the nearest possible combinations are:

$7 \times 8 = 56 \text{ and } 8 \times 9 = 72$

#### Explanation:

The factor exactly in the middle of a list of factors will be the square root.

Any consecutive integers will lie on either side of the square root .

$\sqrt{58} = 7.615 \ldots$

The two nearest integers are $7 \mathmr{and} 8$

However, their product is $7 \times 8 = 56$

The next combination is $8 \times 9 = 72$

There are no consecutive integers with a product of 58.

Further investigation shows that the only factors of 58 are:

$1 , \text{ "2," "29," } 58$