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

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Answer 1 · 2017-01-13T04:18:07

No such pair of integers. Hence, no solution.

As the only factors of $58$ are ${1,2,29,58}$ , 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

$x(x+1)=58$ or $x^2+x=58$ or $x^2+x-58=0$

Using quadratic formula $(-b+-sqrt(b^2-4ac))/(2a)$ , we have

$x=(-1+-sqrt(1^2-4xx1xx(-58)))/2$

= $(-1+-sqrt(1+232))/2$

= $(-1+-sqrt233)/2$

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.

Answer 2 · 2017-01-13T05:09:02

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

$7xx8 = 56" and " 8xx9 =72$

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 .

$sqrt58 = 7.615...$

The two nearest integers are $7 and 8$

However, their product is $7 xx 8 = 56$

The next combination is $8xx9 =72$

There are no consecutive integers with a product of 58.

Further investigation shows that the only factors of 58 are:

$1," "2," "29," "58$