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#