Question

How many pairs of consecutive natural numbers have a product of less than 40000? I am in 5th grade. This is supposed to be easy for me, but it's not. I would greatly appreciate any help at all. Thank you, Anonymous

Answer 1

There are 199 pairs of consecutive natural numbers whose product is less than 40000.

Explanation:

We notice that such statement can be translated into this inequation:

`n \cdot (n+1) < 40000`

Now we solve this inequation to the highest value of
`n` that satisfy the inequation:

`n^(2)+n < 40000`

`n^(2)+n -40000<0`

The Quadratic Formula shows that roots are:

`n_(1,2) = \frac{-1\pm\sqrt{1^(2)-4\cdot (1)\cdot (-40000)}}{2\cdot (1)}`

`n_(1,2) = -(1)/(2)\pm (1)/(2) \cdot √(160001)`

`n_(1) = -(1)/(2)+(1)/(2)\cdot √(160001)`

`n_(1) \approx 199.501`

`n_(2) = -(1)/(2)-(1)/(2)\cdot √(160001)`

`n_(2) \approx -200.501`

Only the first root is valid source to determine the highest possible value of
`n`, which is
`n_(max) = 199`. Each natural number represents an element itself and each pair represents an element as a function of the lowest consecutive natural number. Hence, there are 199 pairs of consecutive natural numbers whose product is less than 40000.