Answer:
There are 199 pairs of consecutive natural numbers whose product is less than 40000.
Step-by-step explanation:
We notice that such statement can be translated into this inequation:
[tex]n \cdot (n+1) < 40000[/tex]
Now we solve this inequation to the highest value of [tex]n[/tex] that satisfy the inequation:
[tex]n^{2}+n < 40000[/tex]
[tex]n^{2}+n -40000<0[/tex]
The Quadratic Formula shows that roots are:
[tex]n_{1,2} = \frac{-1\pm\sqrt{1^{2}-4\cdot (1)\cdot (-40000)}}{2\cdot (1)}[/tex]
[tex]n_{1,2} = -\frac{1}{2}\pm \frac{1}{2} \cdot \sqrt{160001}[/tex]
[tex]n_{1} = -\frac{1}{2}+\frac{1}{2}\cdot \sqrt{160001}[/tex]
[tex]n_{1} \approx 199.501[/tex]
[tex]n_{2} = -\frac{1}{2}-\frac{1}{2}\cdot \sqrt{160001}[/tex]
[tex]n_{2} \approx -200.501[/tex]
Only the first root is valid source to determine the highest possible value of [tex]n[/tex], which is [tex]n_{max} = 199[/tex]. 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.
