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

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:

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

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

$n^{2}+n < 40000$

$n^{2}+n -40000$

The Quadratic Formula shows that roots are:

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

$n_{1,2} = -\frac{1}{2}\pm \frac{1}{2} \cdot \sqrt{160001}$

$n_{1} = -\frac{1}{2}+\frac{1}{2}\cdot \sqrt{160001}$

$n_{1} \approx 199.501$

$n_{2} = -\frac{1}{2}-\frac{1}{2}\cdot \sqrt{160001}$

$n_{2} \approx -200.501$

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.