Question

What is the smallest integer $n$ such that $n\sqrt{2}$ is greater than $20$? (Note: $n\sqrt{2}$ means $n$ times $\sqrt{2}$.)

Answer 1

15

Explanation:

In order to compare $n\sqrt{2}$ to $20$, we can compare the square of $n\sqrt{2}$ to the square of $20.$ We have

\begin{align*}

\left(n\sqrt{2}\right)^2 &= \left(n\sqrt{2}\right)\left(n\sqrt{2}\right) = n^2 \left(\sqrt{2}\right)^2 = n^2\cdot 2= 2n^2,\\

20^2 &= 400.

\end{align*}Therefore, we have $n\sqrt{2} > 20$ whenever $n^2 > 200.$ Since $14^2 = 196$ and $15^2 = 225,$ we know that $\boxed{15}$ is the smallest integer $n$ such that $n\sqrt{2}$ is greater than $20.$

Answer 2

Question:

What is the smallest integer
`$n$` such that
`$n√(2)$` is greater than
`$20$`? (Note:
`$n√(2)$` means
`$n$` times
`$√(2)$`.)

Solution:

• n√2 > 20
• => n > 20/√2
• => n > 4 x 5/√2
• => n > 2 x 2 x 5/√2
• => n > √4 x √4 x √25/√2
• => n > √2 x √4 x √25
• => n > √2 x 4 x 25
• => n > 10√2
• => n > 14.14 (Rounded)

Smallest integer possibility for n is 15.

Hence, the smallest possible integer is 11.