Prove that number $\displaystyle \sqrt{20092010201120122013}$ is not
integer number .
A perfect square never ends in $\displaystyle 3$, therefore $\displaystyle 20092010201120122013$ is not a perfect square, therefore $\displaystyle \sqrt{20092010201120122013}$ is not an integer.
Come on Dhiab, that was too easy, where are the headcracking problems you used to post
PS : the proof for my original statement is quite trivial, a quick Google search will have it done.
Fine, I'll give it
Let $\displaystyle n$ be any integer number. $\displaystyle n^2$ is its square. Let us look at the last digit of $\displaystyle n^2$, using arithmetic modulo $\displaystyle 10$. The following table enumerates all possible cases :
$\displaystyle \begin{array}{|l|l|}
\hline n \mod 10 &n^2 \mod 10 \\
\hline 0 &0 \\
1 &1 \\
2 &4 \\
3 &9 \\
4 &6 \\
5 &5 \\
6 &6
\end{array}$
$\displaystyle \begin{array}{|l|l|}
7 \ \ \ \ \ \ \ \ \ \ \ \ \ &9 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
8 &4 \\
9 &1 \\
\hline
\end{array}$
Sorry, I had to cut the table midway, it was too big for MHF.
This shows that a perfect square may only end by the digits $\displaystyle 0, 1, 4, 5, 6$ and $\displaystyle 9$. If an integer does not terminate by neither of those digits, it cannot possibly be a perfect square.
$\displaystyle \boxed{\mathrm{QED}}$
This implies that if a number terminates by 3, it cannot be a perfect square.
Hello, dhiab!
I assume that the author forgot (or was unaware of) the Final Digit Rule for integer squares.Prove that number $\displaystyle \sqrt{20092010201120122013}$ is not an integer.
If that is true, that number has a special rhythm to it: .$\displaystyle 2009\,2010\,2011\,2012\,2013$
We have: .$\displaystyle N \;=\;10^{16}a + 10^{12}(a+1) + 10^8(a+2) + 10^4(a+3) + (a+4)\;\text{ where }a = 2009$
And I suspect that we are expected to prove algebraically that $\displaystyle N$ is not a square.
Any ideas?