not integer number

May 2009
596
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ALGERIA
Prove that number \(\displaystyle \sqrt{20092010201120122013}\) is not
integer number .
 
Nov 2009
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Wellington
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 :D

PS : the proof for my original statement is quite trivial, a quick Google search will have it done.
 
May 2009
596
31
ALGERIA
Hello : have the demonstration 'A perfect square never ends in '(Talking)(Come on Dhiab, that was too easy)(Evilgrin)
Thanks
 
Nov 2009
927
260
Wellington
Fine, I'll give it (Speechless)

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.
 
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Soroban

MHF Hall of Honor
May 2006
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Hello, dhiab!

Prove that number \(\displaystyle \sqrt{20092010201120122013}\) is not an integer.
I assume that the author forgot (or was unaware of) the Final Digit Rule for integer squares.

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?

 
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May 2009
1,176
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Fine, I'll give it (Speechless)

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.
You do not need to use modulo 10, you can just use modulo 5. This halves the number of calculations you need to perform!
 
May 2009
596
31
ALGERIA
Originally Posted by :Swlabr
"You do not need to use modulo 10, you can just use modulo 5. This halves the number of calculations you need to perform!
Hello :
you know that the rest of division ,by 10, a integer number is first digit.(Clapping)
 
Mar 2010
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And there's always the brute force method:
\(\displaystyle \sqrt{20092010201120122013}=4482411203.93...\)

- Hollywood
 
May 2009
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412
Originally Posted by :Swlabr
"You do not need to use modulo 10, you can just use modulo 5. This halves the number of calculations you need to perform!
Hello :
you know that the rest of division ,by 10, a integer number is first digit.(Clapping)
\(\displaystyle a \equiv 3 \text{ mod } 10 \Rightarrow a \equiv 3 \text{ mod } 5\).
 
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chiph588@

MHF Hall of Honor
Sep 2008
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Champaign, Illinois
And there's always the brute force method:
\(\displaystyle \sqrt{20092010201120122013}=4482411203.93...\)

- Hollywood
Going along these lines you could see that \(\displaystyle 4482411203^2<20092010201120122013 \) and \(\displaystyle 4482411204^2>20092010201120122013 \).
 
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