Problem 10:
The repeat of a natural number is obtained by writing it twice in a row (for example, the repeat of $123$ is $123123$). Find a positive integer (if any) whose repeat is a perfect square.
BdMO National Secondary 2011/10
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Re: BdMO National Secondary 2011/10
The repeat numbers always have divisors 11,101,1001,10001...
It is necessary to have 1,0 to be ''repeat'',otherwise digits won't be same.
To get square either divisors like 11,101,1001.... will be square or their squares will be divisors.
First one isn't possible and while squaring the numbers we find 121,10201,1002001...Here we get '2',so the digits will be multiplied by 2.Digits won't be same.So there is no integers.
If anyone has better,different idea about the problem,please post.Is my solution correct or not?
It is necessary to have 1,0 to be ''repeat'',otherwise digits won't be same.
To get square either divisors like 11,101,1001.... will be square or their squares will be divisors.
First one isn't possible and while squaring the numbers we find 121,10201,1002001...Here we get '2',so the digits will be multiplied by 2.Digits won't be same.So there is no integers.
If anyone has better,different idea about the problem,please post.Is my solution correct or not?
Try not to become a man of success but rather to become a man of value.-Albert Einstein
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Mehfuj Zahir
- Posts:78
- Joined:Thu Jan 20, 2011 10:46 am
Re: BdMO National Secondary 2011/10
You are not correct.Think more carefully.
Re: BdMO National Secondary 2011/10
Hint:
First let the number $n$ has $k+1$ digits,then the repeat of $n,\bar n=(10^{k+1}+1)n$
If $10^{k+1}+1$ is square-free then it must divide $n$ but $n\le 10^{k+1}-1$
So we need to find $k$ so that it is not square-free and Tricky Lemma comes to the rescue
Note that if $k$ even,it is divisible by $11,$so we take $k=10,$then $11^2|10^{11}+1$ from tricky lemma
First let the number $n$ has $k+1$ digits,then the repeat of $n,\bar n=(10^{k+1}+1)n$
If $10^{k+1}+1$ is square-free then it must divide $n$ but $n\le 10^{k+1}-1$
So we need to find $k$ so that it is not square-free and Tricky Lemma comes to the rescue
Note that if $k$ even,it is divisible by $11,$so we take $k=10,$then $11^2|10^{11}+1$ from tricky lemma
One one thing is neutral in the universe, that is $0$.
Re: BdMO National Secondary 2011/10
''Tricky lemma'' is totally a new word to my little knowledge.what's it????
Try not to become a man of success but rather to become a man of value.-Albert Einstein
Re: BdMO National Secondary 2011/10
A very useful one in Number Theory
Let $p$ be an odd prime,$p^a||m$ denotes that $a$ is the largest exponent of $p$ dividing $m,$then the lemma says if $p^a||x-y,p^b||n,p^{a+b}||x^n-y^n$
So if you note that $7,13,11|10^3+1,7^2|10^{21}+1,11^2|10^{33}+1,13^2|10^{39}+1$
Let $p$ be an odd prime,$p^a||m$ denotes that $a$ is the largest exponent of $p$ dividing $m,$then the lemma says if $p^a||x-y,p^b||n,p^{a+b}||x^n-y^n$
So if you note that $7,13,11|10^3+1,7^2|10^{21}+1,11^2|10^{33}+1,13^2|10^{39}+1$
One one thing is neutral in the universe, that is $0$.
Re: BdMO National Secondary 2011/10
Here is a file of Tricky Lemma,I should post this earlier but my pen-drive got fever with virus.So I lost all my books.
Then I collected the following pdf from my freind of Iran-Amir Hossein(how?
)
You may read this
Then I collected the following pdf from my freind of Iran-Amir Hossein(how?
)You may read this
- Attachments
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Lifting The Exponent Lemma - Amir Hossein Parvardi - Version 3(2).pdf- (181.01KiB)Downloaded 397 times
One one thing is neutral in the universe, that is $0$.
Re: BdMO National Secondary 2011/10
I have done this:
Clearly we have to find $n$ such that $10^{n}+1$ is not square free, and we know $10^{11}+1$ is one (Thanks to Masum vai and LTE). Now $\displaystyle\frac{100000000001}{11^{2}}=826446281$. Let us take $a=82644628100=826446281\times 10^{2}$. Then the repetition of $a$,an $11$ digit number, will be
$11^{2}\times 826446281^{2}\times 10^{2}$ which is a square.
Is my solution correct?
Clearly we have to find $n$ such that $10^{n}+1$ is not square free, and we know $10^{11}+1$ is one (Thanks to Masum vai and LTE). Now $\displaystyle\frac{100000000001}{11^{2}}=826446281$. Let us take $a=82644628100=826446281\times 10^{2}$. Then the repetition of $a$,an $11$ digit number, will be
$11^{2}\times 826446281^{2}\times 10^{2}$ which is a square.
Is my solution correct?
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$