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mathslover
1111... (1997 times) 222...(1998 times)5 is a perfect square or not ? Prove.
there is no multiplication. that is a whole digit.
Basically, a perfect square has a prime factorization with even powers of all the prime factors. Time for a pattern:\[1225 =35^2 \\ 112225 = 335^2 \\ 11122225 = 3335^2 \\ \vdots \]
So if there are \(n + 1\) 2's and \(n\) ones, we have a perfect square.
But this is pattern. How will I prove ?
Now you do realize that if a perfect square is ending with \(5\), the number's square root too is.
I am confused. How can one prove that a number is a perfect square? We already know that your number has the square root \(333\cdots (1997~\rm times)5\).
We can just prove it by showing \(333\cdots(\rm 1997~times)5^2 = \rm your ~number\)
Oh, I see the proof.\[\]
We can see that \(1225 = (7\cdot5)^2\). \(112225 = (67\cdot 5)^2\). etc. So your number is definitely in the form \((5n)^2\) which we can see because if it's a perfect square, then the sqrt is a multiple of \(5\).\[25n^2 = 111\cdots(\rm 1997~times)222\cdots(\rm 1998~times)5\]This number is clearly a multiple of \(25\) (I'd explain in the next post).\[n^2 = 444\cdots(1997~\rm times)888\cdots(1996~times)9\]
Yes but this is a question of maths olympiad and when I said to the trainer , he asked that this is not a proof . :(
So \(n = 666\cdots(1996~\rm times)7\).
@mathslover This is clearly a proof.
The most basic proof that a number is a perfect square would be that it has an integer square root, which I did!
@KingGeorge What do you think? :-|
I'm pretty sure it's correct, but I'd like to see a little more justification to be honest.
Another proof is that \(\rm your~number =( 666\cdots(1996~\rm times)7)^2\cdot 5^2\)
Now a number in the form \(a^2 b^2 \) is a perfect square where \(a,b\) are integers.
@mathslover What kind of proof do you want exactly...?
If I may step in, I'm all well and fine with the fact that \(a^2b^2\) is a perfect square, but where is the reasoning that (666...(1996 times)67)\(^2\cdot\)5\(^2\) is the number we want?
See what we are doing is merely a claim...... Can you prove that All numbers of the form 11....11(n times)22...22(n+1times)5 can be written in the form of (5m)^2 for all n belonging to integers and m is also an integer. When you do a claim in Maths Olympiad, you have to later prove it (this is what we were told).
Yes, assuming that your number is a perfect square, it must be in the form of that.
Look,\[(10^{2n+1} + 10^{2n } + 10^{2n -1} +\cdots +10^{n+1} )+2\cdot(10^{n } + 10^{n - 1} +\cdots 10^1) +5\]your number is the case where \(n = 1997\)
I don't even know what I am doing :-|
And yes, @KingGeorge, we can find a pattern and just assert that this is the square root.
@mathslover Try math.stackexchange
I'm going to try to work some things out on some scratch paper. I've got no great ideas to prove this is a perfect square right now though.
Shubham is here to save the day :-)
I think I have a proof through direct multiplication (by hand). It's REALLY ugly, but it works.
It also relies on a weird multiplication trick that's kind of difficult to explain.
And some notationally difficult proofs by induction.
Let there be a number : 1111..(n times) 2222.(n+1 times) 5 this can be written as 5+2*(10 + 10^2 ..10^n+1) + 1*(10^n+2 +10^n+3.. 10^2n+1) 5+(20/9)(10^n+1 -1) + (10^n+2 /9)(10^n -1) => 5+(1/9) (20 .10^(n+1) -20 + 10^(2n+2) - 10^(n+2) ) => 5 + (1/9) (2 .10^(n+2) - 20 + 10^2(n+1) -10^(n+2) ) => 5+(1/9) ( 10^2(n+1) + 10^(n+2) - 20 ) => 5+(1/9) (100. 10^2n + 100 .10^n -20) => 5 +(1/9) (99.10^2n + 10^2n + 99.10^n + 10^n - 20 ) =>5+ 11(10^2n + 10^n)+ (1/9)(10^2n + 10^n) -20/9 =>5 +(100/9) (10^2n + 10^n) - 20/9 =>(25/9) + (100/9)(10^2n + 10^n) =>(25/9) ( 1 + 4.(10^2n + 10^n) ) =>(25/9) ( 1+ (2. 10^n)^2 + 2. (2.10^n) ) =>(25/9)( 2.10^n + 1)^2 =>[ (5/3)( 2.10^n +1) ]^2 We just have to prove (2.10^n +1) is always divisible by 3 This can either be proven from induction, or any other legit way. Also it can easily be understood as sum of digits of 2.10^n +1 is always 3 HENCE PROVED
You can easily verify by putting n=1,2 etc , I just did, it matches! ^_^
One question. I follow everything, except for some reason, I can't figure out why you can pull out a 1/9. Could you explain that to me?
Wait, nevermind. I think I see why.
In this case we dont even need to prove for all n. JUST prove when n=1997
when n=1997 2*10^n +1=200000.....00001 Since sum of digits is 3 it is divisible by 3
The proof that I have, while terribly messy and involving obscure multiplication tricks, is specific to this n, but easily generalized to any \(n\ge2\).
The video link gives a nice explanation of how I did the multiplication. With a couple proofs by induction, I was able to do this with 333...(1997 times)5 squared to get the desired solution. The only time I needed the specific n was when showing that the first 1997 digits are all 1. http://www.youtube.com/watch?v=kvLjpQ0XGao
If requested, I can type up most of my proof tomorrow.
I claim that 333...(1997 times)5 squared is 11...(1997 times)22...(1998 times)5. We build our number starting from the right. 5x5=25, so 5 is our first digit, and we carry the 2. Now, I claim that for digits 2 through 1998, we place a two, and carry n+1 to the next digit where n is the digit number. I will prove this through induction. Base: Second digit is 5x3+3x5+2=32, so we place a two and carry 3. So this works. Now we assume this is true up to \(k-1>2\). Then the \(k\)-th digit will be formed from the number \(5\cdot6+9\cdot(k-2)+k=12+10k\). Taken modulo 10, this is 2, so we place a 2. Also, \(12+10k=10\cdot(k+1)+2\), so we carry \(k+1\) to the next digit. So we're done with the proof. We continue until \(n=1998\) (once for each 3, and the first step). So far, I've shown that 333...(1997 times)5 squared is __..._22...(1997 times)5, where _ marks an unknown digit.
For the 1999-th digit, we carried over 1999 from the previous number. However, we no longer factor in the 5's when finding this digit. So the 1999-th digit is given by \(1999+(9*1997)=19972\) so we place a 2 and carry 1997. The next part was the hardest part for me.
I claim that for \(n\ge2000\), we place a 1 in the n-th digit, and carry \(3996-n\) to the next digit. Again, we will prove by induction. Base: 2000-th digit is obtained by \(1997+9*1996=19961\) so you place a 1, and carry 1996. So the base case works. Now we assume this is true up to some \(k-1\ge2000\). Then, the \(k\)-th digit is obtained from \[3996-(k-1)+9\cdot(3996-(k))=39961-10k=10\cdot(3996-k)+1.\]So we place a 1, and carry 3996-k. When \(n=3996\), we place 1 in the 3996-th spot, and we carry nothing. Then we've finished since there are no more multiplications that need to be done. This concludes the proof that 333...(1997 times)5 squared is 11...(1997 times)22...(1998 times)5.