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mathslover
 3 years ago
1111... (1997 times) 222...(1998 times)5
is a perfect square or not ? Prove.
mathslover
 3 years ago
1111... (1997 times) 222...(1998 times)5 is a perfect square or not ? Prove.

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mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0there is no multiplication. that is a whole digit.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Basically, a perfect square has a prime factorization with even powers of all the prime factors. Time for a pattern:\[1225 =35^2 \\ 112225 = 335^2 \\ 11122225 = 3335^2 \\ \vdots \]

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0So if there are \(n + 1\) 2's and \(n\) ones, we have a perfect square.

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0But this is pattern. How will I prove ?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Now you do realize that if a perfect square is ending with \(5\), the number's square root too is.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0I am confused. How can one prove that a number is a perfect square? We already know that your number has the square root \(333\cdots (1997~\rm times)5\).

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0We can just prove it by showing \(333\cdots(\rm 1997~times)5^2 = \rm your ~number\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, I see the proof.\[\]

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0We 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).\[n^2 = 444\cdots(1997~\rm times)888\cdots(1996~times)9\]

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0Yes but this is a question of maths olympiad and when I said to the trainer , he asked that this is not a proof . :(

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0So \(n = 666\cdots(1996~\rm times)7\).

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0@mathslover This is clearly a proof.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0The most basic proof that a number is a perfect square would be that it has an integer square root, which I did!

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0@KingGeorge What do you think? :

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I'm pretty sure it's correct, but I'd like to see a little more justification to be honest.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Another proof is that \(\rm your~number =( 666\cdots(1996~\rm times)7)^2\cdot 5^2\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Now a number in the form \(a^2 b^2 \) is a perfect square where \(a,b\) are integers.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0@mathslover What kind of proof do you want exactly...?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1If 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?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0See 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).

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, assuming that your number is a perfect square, it must be in the form of that.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Look,\[(10^{2n+1} + 10^{2n } + 10^{2n 1} +\cdots +10^{n+1} )+2\cdot(10^{n } + 10^{n  1} +\cdots 10^1) +5\]your number is the case where \(n = 1997\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0I don't even know what I am doing :

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0And yes, @KingGeorge, we can find a pattern and just assert that this is the square root.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0@mathslover Try math.stackexchange

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I'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.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0Shubham is here to save the day :)

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I think I have a proof through direct multiplication (by hand). It's REALLY ugly, but it works.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1It also relies on a weird multiplication trick that's kind of difficult to explain.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1And some notationally difficult proofs by induction.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.4Let 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

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.4You can easily verify by putting n=1,2 etc , I just did, it matches! ^_^

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1One 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?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Wait, nevermind. I think I see why.

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0In this case we dont even need to prove for all n. JUST prove when n=1997

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.0when n=1997 2*10^n +1=200000.....00001 Since sum of digits is 3 it is divisible by 3

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1The proof that I have, while terribly messy and involving obscure multiplication tricks, is specific to this n, but easily generalized to any \(n\ge2\).

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1The 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

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1If requested, I can type up most of my proof tomorrow.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I 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 \(k1>2\). Then the \(k\)th digit will be formed from the number \(5\cdot6+9\cdot(k2)+k=12+10k\). 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1For the 1999th digit, we carried over 1999 from the previous number. However, we no longer factor in the 5's when finding this digit. So the 1999th 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I claim that for \(n\ge2000\), we place a 1 in the nth digit, and carry \(3996n\) to the next digit. Again, we will prove by induction. Base: 2000th 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 \(k1\ge2000\). Then, the \(k\)th digit is obtained from \[3996(k1)+9\cdot(3996(k))=3996110k=10\cdot(3996k)+1.\]So we place a 1, and carry 3996k. When \(n=3996\), we place 1 in the 3996th 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.
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