Question

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

Answer 1

there is no multiplication. that is a whole digit.

Answer 2

^ number

Answer 3

Basically, a perfect square has a prime factorization with even powers of all the prime factors. Time for a pattern:\[1225 =35^2 \\ 112225 = 335^2 \\ 11122225 = 3335^2 \\ \vdots \]

Answer 4

So if there are \(n + 1\) 2's and \(n\) ones, we have a perfect square.

Answer 5

But this is pattern. How will I prove ?

Answer 6

Now you do realize that if a perfect square is ending with \(5\), the number's square root too is.

Answer 7

I am confused. How can one prove that a number is a perfect square? We already know that your number has the square root \(333\cdots (1997~\rm times)5\).

Answer 8

We can just prove it by showing \(333\cdots(\rm 1997~times)5^2 = \rm your ~number\)

Answer 9

Oh, I see the proof.\[\]

Answer 10

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).\[n^2 = 444\cdots(1997~\rm times)888\cdots(1996~times)9\]

Answer 11

Yes but this is a question of maths olympiad and when I said to the trainer , he asked that this is not a proof . :(

Answer 12

So \(n = 666\cdots(1996~\rm times)7\).

Answer 13

@mathslover This is clearly a proof.

Answer 14

The most basic proof that a number is a perfect square would be that it has an integer square root, which I did!

Answer 15

@KingGeorge What do you think? :-|

Answer 16

I'm pretty sure it's correct, but I'd like to see a little more justification to be honest.

Answer 17

Another proof is that \(\rm your~number =( 666\cdots(1996~\rm times)7)^2\cdot 5^2\)

Answer 18

Now a number in the form \(a^2 b^2 \) is a perfect square where \(a,b\) are integers.

Answer 19

@mathslover What kind of proof do you want exactly...?

Answer 20

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?

Answer 21

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).

Answer 22

Yes, assuming that your number is a perfect square, it must be in the form of that.

Answer 23

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 \(n = 1997\)

Answer 24

I don't even know what I am doing :-|

Answer 25

And yes, @KingGeorge, we can find a pattern and just assert that this is the square root.

Answer 26

@mathslover Try math.stackexchange

Answer 27

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.

Answer 28

:-(

Answer 29

Shubham is here to save the day :-)

Answer 30

I think I have a proof through direct multiplication (by hand). It's REALLY ugly, but it works.

Answer 31

It also relies on a weird multiplication trick that's kind of difficult to explain.

Answer 32

And some notationally difficult proofs by induction.

Answer 33

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

Answer 34

phew!

Answer 35

You can easily verify by putting n=1,2 etc , I just did, it matches! ^_^

Answer 36

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?

Answer 37

Which step sir ?

Answer 38

Wait, nevermind. I think I see why.

Answer 39

In this case we dont even need to prove for all n.
JUST prove when n=1997

Answer 40

when n=1997
2*10^n +1=200000.....00001
Since sum of digits is 3 it is divisible by 3

Answer 41

The proof that I have, while terribly messy and involving obscure multiplication tricks, is specific to this n, but easily generalized to any \(n\ge2\).

Answer 42

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

Answer 43

If requested, I can type up most of my proof tomorrow.

Answer 44

Hmmm

Answer 45

Plz @KingGeorge

Answer 46

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.

Answer 47

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.

Answer 48

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.