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Prove that the numbers 49, 4489, 444889, ... obtained by inserting 48 into the middle of the preceding number are squares of integers.

2007-04-05 11:39:45 · 4 answers · asked by Nidhogg 1 in Science & Mathematics Mathematics

4 answers

let a(n) n th number = 4(ntimes)8(ntimes) + 1
= 10^n(4 nitmes) + (8 ntimes) + 1
= 10^n*4/9(9 n times ) + 8/9(9 ntimes ) + 1
= 10^n*4/9(10^n-1) +8/9(10^n-1) +1
= 4/9(10^n(10^n-1) + 2(10^n-1)) +1

let 10^n-1 = t

= 4/9(t(t+1) + 2t) + 1
= 4/9(t^2 + 3t) + 1
= 4/9(t^2+3t+(9/4))
= 4/9(t+3/2)^2
= (2/3)^2(10^n+1/2)^2
= (1/3)^2(2*10^n+1)^2

now 2*10^n+1 mod 3 = 0

so we get = ((2*10^n+1)/3)^2
which is a perfect square

2007-04-05 19:48:34 · answer #1 · answered by Mein Hoon Na 7 · 2 1

(I got it! It was tricky though.)

For terminology, let the k'th number be N(k), and note it has 2k digits.

General formula for N(k):
N(k) = 4*10^(2k-1) + 4*10^(2k-2) +... + 4*10^(k) ...
+ 8*10^(k-1) + 8*10^(k-2) + ... + 8*10^(1) + 9

I can see several different ways to attack it:
a) Maybe the most promising is the simplifying observation that:
M(k) = [2*10^k + 1 ]/ 3
and this gives us the closed-form and recurrence relations for the sequence
M(k) = 10*M(k-1) - 3


b) Prove by induction or direct proof that
the number M(k) = 6...67^2 where the 6 digit is continued (k-1 times)
c) Prove by induction or direct proof that the number N(k) is a square (don't actually need to show what it is the square of).

So here's a)
M(k) = [2*10^k + 1 ]/ 3
M(k) = 10*M(k-1) - 3

Inductive step:
M(k-1) is true i.e. [ M(k-1) ]² = N(k-1) as defined above
Write √N(k-1) = M(k), and note it is always an integer

Then:
M(k) = 10*M(k-1) - 3
= 10*√N(k-1) - 3

So:
[ M(k) ]² = [ 10*M(k-1) - 3 ]²
Noting:
[ M(k) ]² = [ 10*√N(k-1) - 3 ]²
and since we know √N(k-1) is an integer, this easily proves that M(k) is a square.
To prove that it is the square we want, is slightly harder:

[ M(k) ]² = [ 10*M(k-1) ]² - 60*M(k-1)+ 9
[ M(k) ]² = 100* [ M(k-1) ]² - 60*M(k-1)+ 9
[ M(k) ]² = 100*N(k-1) - 60*M(k-1)+ 9
[ M(k) ]² = 100*N(k-1) - 60*[2*10^k + 1 ]/3 + 9
[ M(k) ]² = 100*N(k-1) - 4*10^(k+1) -20 + 9

...
"100*N(k-1)" term means the digits of N(k-1) are all shifted 2 left, except for...
the (k+1)'th ..2nd digits, which is reduced by 4 (*10^(k+1))
the 1st ("tens") digit is reduced by 20
the 0th ("ones") digit becomes 9

(you can crunch it all out inductively by letting the decimal digits of
N(k) = d_k, d_(k-1),.. d_0 )

____________________________________

Other not-so-elegant tries: as to doing b):
M(k) = 6...67^2 where the 6 digit is continued (k-1 times)
= Σ_i=k-1_i=0 [ 6*10^i ] + 1

Then [M(k)]² =
= ( Σ [ 6*10^i ] + 1 )²
= ( Σ [ 6*10^i ] )² + 2 Σ [ 6*10^i ] + 1
= ( Σ [ 6*10^i ] Σ [ 6*10^i ] ) + Σ [ 12*10^i ] + 1
= ( Σ [ 6*10^i ] Σ [ 6*10^i ] ) + Σ [ 12*10^i ] + 1 ...

probably it can be done directly, but induction looks easier:

Inductive assumption at step k:
[M(k)]² = [ 6*10^(k) + M(k-1) ]² = [6...67]²
and:
M(k+1) = 6*10^(k+1) + M(k) ; for k>=1
Then:
[M(k+1)]² = [ 6*10^(k+1) + M(k) ]²
= [ 6*10^(k+1) + M(k) ]²
= 36*10^(2k+2)) + 2M(k)*6*10^(k+1) + [ M(k) ]²
= 36*10^(2k+2)) + 12*10^(k+1)*M(k) + [ M(k) ]²

Plug in [M(k)]² = [ 6*10^(k) + M(k-1) ]²
= 36*10^(2k+2)) + 12*10^(k+1)*M(k) + [ M(k) ]²

and so on.

2007-04-05 20:16:09 · answer #2 · answered by smci 7 · 0 1

Let us look at the case of 8 digits, so that we have

(6666)² + 2(6666) + 1 = (4444)(10002) + 1

which happens to be true. Note that the left side is a perfect square. Now, after eliminating 1 from both sides, the left side can be re-expressed as:

(9999)(4444) + (3)(4444)

So, dividing both sides by 4444 gets us:

9999 + 3 = 10002

which is obviously true. It's easy to generalize this for larger numbers of 10, 12, 14, etc. digits.

2007-04-05 21:23:39 · answer #3 · answered by Scythian1950 7 · 0 1

Tried to work on the problem but got nowhere. Tried to find a relationship between the numbers in the sequence, and got

7^2, 67^2, 667^2, 6667^2, and so forth.

Tried to relate a term with its previous term and create a sequence from it, but failed. Excellent question though! That's why I starred it so I can find out the answer.

2007-04-05 19:06:55 · answer #4 · answered by Puggy 7 · 0 1

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