My regular puzzle...

Well I've been pretty busy recently and I haven't been on the board as much as I would like - particularly to answer some of the questions recently posted.

But anyway, I received this puzzle from someone the other day. Take a look:

An absent-minded mathematician has been asked to call at a house in a street (in which the houses are numbered consecutively from 1 upwards in the usual fashion), only he has forgotten its number. What he does remember, though, is that the sum of the house numbers before this house happens to be equal to the sum of the house numbers after this house, and also that the house he is looking for has a two-digit number.

What is the number of the house he is looking for? And how many houses are there in the street?

Only pencil and paper are permitted.

I don't see an easy way around it yet, but ultimately it comes down to plugging various values of k, 9 < k < 100, k being the address, in the following equation for n, where n is the number of houses on the street.

http://www.vbforums.com/attachment.php?s=&postid=937992

I do know the answer, but you didn't allow calculaters, so I won't say a word.

But, let me play with the Radical a little more, see if I can factor it, chunk it up a little.

I cheated as well, used excel.

An interesting puzzle, if the mathematician is unsure whether it was two or three digits in the house number then I can offer him just one other address to check.

Good work NotLKH, you've got the right equation but there is a big shortcut. Try seeing what happens if K is made up of two factors p and q so n=pq.

I'll post the solution soon if no-one finishes it off.

Starman, it'd be interesting to find a rough relationship between number of digits and possible solutions?

2 for 4 digits

1 about half way up the fives and then excel ran out of lines - I didn't go any further.

The uniqueness is interesting and I'm looking forward to your solution.

Finished!
With only Pen & Paper,
the Address = 35, and there are 49 houses on the street.

Created a proof for the generic equation seen in the beginning of this image, and the rest explains the answer.
http://www.vbforums.com/attachment.php?s=&postid=943056

And, attached in the zip is a pdf version of My proof of the validity of the series expression given above.

As NotLKH put it...

N(N+1) / 2 = K^2

is the derived condition.

Now since N and K are both integers and N(N+1)/2 is a perfect square...

Either N or (N+1) must be Perfect square.

So the easiest way to find N and K

is choose the perfect squares ie.. 4,9,16,25,36,49,64,81...

add 1 or subtract 1 and multiply with it and divide it by 2, if it is a perfect square you got a valid door number.

Say suppose you chose 9, then find out

if [(9-1)*9 / 2] or [9*(9+1)/2] is a perfect square...

in this case (8 * 9)/2 = 36 is a perfect square.

so Door number is Sqrt(36) = 6


Next if we chose 16

Neither of ((16-1)*16)/2 and (16* (16+1) ) /2 is a perfect square..so we skip them...


When we choose 49

49 * (49+1) / 2 = 1225 = 35^2 is perfect square

so we have a valid door number = 35 (2-digit)


If you need to calculate more ..here is the vb code

VB Code:

Dim i As Long
Dim pre As Double
Dim pos As Double
 
For i = 2 To 100000
    pre = Sqr(((i ^ 2) - 1) * (i ^ 2) / 2)
    pos = Sqr((i ^ 2) * ((i ^ 2) + 1) / 2)
    If pre = Int(pre) Then
        Debug.Print "Door Number:" & pre & " , No. of Houses: " & (i ^ 2) - 1
    ElseIf pos = Int(pos) Then
        Debug.Print "Door Number:" & pos & " No. of Houses: " & i ^ 2
    End If
Next

Have fun !!
:D

TT,
Heres one for you,
using only paper and pen, with a simple calculator, but no programming,

Find the first 5 digit whole value of Y such that X is a whole value, where X and Y satisfy
45X^2 + 4 = Y^2

-Lou

Well done both NotLKH and TT, I'm really impressed! Can I have a go at that prob NotLKH?

Go ahead! Its really pretty simple. I'm pretty sure it would take a
little while with TT's method, but I'd like to see how you do it.

:)