"The House where she lives"

[zenith20]

"The House where she lives"

It was at a cocktail party in New York that I met Stephanie. We exchanged our phone numbers and decided to meet each other soon.
When she rang up and invited me to her house this is how she gave me the number of her house:
'I live in a long street. Numbered on the side of my house are the houses one, two, three and so on. All the numbers on one side of my house add up to exactly the same as all the numbers on the other side of my house. I know there are more than fifty houses on that side of the street, but not so many as five hundred.

Can you find Stephanie's house number?

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solution:
The numbers of the houses on each side will add up alike if the number of the house be 1 and there are no other houses, and if the number be 6 with 8 houses in all, if 35 with 49 houses, if 204 with 288 houses, if 1189 with 1681 houses and so on. But we know that there are more than 50 and less than 500 houses, and so we are limited to a single case.

The number of the house must be 204.
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i got the whole question and also the answer! but my problem is to get the formula from which it's concluded the number of the house and the total numbers in each case. (e.g. 204 with 288 houses)

thank you in advance,
Zenith
(p.s the puzzle is again one of Shakuntala Devi's)

 

[mmm4444bot]

The numbers 1, 6, 35, 204, 1189, … are called Perfect Meridians because the sums of the remaining elements on each side (in their respective sequences) are equal.

There is a recursive formula for the sequence of Perfect Meridians:

a[sub:1v625g42]n[/sub:1v625g42] = 6 * a[sub:1v625g42]n-1[/sub:1v625g42] - a[sub:1v625g42]n-2[/sub:1v625g42]

for n > 1

with a[sub:1v625g42]0[/sub:1v625g42] = 0 and a[sub:1v625g42]1[/sub:1v625g42] = 1

n = 2: a[sub:1v625g42]2[/sub:1v625g42] = 6 * 1 - 0 = 6

n = 3: a[sub:1v625g42]3[/sub:1v625g42] = 6 * 6 - 1 = 35

n = 4: a[sub:1v625g42]4[/sub:1v625g42] = 6 * 35 - 6 = 204

n = 5: a[sub:1v625g42]5[/sub:1v625g42] = 6 * 204 - 35 = 1189

n = 6: a[sub:1v625g42]6[/sub:1v625g42] = 6 * 1189 - 204 = 6930

n = 7: et cetera …

There's also a formula for the respective nth elements (1, 8, 49, 288, 1681, …).

a[sub:1v625g42]n+1[/sub:1v625g42] = 6 * a[sub:1v625g42]n[/sub:1v625g42] - a[sub:1v625g42]n-1[/sub:1v625g42] + 2

for n > 0

with a[sub:1v625g42]0[/sub:1v625g42] = 0 and a[sub:1v625g42]1[/sub:1v625g42] = 1

n = 1: a[sub:1v625g42]2[/sub:1v625g42] = 6 * 1 - 0 + 2 = 8

n = 2: a[sub:1v625g42]3[/sub:1v625g42] = 6 * 8 - 1 + 2 = 49

n = 3: a[sub:1v625g42]4[/sub:1v625g42] = 6 * 49 - 8 + 2 = 288

n = 4: a[sub:1v625g42]5[/sub:1v625g42] = 6 * 288 - 49 + 2 = 1681

n = 5: a[sub:1v625g42]6[/sub:1v625g42] = 6 * 1681 - 288 + 2 = 9800

n = 6: et cetera …

 

[mmm4444bot]

Using the famous formula for the sum of the first n Natural numbers, I derived another formula for the Perfect Meridians, in terms of the nth element in the sequence.

In other words, if it's known that a sequence of the first n Natural numbers contains a Perfect Meridian (M), then:

M = sqrt[n(n + 1)/2]

When n = 8: M = sqrt[8(8 + 1)/2] = 6

When n = 49: M = sqrt[49(49 + 1)/2] = 35

When n = 288: M = sqrt[288(288 + 1)/2] = 204

et cetera.

 

[zenith20]

Dear mmm4444bot, that was SO AMAZING!
thank you soooooooooo much for such a formula, now i'm more curious to know how to get this formula! any detailed proves for that?
i'd be really grateful for that

thanks again and again

 

[mmm4444bot]

I do not know who discovered Perfect Meridians or how the recursive formulas were derived.

You will learn about stuff like this, if you study Number Theory.

For the formula that I dervied for M in terms of n, I used the summation formula for the first n Natural numbers:

Sum = n(1 + n)/2

For the sequence of n elements containing a Perfect Meridian M, we have:

Left Sum = (M - 1)(1 + M - 1)/2

Right Sum = (n - M)(M + 1 + n)/2

Set these sums equal, and solve for M.

 

[zenith20]

i know a bit about recursive functions in programming, and i'm so eager to learn more about them.

anyway, my friend, you're one of the BESTs

 

[mmm4444bot]

I like patterns and phenomena within the Naturals, too.

Who would have thought (at first glance), that adding up all the house numbers and taking the square root of their sum would provide the Perfect Meridian ?

Neat !

 

[zenith20]

So, without knowing about the recursive formula, one can solve such a problem like this:
starting from n=1 and checking the following formula for M which has to be an Integer:
M = sqrt[n(n + 1)/2]

then those Perfect Meridians would be dervied and the correspondent n's would be the number of the total houses then, yes?

 

[mmm4444bot]

Not all Natural-number sequences contain a Perfect Meridian.

The formula that I derived for M requires that you first know n.

In other words, if you know in advance that the sequence of the first 288 Natural numbers contains a Perfect Meridian, then you can set n = 288 in that formula and evaluate M = 204.

If you do not know a suitable value for n, then you'll need to use the recursive formula (or some other method) to first find one.

 

[zenith20]

Thank you very much for this final tip

 

[soroban]

Hello, zenith20!

I solved this quickly, but only because I knew some trivia about Triangular Numbers.

It was at a cocktail party in New York that I met Stephanie.
We exchanged our phone numbers and decided to meet each other soon.
When she rang up and invited me to her house this is how she gave me the number of her house:
'I live in a long street. Numbered on the side of my house are the houses 1, 2, 3, and so on.
All the numbers on one side of my house add up to exactly the same as all the numbers on the other side of my house.
I know there are more than fifty houses on that side of the street, but not so many as five hundred."

Can you find Stephanie's house number?

Triangular Numbers: .\(\displaystyle T_n\)

. . \(\displaystyle \begin{array}{cccc} n && T_n \\ \hline 1 &= & 1 \\ 1+2 &=& 3 \\ 1+2+3 &=& 6 \\ 1+2+3+4 &=& 10 \\ \vdots && \vdots \\ 1+2+3+\hdots + n &=& \frac{n(n+1)}{2} \end{array}\)


Let \(\displaystyle S\) = Stephanie's house number.
Let \(\displaystyle N\) = number of last house on her street.


\(\displaystyle \text{We have: }\;\underbrace{1,\;2,\;3,\;\hdots\;,(S-1)}_{\text{one si\!de}},\;S,\;\underbrace{(S+1),\s+2),\;\hdots\;N}_{\text{other si\!de}}\)

\(\displaystyle \text{The sum of the numbers on one side is: }\;\frac{S(S-1)}{2}\) .[1]

\(\displaystyle \text{The sum on the other side is: }\;\frac{N(N+1)}{2} - \frac{S(S+1)}{2}\) .[2]


Equate [2] and [1]:

. . \(\displaystyle \frac{N(N+1)}{2} - \frac{S(S+1)}{2} \;=\;\frac{S(S-1)}{2} \quad\Rightarrow\quad \frac{N(N+1)}{2} \:=\:S^2\)

Hence, we have a Triangular Number which is also a Square.

These have the recursive form: .\(\displaystyle S_1 = 1,\;S_2 = 6,\;S_n \:=\:6S_{n-1} - S_{n-2}\)

Hence, we have the sequence: .\(\displaystyle 1,\;6,\;35,\;204,\;1189,\;\hdots\)


Therefore, Stephanie's house is #204
. . (and there are 288 houses on her side of the street.)

 

[zenith20]

Dear Soroban,
Thanks for this explanation, every detail helps me get to it more and more.