[MOVED] What number does Mr Jones live at, and...?

[cclaire]

Mr Jones lives on a long street, the sum of numbers up to his own house, but excluding it, equal the sum of numbers of his house to the end of the street. If the houses are numbered consecutively from 1, what number does Mr Jones live at and how long is the street?

Found this in a book which gives the solution but not how to get to it - any help??

 

[stapel]

Found this in a book which gives the solution but not how to get to it - any help??

What have you attempted so far? This clearly involves arithmetic series. How have you applied those formulas?

Please be specific. Thank you!

Eliz.

 

[cclaire]

Yeah I've applied arithmetic series coming up with the equation:

2n(n-1) = m (m+1) where n = mr jones house number, m = number of houses in street.

Obviously I've too many unknowns, so I need another equation to solve with this one..... just not sure what it would be.

 

[cclaire]

That's sort of how I feel - that there must be info missed out the question - but the question is straight from a book. Just thought I might be over looking something. Never mind

 

[soroban]

Hello, cclaire!

Mr Jones lives on a long street.
The sum of numbers up to his own house, but excluding it,
equal the sum of numbers of his house to the end of the street.
If the houses are numbered consecutively from 1,
what number does Mr Jones live at and how long is the street?

There is nothing wrong with the problem nor with your equation:

We are expected to solve the Pellian equation: \(\displaystyle \:2n(n\,-\,1) \:=\:m(m\,+\,1)\)
. . where \(\displaystyle m\) and \(\displaystyle n\) are positive integers.

Without going into the procedure, here are the answers: \(\displaystyle \:n \,=\,15,\;m\,=\,20\)


Therefore: there are 20 houses on this street and Mr. Jones lives at #15.

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Check

\(\displaystyle \begin{array}{ccc}1\,+\,2\,+\,3\,+\,4\,+\,\cdots\,+\,14 & \;=\;& 105 \\ \\ \\
15\,+\,16\,+\,17\,+\,18\,+\,19\,+\,20 & \;=\; & 105\end{array}\)

 

[Denis]

Street can be real short: 1 2 3
Jones lives at 3; numbers before = 1+2 = 3; numbers from his and on = 3

Can be a little longer:
1,2,3,4,5,6,7,8,9,10,11,12,13,14,[15:Jones],16,17,18,19,20
1 to 14 = 105; 15 to 20 = 105

If Jones' number is to be excluded entirely: 1,2,3,4,5,[6],7,8 ; 15 each side

As already indicated, you need more info to make answer unique.

 

[cclaire]

Thanks guys

Googled Pellian eqns and I'm learning - however I still feel there's some trial and error involved in determining the square number soln inside my square root - is there a more sensible way than using pythagorus on

2(m^2)+2m+1=l^2 where l = integer

 

[Denis]

however I still feel there's some trial and error involved in determining the square number soln inside my square root - is there a more sensible way than using pythagorus on

2(m^2)+2m+1=l^2 where l = integer

What are you asking? What's Big Pete Pythagoras got to do with this?
What's a "square number soln inside my square root"?

You know that 2(m^2)+2m+1 is odd, right?

Changing your | to x:
2m^2 + 2m + 1 - x^2 = 0; quadratic formula gives:
m = [sqrt(2x^2 - 1) - 1] / 2

First 4 solutions are m = 3,20,119,696.