### Brain Teasers

# Mad Ade's House Number

Mad Ade lives on a long street In Sunny Madadia and he has noticed that the sum of the house numbers up to his own house, "The Kebabers rest", but excluding it, equals the sum of the numbers of his house to the end of the road. If the houses are numbered consecutively, starting from 1, what are the possible numbers of Mad Ade's house?

There are less than 1000 houses on the road.

There are less than 1000 houses on the road.

### Answer

There are four possible answers if the total number of Houses is no more than 1000.Mad Ade's house could be #3 out of 3, #15 out of 20, #85 out of 119, or #493 out of 696.

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## Comments

How about #35 out of 49?

I don't think so Jack. Mad, I voted this one serious fun but judging from the average response both of us must be enjoymentally challenged. How's this for a solution. Since the sum to nextdoor equals the sum from Kebabpalace to the end then the first sum is half the total street sum. If your house is house a and there are b houses in the street then 2(0.5(a-1)(a) = 0.5(b)(b+1) by series summation formulae. Rearranging as a quadratic equation gives 2a^2 -2a-b(1+b)=0. Solving for a by the quadratic formula gives a=0.5(1+/- sqrt(1+2b(b+1))).

To have an integer solution 2b^2+2b+1 must be a square number. Thus by trial and error when b= 3 for example it becomes (18+6+1) = 25 so a=0.5(1+5) = 3 also. Further trial and error on the above expression will find other integer solutions.

To have an integer solution 2b^2+2b+1 must be a square number. Thus by trial and error when b= 3 for example it becomes (18+6+1) = 25 so a=0.5(1+5) = 3 also. Further trial and error on the above expression will find other integer solutions.

Nice teaser! Really nice start on the analysis Jimbo!

What you're looking for is a pair of triangular numbers x and y where 2x = y. The house number is 'a' where t(a) = x is a triangular number and the number of houses on the street is 'b' where t(b) = y is also a triangular number.

t(n) = n*(n+1)/2

As Jimbo shows, this gives:

2 * (a * (a-1)/2) = b * (b-1)/2

or

2a * (a-1) = b * (b-1)

Jimbo then solves for a:

a = 0.5(1+/- sqrt(1+2b(b+1)))

Here you can simplify to only take the positive root:

a = (1 + sqrt(1 + 2b * (b+1))) / 2

With a fair amount of effort, you can take this further to generate the nth value for a as:

a(n) = (((1+sqrt(2))^(2n-1) - (1-sqrt(2))^(2n-1)) / sqrt(+1) / 2

Easier is to generate the next value for a(n) from the previous values as:

a(n) = 6*a(n-1) - a(n-2) - 2

where a(1)=3 and a(2)=15.

a(3) = 6*15 - 3 - 2 = 85

a(4) = 6*85 - 15 - 2 = 493

a(5) = 6*493 - 85 - 15 - 2 = 2871

a(6) = 6*2871 - 493 - 85 - 2 = 16731

...

A nearly identical generating function can be used to generate the values of b:

b(n) = 6*b(n-1) - b(n-2) + 2

where b(1)=3 and b(2)=20.

b(3) = 6*20 - 3 + 2 = 119

b(4) = 6*119 - 20 + 2 = 696

b(5) = 6*696 - 119 + 2 = 4059

b(6) = 6*4059 - 696 + 2 = 23661

...

What you're looking for is a pair of triangular numbers x and y where 2x = y. The house number is 'a' where t(a) = x is a triangular number and the number of houses on the street is 'b' where t(b) = y is also a triangular number.

t(n) = n*(n+1)/2

As Jimbo shows, this gives:

2 * (a * (a-1)/2) = b * (b-1)/2

or

2a * (a-1) = b * (b-1)

Jimbo then solves for a:

a = 0.5(1+/- sqrt(1+2b(b+1)))

Here you can simplify to only take the positive root:

a = (1 + sqrt(1 + 2b * (b+1))) / 2

With a fair amount of effort, you can take this further to generate the nth value for a as:

a(n) = (((1+sqrt(2))^(2n-1) - (1-sqrt(2))^(2n-1)) / sqrt(+1) / 2

Easier is to generate the next value for a(n) from the previous values as:

a(n) = 6*a(n-1) - a(n-2) - 2

where a(1)=3 and a(2)=15.

a(3) = 6*15 - 3 - 2 = 85

a(4) = 6*85 - 15 - 2 = 493

a(5) = 6*493 - 85 - 15 - 2 = 2871

a(6) = 6*2871 - 493 - 85 - 2 = 16731

...

A nearly identical generating function can be used to generate the values of b:

b(n) = 6*b(n-1) - b(n-2) + 2

where b(1)=3 and b(2)=20.

b(3) = 6*20 - 3 + 2 = 119

b(4) = 6*119 - 20 + 2 = 696

b(5) = 6*696 - 119 + 2 = 4059

b(6) = 6*4059 - 696 + 2 = 23661

...

Of course where it displays

a(n) = (((1+sqrt(2))^(2n-1) - (1-sqrt(2))^(2n-1)) / sqrt(+1) / 2

in my previous post, I actually typed:

a(n) = (((1+sqrt(2))^(2n-1) - (1-sqrt(2))^(2n-1)) / sqrt( 8 )+1) / 2

a(n) = (((1+sqrt(2))^(2n-1) - (1-sqrt(2))^(2n-1)) / sqrt(+1) / 2

in my previous post, I actually typed:

a(n) = (((1+sqrt(2))^(2n-1) - (1-sqrt(2))^(2n-1)) / sqrt( 8 )+1) / 2

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