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Last house on the street (Posted on 2022-05-31) Difficulty: 3 of 5
The great Dudeney, bless his soul, created, inter alia, the following puzzle:

There was a man who lived on a XXXX street numbered 1,2,3 … etc. on his side and all the following numbers on the other side of him. The numbers on both sides added to exactly the same amount. What was the number of the last house?

I have slightly changed the original wording and also took the liberty of coding a single 4 letter word of his text to avoid getting a trivial solution.
I ask for a solution which is over 45. The upper limit stays open. So:

a. Provide an answer (or answers) 46 and up.
b. What word was replaced by XXXX?

Bonne chance!

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution my solution | Comment 4 of 7 |
Might as well find as many solutions as possible,then weed out the ones disallowed:

I assumed "his side" was his side of the street. The side between 1 and him doesn't constitute the whole of his side. ... and his side includes his own house

for n=2:10000
    for i=1:n-1
       if sum(1:i)==sum(i+1:n)
           disp([i+1 n])
       end
    end
end

finds

  other side
  ----------
  start  end
     3     3
    15    20
    85   119
   493   696
  2871  4059
  
The trivial solution would have been 3 houses on the block, with 1+2 on one side totaling to house 3's number.

The case of 20 houses on the block with 1 through 14 on one side and 15 through 20 on the other is not really trivial, but it doesn't have the required minimum of 46 houses.

There could be 119, or 696, or 4059 houses on the street, with the first house on the "other" side numbered 85, 493, or 2871 respectively.

I would think the smaller numbers, especially the disallowed 20 houses on the street would be more realistic. There would be 14 houses on one side and 6 on the other on a street that was shaped like a U, with the smaller number on the inner side and a larger number on the outer side. The larger numbers would make that discrepancy less likely, with just a bend in the road.

We could have asked for i^2+i == (n^2 + n)/2 equivalently, as the LHS is twice the sum of the first i whole numbers and the RHS is the sum of the first n numbers.

I wouldn't know what the XXXX represents.

  Posted by Charlie on 2022-05-31 11:58:59
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