Answer Posted / guest

There are 288 houses and Jessy's house number is 204.

Let's assume that in the "Friends Society" there are total N
houses numbered from 1 to N and Jessy's house number is X.

Now it is given that all the house numbers on the left side
of Jessy's house add up exactly the same as all the house
numbers on the right side of her house. Hence,

1 + 2 + 3 + ..... + (X-1) = (X+1) + (X+2) + (X+3) + ..... + N

Both the sides of the above equations are in A.P. Hence,
using A.P. summation formaula,

[(X-1)/2][2*(1) + (X-1-1)] = [(N-X)/2][2*(X+1) + (N-X-1)]

[X-1][(2) + (X-2)] = [N-X][(2X+2) + (N-X-1)]

(X-1)(X) = (N-X)(N+X+1)

X2 - X = N2 + NX + N - NX - X2 - X

X2 = N2 + N - X2

2X2 = N2 + N

X2 = (N2 + N)/2

X2 = N(N+1)/2

Now, using Trial and Error method to find values of N and X
such that above equation is satisfied, we get

1. N = 8, X = 6

2. N = 49, X = 35

3. N = 288, X = 204

4. N = 1681, X = 1189

5. N = 9800, X = 6930

But we require minimal possible answer and it is given that
John's house number is 109. It means that there are atleast
109 houses. Hence, first two are not possible. And the
answer is : there are 288 houses and Jessy's house number is
204.

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