Permalink Submitted by Anonymous on April 22, 2013

Assuming twins for simplicity: let the children's ages be x, x and y.
Clue 1: x.x.y=36
Or rewritten: y=36/(x^2) x^2 denotes x squared!
but 36 is a square number and can be written as 6.6 or 6^2.

Now we get y=6^2/x^2 = (6/x)^2
Get the square root of both sides: sqrt(y) = sqrt((6/x)^2) = 6/x or x=6/sqrt(y)
so the age of the eldest child, y>x and must be a square number i.e. 1,4,9,16,25,36,...
let y=1 then x=6 but this does not satisfy the eldest child clue (y>x) so discard.
let y=4 then x=3 accept. So 3+3+4=10 the number on the house next door.
let y=9 then x=2 accept. So 2+2+9=13 and there is no way anyone lives in a number 13 house so discard!
let y=16 then x=6/4 which is not a whole number.
let y=25 then x=6/5 which is not a whole number.
let y=36 then x=1 which is possible also but unlikely lets be honest!

So I reckon I have to lean towards the twins being 3 with the eldest being 4!

## lucky for some

Assuming twins for simplicity: let the children's ages be x, x and y.

Clue 1: x.x.y=36

Or rewritten: y=36/(x^2) x^2 denotes x squared!

but 36 is a square number and can be written as 6.6 or 6^2.

Now we get y=6^2/x^2 = (6/x)^2

Get the square root of both sides: sqrt(y) = sqrt((6/x)^2) = 6/x or x=6/sqrt(y)

so the age of the eldest child, y>x and must be a square number i.e. 1,4,9,16,25,36,...

let y=1 then x=6 but this does not satisfy the eldest child clue (y>x) so discard.

let y=4 then x=3 accept. So 3+3+4=10 the number on the house next door.

let y=9 then x=2 accept. So 2+2+9=13 and there is no way anyone lives in a number 13 house so discard!

let y=16 then x=6/4 which is not a whole number.

let y=25 then x=6/5 which is not a whole number.

let y=36 then x=1 which is possible also but unlikely lets be honest!

So I reckon I have to lean towards the twins being 3 with the eldest being 4!