Three Daughters/Solution
Classic Riddles
One of my neighbours has $3$ daughters.
"How old are they?" I asked.
"The product of their ages is $36$," he replied.
I said, "That's not a great deal of help. Throw me a bone, here."
"Okay," he said, "the sum of their ages equals our house number."
I went outside to look at the front door to remind myself what that was, and said, "Nope, still can't work it out."
"My oldest daughter is the only one in the family who has green eyes."
"Aha," I replied, "got it. Their ages are ..."
and I told him.
Solution
The ages of the neighbour's three daughters are $9$, $2$ and $2$.
Proof
There are $8$ ways of selecting $3$ positive integers whose product is $36$:
| \(\ds 36\) | \(=\) | \(\ds 1 \times 1 \times 36\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 \times 2 \times 18\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 \times 3 \times 12\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 \times 4 \times 9\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 \times 6 \times 6\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \times 2 \times 9\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \times 3 \times 6\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 3 \times 3 \times 4\) |
Indeed, it is not possible to know what the ages are from just knowing the product of their ages.
Now we learn that the sum of their ages equals the house number at which they live.
Let us investigate what those house numbers could be, by examining the $8$ possibilities above:
| \(\ds 1 + 1 + 36\) | \(=\) | \(\ds 38\) | ||||||||||||
| \(\ds 1 + 2 + 18\) | \(=\) | \(\ds 21\) | ||||||||||||
| \(\ds 1 + 3 + 12\) | \(=\) | \(\ds 16\) | ||||||||||||
| \(\ds 1 + 4 + 9\) | \(=\) | \(\ds 14\) | ||||||||||||
| \(\ds 1 + 6 + 6\) | \(=\) | \(\ds 13\) | ||||||||||||
| \(\ds 2 + 2 + 9\) | \(=\) | \(\ds 13\) | ||||||||||||
| \(\ds 2 + 3 + 6\) | \(=\) | \(\ds 11\) | ||||||||||||
| \(\ds 3 + 3 + 4\) | \(=\) | \(\ds 10\) |
Having looked at the house number, I still could not tell the ages of his daughters.
For all the sums except for $13$, those sums are the only way you can make that number from $3$ positive integers whose product is $36$
So the house number must be $13$.
Finally, I was told some information about the oldest daughter.
If the ages were $1$, $6$ and $6$, technically speaking there is no "oldest" daughter, as the $6$-year-olds are twins.
So for there to be an "oldest daughter", ages must be $2$, $2$ and $9$.
$\blacksquare$
Quibbles
The assumption is that as they are both $6$ years old, the daughters of the $1$, $6$, $6$ family are twins.
However, by the convention that describing a child's age as $6$ means they can be anywhere between $6$ and $7$, it is possible that their ages are just-$6$ and nearly-$7$, and that the mother (poor woman) fell pregnant less than $3$ months after giving birth, or that the second child was severely premature.
Hence one of the $6$-year-olds could easily be referred to as the "oldest daughter".
Besides, it is of course quite possible for the neighbour to refer to the "older twin" as the "oldest daughter".