# How old are the kids?

Two men are sitting in the park:

-I have 2 kids under school age. - one said.

-How old are they? - Asked the other.

-The product of their ages is equal to the number of pigeons that you see around us.

-This information is insufficient for me.

-The elder has green eyes.

-Now I see.

How old are the kids? (assuming a child can start school being 6 )

The children are aged $4$ and $1$.

$4$ is the only number which can be a product of two different combinations of two numbers below $6$, where one of these combinations consists of two identical numbers.

We need two combinations because knowing the product of their ages wasn't enough information.

One of these combinations has to consist of two identical numbers because the information that there is one child that is older than the other one has to eliminate that combination.

The only product this can refer to is $4$ with the combinations $2\times2$ and $1\times4$.

• the only way to get 20(as in your example) as a product would be with 4 and 5 in which case the second man would have already known the answer after counting all the pidgeons. – The Dark Truth Oct 1 '15 at 8:47
• But what about non-identical twins (older by a few minutes) or one being 2 years and zero months old, the other 2 years and 11 months? – Ken Y-N Oct 1 '15 at 9:34
• In that case I would say that they are the same age and not one is older than the other(people usually count age in years). And the problem would not be solvable otherwise. – The Dark Truth Oct 1 '15 at 9:40
• @TheDarkTruth If two siblings (or any other two persons) are born close enough to have the same 'numeric' age, it is IMHO very reasonable to refer to the elder of the two person as 'the elder'. – jarnbjo Oct 1 '15 at 9:43
• @jwodder, I know it's not your puzzle, but does this assume that one twin is not older than the other? – user1717828 Oct 1 '15 at 14:15

The logic offered here follows the deduction process.

The product must be some number $1 \leq p \leq 25$, where $p = a_1 a_2$ with $1 \leq a_1 \leq a_2 \leq 5$. The $a_i$ are the ages.

If $p$ were prime, then we'd have $a_1 = 1$ and $a_2 = p$ so $p$ may not be prime, because the product information is insufficient. We thus know $p$ may not be $2,3,5,7,11,13,17,19,23$. We may also exclude product values that, when factorized, necessarily include a factor greater than 5. This way, we also exclude $14,18,21,22,24$.

Because the product value $p$ is at first insuficcient, it must admit more than one factorization in factors no greater than 5. This excludes $6, 8, 9, 10, 12, 15, 16, 20, 25$. The only possibility is thus $4 = 2\times 2 = 4\times 1$. Since there must be an elder, they may not be equal factors, and the ages are $1$ and $4$.

EDIT: Welp, a little too late.

• nice explanation, sorry that only 1 answer can be accepted (so i'll keep the first one) – Yavpolnom Shoke Oct 1 '15 at 7:44
• Correct answer, but I think incomplete: There could be zero pigeons (p=0). In that case the 'other person' would only know after the first hint that one of them is 0 years old, and has no clue about the other's age. Of course, given that after the second hint the 'other person' knows the ages, this possibility is then excluded. – user10445 Oct 1 '15 at 9:28

Perhaps a simpler way to figure this out:

Knowing the product of the two ages was not enough for the second man to deduce the ages, which means the product has at least two distinct factorizations using numbers less than 6.

Once he knew that the ages were different (one child is the elder), he could figure it out, so obviously one of the factorizations was into 2 identical numbers. That means that the product of the ages is a square number.

Knowing this, it is trivial to check the squares of numbers less than 6 to determine which of them has a second factorization:

$1^2 = 1 \leftarrow$ no other factorization
$2^2 = 4 \leftarrow$ can also be written as $1\times4$
$3^2 = 9 \leftarrow$ no other factorization using numbers under 6
$4^2 = 16 \leftarrow$ no other factorization using numbers under 6
$5^2 = 25 \leftarrow$ no other factorization using numbers under 6

Thus the product is obviously $4$, and the children are $1$ and $4$ years old.

That the father had to imply one is older than the other means the product is a perfect square that could also be written as the product of 1 and itself. Even then it has to be less than 6, so it can only be 4 while allowing 2 solutions at first (2*2 and 1*4). The elder one is 4 and the other is 1.

Green is the fourth color in the order 'VIBGYOR'.

So the age of the elder one is 4 and other kid is 1.

• Can you explain your thinking? – AJL Oct 1 '15 at 10:30
• Ah, the rainbow colour acronym, nice approach! Though you haven't explained where 1 comes from... – ʰᵈˑ Oct 1 '15 at 10:40
• Also, who orders the colors of the rainbow this way? Has nobody introduced you to our friend, Roy G. Biv? – dberm22 Oct 1 '15 at 12:38
• @dberm22 - green is #4 in either direction... – Floris Oct 1 '15 at 15:04
• @Floris I'm aware, but it's the principle of the matter – dberm22 Oct 1 '15 at 20:06