Children

An easier number puzzle is as follows. Two friends are chatting:
- Peter, how old are your children?
- Well Thomas, there are three of them and the product of their ages is 36.
- That is not enough ...
- The sum of their ages is exactly the number of beers we have drunk today.
- That is still not enough.
- OK, the last thing is that my oldest child wears a red hat.
How old were each of Peter's children?

Solution:
Let’s start with the known product – 36. Write on a sheet of paper the possible combinations giving the product of 36. Knowing that the sum is not enough to be sure, there are two possible combinations with the same sum (1-6-6 a 2-2-9). And as we learned further that the oldest son wears a hat, it is clear that the correct combination of ages is 2-2-9, where there is exactly one of them the oldest one.

A Riddle

Fifteen score and more I weigh
When reckoned by the brood of Re
But less than one score you would quoth
If counting true by rule of Thoth.

I was born by Father Greg
When Famous Bill to Anne bent leg
And hail�d by Rome so full of pride
But on my birth great Caesar died.

I keep account and deft compute
For every man, genteel or brute
But don�t correct me when I sum
And twelve of thirty�s less than one

I�ve been named for Gods and Kings
And kin cavort with Maya and Ming
But sheep and fish define my frame
Tell me now what is my name?

Answer:calendar

Old Riddle

Born long ago, but made yesterday.
and only employed when others sleep.
What few would wish to give away,
But none would wish forever to keep.

Answer: Bed

Roses

I have a number of roses for sale. The first buyer bought half of my roses then I gave him additional one for free. The second buyer bought half of the remaining roses then I gave him additional one also for free. The third buyer also bought half of the remaining roses then I gave him additional one also for free, this time all of my roses has been sold out.

How many roses do I have?

Solution: 14

Savoury

A teacher says: I'm thinking of two natural numbers bigger than 1. Try to guess what they are.
The first student knows their product and the other one knows their sum.
First: I do not know the sum.
Second: I knew that. The sum is less than 14.
First: I knew that. However, now I know the numbers.
Second: And so do I.
What were the numbers?

Solution:

The numbers were 2 and 9. And here comes the entire solution.
There shall be two natural numbers bigger than 1. First student knows their product and the other one knows their sum.
The sum is smaller than 14 (for natural numbers bigger than 1), so the following combinations are possible:
2 2 ... NO - the first student would have known the sum as well
2 3 ... NO - the first student would have known the sum as well
2 4 ... NO - the first student would have known the sum as well
2 5 ... NO - the first student would have known the sum as well
2 6
2 7 ... NO - the first student would have known the sum as well
2 8
2 9
2 10
2 11 ... NO - the first student would have known the sum as well
3 3 ... NO - the first student would have known the sum as well
3 4
3 5 ... NO - the first student would have known the sum as well
3 6
3 7 ... NO - the first student would have known the sum as well
3 8 ... NO - the product does not have all possible sums smaller than 14 (eg. 2 + 12)
3 9 ... NO - the first student would have known the sum as well
3 10 ... NO - the product does not have all possible sums smaller than 14
4 4
4 5
4 6 ... NO - the product does not have all possible sums smaller than 14
4 7 ... NO - the product does not have all possible sums smaller than 14
4 8 ... NO - the product does not have all possible sums smaller than 14
4 9 ... NO - the product does not have all possible sums smaller than 14
5 5 ... NO - the first student would have known the sum as well
5 6 ... NO - the product does not have all possible sums smaller than 14
5 7 ... NO - the first student would have known the sum as well
5 8 ... NO - the product does not have all possible sums smaller than 14
6 6 ... NO - the product does not have all possible sums smaller than 14
6 7 ... NO - the product does not have all possible sums smaller than 14

So there are the following combinations left:
2 6 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14 (it is impossible to create a pair of numbers from sum 8, so that the product would have an alternative sum bigger than 14 ... eg. if 4 and 4, then there is no sum – created from their product 16 – bigger than 14 – eg. 2 + 8 = only 10)
2 8
2 9
2 10
3 4 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14
3 6 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14
4 4 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14
4 5 ... NO – it is impossible to create any pair of numbers from the given sum, where there would be at least one sum (created from their product) bigger than 14

The second student (knowing the sum) knew, that the first student (knowing the product) does not know the sum and he thought that the first student does not know that the sum is smaller than 14.

Only 3 combinations left:
2 8 ... product = 16, sum = 10
2 9 ... product = 18, sum = 11
2 10 ... product = 20, sum = 12

Let’s eliminate the sums, which can be created using a unique combination of numbers – if the sum is clear when knowing the product (this could have been done earlier, but it wouldn’t be so exciting) - because the second student knew, that his sum is not created with such a pair of numbers. And so the sum can not be 10 (because 7 and 3) – the second student knew, that the first student does not know the sum – but if the sum was 10, then the first student could have known the sum if the pair was 7 and 3.
The same reasoning is used for eliminating sum 12 (because 5 and 7).
So we have just one possibility – the only solution – 2 and 9.
And that’s it.

Trains

A train leaves New York for Boston. Five minutes later another train leaves Boston for New York, at double the speed. Which train will be closer to New York when they encounter?

Solution: Of course, when the trains encounter, they will be approximately the same distance away from New York. The New York train will be closer to New York by approximately one train length because they're coming from different directions. That is, unless you take "meet" to mean "perfectly overlap".

Adam and Eve

Adam has none, Eve has two, Everyone else has five. What is it that they have?

Solution:

The Letter 'e' .....Adam has no 'e', Eve has two 'e's and everyone else has 5 'e's