[rookie1ja]

Complex Deduction - Back to the Number Puzzles
This is definitely one of the harder number puzzles on this site.
A teacher says: I'm thinking of two natural numbers greater 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?

Spoiler for Solution:

Savoury - 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.

Spoiler for old wording:

This is definitely one of the hardest number puzzles on this site.
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?

[shiang]

lol, this one brings back memories and I figured this out in like 10 min when I was a sophmore in HS (teacher was preaty impressed), I actually found this one preaty eazy as its a process of elimination. The 2nd stament "I knew that, and the sum is less than 14" forces the sum to be 11 because it is the only number where all the sets of the sums have have to have 1 non-prime number and have more than one possible set of sums.

2-9
3-8
4-7
5-6

this means the product has to be 18,24,28, or 30 and of those numbers the only one where all possible sums of two numbers gives a product less than 14 is 18.

2+9 < 14
2+12 = 14
2+14 > 14
2+15 > 14

so the product is 18 and the sum is 11, the two numbers are 2 and 9.

I think my solution is slightly more elegant, ...ok my ego got the best of me again.

*if two natural numbers x*y = K and x<y then (x+1)(y-1) >= K because if multiplied out it would be xy+y-x-1 so if x<y then y-x-1 >= 0. I just figured this on the spot so I thing it's right but I'm not 2 sure if there are exceptions. Keep in mind the initial conditions.

[shiang]

accidental repeat

[milkplus71]

was teacher preaty impressed with your spelling?

[walkingagh]

I came up with the same solution as you did. Your equation with K was interesting, but trivial.

The largest the product of any two numbers that sum to X can be is if each of the numbers is x/2. Geometrically this is saying that a square has more area per outer edge than a non-square rectangle. Your statement is a corollary to this much more powerful theorem.

[eikonoklaste]

hahaha... right; one of the hardest on the site, please.

[stormpool]

Read the question again and assume that each students responds to the other instantly as I thought.

They are geniuses..

[vebshid]

"3 10 ... NO - the product does not have all possible sums smaller than 14"
++++++++++++++++++++++++++++++++++++++++++++++
Why the the product should have all possible sums smaller than 14?
and
what's the meaning of "First: I knew that. However, now I know the numbers."
what does the First "knew"? I can't understand from the text, maybe that's because i'm not a English native people~

[1stone]

You don't have to take all combinations to work this one out. It is based on the following:

If a number is the product of two primes, then you know their sum.

Clues:
A: I don't know the sum. So the number is not the product of two primes. One of the numbers must be 4, 6, 8, 9 or 10.

B: I knew that. This is the biggest clue! If the sum can also be a sum of two primes, then B would not be absolutely sure that A does not know the numbers eg if B's sum was 6, then it could be that A's product is 9=3x3, from which A could immediately deduce the two numbers. So the fact that B knew that A did not know the sum limits the sum to be 11. Every other number under 14 is the sum of two primes: 4=2+2, 5=2+3, 6=3+3, 7=3+4, 8=3+5, 9=2+7, 10=3+7, 12=5+7, 13=2+11.

So the possibilities are 2 9, 3 8, 4 7, or 5 6.

B: The sum is less than 14
A: I knew that. This implies that the product cannot be factored out into two numbers whose sum is larger or equal to 14. Now the largest sum that the factors can add to is when we take a large and a small factor. This eliminates 3x8=24=2x12, which allows 2+12=14;v or 4x7=28=2x14 with 2+14>14; or 5x6=30=2x15, 2+15>14; leaving only 2x9=18=3x6. So the numbers are 2 and 9.

[hansonjw]

Can someone explain to me why it is not 3 and 4?