Attention math geniuses: Solve these age puzzles and include the solution.
A teacher (L) is talking to his colleague (K):
L: "You have three children. How old are your children?"
K: "If you add up their ages (in whole numbers), you get 13."
L: "That doesn't answer my question."
The colleague tells the teacher the product of the three ages. This is greater than zero.
L: "Even with that, I can't determine the ages of your three children."
K: "My oldest child is at least 1 year older than any of his siblings."
L: "Ahhh. Now I know how old your three children are."
Determine the ages of the three children.
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We know: The sum of the three figures is 13.
Now there is the product we do not know, we only know that it is not 0, that is, none of the children is 0 years than.
We have the following combinations:
11, 1, 1
10, 2, 1
9, 3, 1
9, 2, 2
8, 4, 1
8, 3, 2
7,5,1
7,4.2
7,3,3
6.6.1
6,5,2
6,4,3
5,5.3
5,4,4
Now L says that by specifying the product, it cannot determine how old the children are. We look at the product for the above combinations:
11, 1, 1, 11
10, 2, 1: 20
9, 3, 1: 27
9, 2, 2: 36
8, 4, 1:32
8, 3, 2:48
7,5,1.35
7,4,2:56
7,3,3: 63
6.6,1:36
6,5,2:60
6,4,3:72
5,5,3:75
5,4,4: 80
So if the product was, for example, 48, then L now knows that the children are 8,3,2 as, because only this combination gives 48. It must therefore be a combination in which the product is not unambiguous, so the product must occur more than once. This is only the case for 36, because only for 36 there are two possible combinations, namely 9,2,2 and 6,6,1. With the last info (the oldest child is at least one year older than the two siblings) you now know that the children are 9, 2, 2 years old.
We know that only through the additional information “there is a clearly oldest child” the task was clearly solved. In this way, there was a conceivable solution in which there were at least two equal “oldest” children. There are not so many of them, because these oldest children must not be older than 6 (because 7+7 would be more than 13):
And that’s it. Price question: Which of the two combinations was conceivable? It must be one in which there is another combination in which the same product and the sum 13 come out.
Consider this for the second case: 75 = 5 * 5 * 3 is already the prime factor decomposition. The only other way to get to the same product with 3 factors is when a child has age 1. Then the other ages would have to picture the missing prime factor:
But in all these cases, the oldest child would be too old to allow a sum of 13.
That is, the product is 36.
The age of each child is therefore a divider of 36. We think that the oldest child must be at least 5 years old (because 4 + 4 + 4 would be too small). Conversely, age must not be above 13. The following dividers are left for the oldest child:
12 or 9 or 6.
12 is too high, only the solution (12,1,0) would allow the sum 13 but its product is not 36.
Let us think about 6 and the possibilities to get to the sum 13: (6.6,1) is not the solution because there is no clear oldest child. (6,5,2) and (6,4,3) do not go because their product is not 36.
Well, there’s no more. So the oldest child is not 6.
Therefore, the oldest child is 9. We distribute the remaining factor of 4 to the other two children:
The task is not reproduced correctly or not completely, as there is no clear solution:
1+2+10, 2+4+7, …
Yes. There is a clear solution. The statement by L that, after the indication of the product, it does not know how old the children are, is important – so the product must not be unambiguous, because otherwise it could derive this from it. And from this knowledge one can find the clear solution.
However, other solutions are possible. What is missing is the product a.
It’s all about paying
would also be a valid solution.
Ah, okay, that’s right.
You have the additional requirement:
With the product, in all valid combinations with sum 13, it would only be in difficulty with product 36, since the combinations “9 2” and “6 6 1”. The latter is then excluded by the rule that the oldest is at least one year older than the others.
Thus the correct solution would be:
No. If 11.1 were the solution, the product would be 11. Then L could have concluded from the indication of the product that the children 11, 1 and 1 are, because there is no other combination of three natural numbers whose sum is 13 at which the product 11 is. However, L could not close it, so is 11, 1, 1 NO valid solution.