Could you explain "The sum of two of the numbers is equal to the third"??
Call the three numbers a, b, and c. This means c = a + b, but we still don’t know to which person each number belongs.
When person 1 (p1) is asked what his number is, he has no way to know whether he has a, b, or c, so he says he doesn’t know. Same goes for p2 and p3. Clearly p1 somehow gains information by p2 and p3 passing. Either he realizes that he must be either a or b, and such his number is the difference between p2 and p3’s numbers, or he realizes that he must be c and so his number is the sum of p2 and p3’s numbers.
That’s all I have so far. Anyone have other ideas?
If p1 KNOWS that he’s the largest then he has to have gained some other piece of information. Say the numbers he sees are 32 and 33. His number would have to be either 1 or 65. If p1 was 1 then the other two would have known p1 couldn’t be the sum of the other two
If p2 sees 1 and 33, s/he would wonder if s/he is 32 or 34.
P3 would consider 31 or 33.
P1 knows that P2 and P3 are not equal. So they know that the set isn't [2A, A, A].
P2 knows that P1 and P3 are not equal. So they know that the set isn't [A, 2A, A]. They also know that if P1 doesn't know, then they were able to make the same deduction. So they now know that both [2A, A, A] and [A, 2A, A] aren't correct. Since they know that [2A, A, A] isn't correct, they can also know that [2A, 3A, A] isn't correct either. Because they'd be able to see if P1 = 2A and P3 = A, and if that were true and P1 doesn't know their number, it would have to be because P2 isn't A. And if P2 isn't A, they'd have to be 3A.
P3 knows that P1 and P2 aren't equal. Eliminates [A, A, 2A]. Knows that [2A, A, A], [A, 2A, A], and [2A, 3A, A], are eliminated. Using the same process as P2, they can eliminate [2A, A, 3A], [A, 2A, 3A], and also [2A, 3A, 5A]. Because they can see the numbers and they know if P1 is 2A and P2 is 3A.
Now we're back at P1. Who now knows.
So P2 and P3 are in the eliminated sets. Which means we're one of these
[2A, A, A]; [3A, 2A, A]; [4A, 3A, A]; [3A, A, 2A]; [4A, A, 3A]; [5A, 2A, 3A]; [8A, 3A, 5A]
We know his number is 65. To find the set, we can factor 65: (5 * 13). We can check the other numbers 2(13) = 26. 3(13) = 39. And technically, you don't need to find the other numbers. The final answer is 5A * 2A * 3A or (A^3) * 30.
Why? Couldn't it be an infinite number of 3 size arrays comprised of A where two elements sum to the third? [24A, 13A, 11A]? How did we deduce this set of arrays?
EDIT: Solved from another reddit comment. Tuples without a common factor like the one above are considered as a=1.
"They're not eliminated; they correspond to a = 1."
You should be able to generate an infinite number of these problems just by multiplying the first formula factor by a prime number. Like the same question but the person answers '52' restricts you to either [4a, 3a, a] or [4a, a, 3a]. Since the question only asks for the product of all the terms the answer is 4 * 13 + 3 * 13 + 13 = 104.
So A + B = C and A + C = B. But we know that A + B = C, so we can replace C with (A + B). So we know that A + A + B = B.
So 2A + B = B. Or 2A = 0.
And this holds any way you slice it.
Even if you were to try and brute force it.
A = 1
B = 2
Then C = 3. But A + C has to equal B. That's 1 + 3 = 2? That's not true.
I don't see a case where you can add to the sum of two numbers one of the numbers and get the other number.
I'm guessing that's a misreading of the problem. Because it looks like the third number is the sum of the first two.
The original problem is a little ambiguously worded. You could say "one of their numbers is the sum of the other two" and it would be a little clearer.
No it isn't. If it said "the sum of any two of the numbers is equal to the third", that would be a contradiction. What it says is "the sum of two of the numbers is equal to the third".
Buying two of the items gets you the third for free.
The implication is any two.
It’s ok that it’s ambiguous. It happens. In most cases, we clarify and move on. There’s no need to defend it.
What's especially strange here is, they repeatedly demonstrate if you interpret it that way, the problem is obviously, trivially, unsolvable, in a way that a beginner in algebra could intuit. (roughly 12 years old, at least, we started touching algebra in 7th grade)
I really don't get it.
When I've seen this sort of thing play out this way, the talking-down is usually for the benefit of demonstrating something to an observer (i.e. I am smart look at this thing I figured out; I can hold my own when the haters chirp; look they say $INTERLOCUTOR is a thinker but they can't even understand me!), but ~0 of that would apply here, at least traditionally.