8 ms·

Could you explain "The sum of two of the numbers is equal to the third"??

by bbstats 2y ago

deleted 2y ago
[deleted]
malisper 2y ago
if the three numbers are a, b, and c, then either a+b=c, a+c=b, or b+c=a
- bena 2y ago
  And they must all be positive integers. 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.
  - refulgentis 2y ago
    One of the cases has to be true, not all 3. (as you show, they're mutually exclusive for positive integers) i.e. "either" is important in the parent comment.
    - bena 2y ago
      Which is why I indicated that it would be a misreading of the problem. 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.
      - deleted 2y ago
        [deleted]
      - thaumasiotes 2y ago
        > The original problem is a little ambiguously worded. 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".
        bena 2y ago
        I have three items. 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.
        thaumasiotes 2y ago
        Why look for ambiguity that isn't there?
        refulgentis 2y ago
        There's a certain mind that either doesn't realize they're sidestepping the problem and turning it into a editing review, or realizes it, and doesn't understand why it seems off-topic/trivial to others. 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.
        bena 2y ago
        One often doesn't look for ambiguity. It is there. It is fine.
refulgentis 2y ago
Given #s x,y, and z, either x + y = z, x + z = y, or y + z = x.
rappatic 2y ago
I think: 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?
- deleted 2y ago
  [deleted]
- aardvarkr 2y ago
  I think it has something to do with applying the lower bound of 1. 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
  - oezi 2y ago
    But p2 and p3 don't yet know what they are themselves just because they see a 1: If p2 sees 1 and 33, s/he would wonder if s/he is 32 or 34. P3 would consider 31 or 33.
- bena 2y ago
  The answer is online and it's clever. 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.

Could you explain "The sum of two of the numbers is equal to the third"??

deleted 2y ago

[deleted]

malisper 2y ago

if the three numbers are a, b, and c, then either a+b=c, a+c=b, or b+c=a

bena 2y ago
And they must all be positive integers. 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.
- refulgentis 2y ago
  One of the cases has to be true, not all 3. (as you show, they're mutually exclusive for positive integers) i.e. "either" is important in the parent comment.
  - bena 2y ago
    Which is why I indicated that it would be a misreading of the problem. 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.
    - deleted 2y ago
      [deleted]
    - thaumasiotes 2y ago
      > The original problem is a little ambiguously worded. 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".
      - bena 2y ago
        I have three items. 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.
        thaumasiotes 2y ago
        Why look for ambiguity that isn't there?
        refulgentis 2y ago
        There's a certain mind that either doesn't realize they're sidestepping the problem and turning it into a editing review, or realizes it, and doesn't understand why it seems off-topic/trivial to others. 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.
        bena 2y ago
        One often doesn't look for ambiguity. It is there. It is fine.

refulgentis 2y ago

Given #s x,y, and z, either x + y = z, x + z = y, or y + z = x.

rappatic 2y ago

I think: 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?

deleted 2y ago
[deleted]
aardvarkr 2y ago
I think it has something to do with applying the lower bound of 1. 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
- oezi 2y ago
  But p2 and p3 don't yet know what they are themselves just because they see a 1: If p2 sees 1 and 33, s/he would wonder if s/he is 32 or 34. P3 would consider 31 or 33.
bena 2y ago
The answer is online and it's clever. 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.