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## Question

I couldn't find this one on the site...so I figured I'd post it...if it has already been done, I apologize...

There was once a king who wanted to determine who the smartest person in the land was (funny how that seems to happen a lot in these forums). After a bunch of tests, it was down to two men (P and S), who claimed to be "perfect" logicians (meaning there was never any flaw in their logic).

The king then told them their final test:

"I am thinking of two different numbers greater than 1, whose sum is less than 100. I will tell P the product of my two numbers, and I will tell S the sum of the two numbers," the kind stated.

The king whispers in P's ear the product of his two numbers, and whispers to S the sum of the two numbers...The following is the conversation that the logicians had:

P: I don't know the two numbers...

S: I know you don't

P: Ah...now I know them...

S: So do I

Together, both men then stated the two numbers to the king, and he stared in disbelief...

What were the two numbers?

(NOTE: I actually wrote a program to solve this puzzle...however, I know it can be done without the use of any computer...)

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As for this variation the answer could be:

The pair of numbers could be 21 and 10. Their product 210 and the sum 31.

Initially, P has two choices: 15*14 and 21*10. However, if the numbers were 15 and 14, then S would guess after P didn't, since the sum of 29 allows only one such pair with non-unique product.

To find whether this is the only possible solution, you'd still save a lot of time by writing a computer program. I suspect, there are other solutions.

As far as puzzles go, this variation is more fair than the one posted before. Because solving the other one without a computer program was not a feasible task. I would change the question of the puzzle "What could be the numbers?"

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As for this variation the answer could be:

The pair of numbers could be 21 and 10. Their product 210 and the sum 31.

Initially, P has two choices: 15*14 and 21*10. However, if the numbers were 15 and 14, then S would guess after P didn't, since the sum of 29 allows only one such pair with non-unique product.

To find whether this is the only possible solution, you'd still save a lot of time by writing a computer program. I suspect, there are other solutions.

As far as puzzles go, this variation is more fair than the one posted before. Because solving the other one without a computer program was not a feasible task. I would change the question of the puzzle "What could be the numbers?"

After looking at the other problem posted, I'm not sure if this one is actually any easier...in fact I could solve the other one easier than this one (however that could just be the way I think)...

Second, this is not at all what I got for my answer...I'm confused as to how you made the statement that P has two choices and why those were the only 2 choices P had. Unless I have misstated the problem, you over-simplified it...maybe if you walk through how you got the answer...because from what I have found there is only one possible answer to this problem, and it is not the one that you stated...

Initially, P says that he doesn't know the two numbers, which means that his product cannot be the product of 2 distinct factors of the number he was told whose sum is less than 100 (so, they can't be 2 prime numbers...his number cannot be the cube of a prime...or the 4th power of a prime...)

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• 0 As for this variation the answer could be:

The pair of numbers could be 21 and 10. Their product 210 and the sum 31.

Initially, P has two choices: 15*14 and 21*10. However, if the numbers were 15 and 14, then S would guess after P didn't, since the sum of 29 allows only one such pair with non-unique product.

Actually that's not an answer. Here's why:

the key is in the second statement: I know that you do not know. That means that every possible combination of numbers that add up to 31 (i.e. 2+29, 3+28, 4+27...) when multiplied together must have multiple factors. 2*29 = 58 has only 1 possible factorization. Therefore, if S is given 31, he can't say "I know you don't know"

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After looking at the other problem posted, I'm not sure if this one is actually any easier...in fact I could solve the other one easier than this one (however that could just be the way I think)...

Second, this is not at all what I got for my answer...I'm confused as to how you made the statement that P has two choices and why those were the only 2 choices P had. Unless I have misstated the problem, you over-simplified it...maybe if you walk through how you got the answer...because from what I have found there is only one possible answer to this problem, and it is not the one that you stated...

Initially, P says that he doesn't know the two numbers, which means that his product cannot be the product of 2 distinct factors of the number he was told whose sum is less than 100 (so, they can't be 2 prime numbers...his number cannot be the cube of a prime...or the 4th power of a prime...)

Sorry, my mistake. I didn't pay due attention to the statement of the problem. I just saw the names P and S, and assumed this problem was the same. (The other puzzle had two two-digit numbers.) So scratch my previous posts.

However, I don't see, why product couldn't be a 4th power of a prime. Say, 81 could be 9*9, or it could be 3*27.

Edited by Prime

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Sorry, my mistake. I didn't pay due attention to the statement of the problem. I just saw the names P and S, and assumed this problem was the same. (The other puzzle had two two-digit numbers.) So scratch my previous posts.

However, I don't see, why product couldn't be a 4th power of a prime. Say, 81 could be 9*9, or it could be 3*27.

can't be 9*9 since it says they are distinct numbers...so that means it could only be 3 * 27...and therefore he would know the answer just by that

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Here the first statement by S "I know, you don't" gives me problems. (I assume, by that he meant that he knew that P couldn't tell the numbers without him saying so.)

The numbers could be 2 and 6.

The sum of 8 gives S three possibilities to choose from:

2+6; 3+5; and 4+4.

If it were 3 and 5 with the product of 15, P would guess immediately.

If it were 4 and 4 with the product of 16, then possible numbers and their sums (as far as P knows) are 8+2=10, or 4+4=8, both of which would stump S even after P did not guess.

But 2+6 with the product of 12 gives P the choice between 3*4 and 2*6. If it were 3*4, then 3+4=7 gives S a choice between 2+5 and 3+4. If it were 2 and 5, S knows P would have guessed on his first try. So S would guess immediately after P didn't. Since that didn't happen, P knows S must have 8 and therefore the numbers are 2 and 6.

However, how would S "know" that P didn't? (The sum of 8 does offer a unique product of 3*15.)

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4 and 13

Correct...It's a great riddle...I'm assuming you figured out how to get the answer and are leaving that up for everyone else to figure out And yes, that is the only answer to the puzzle...

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I agree with Prof. Templeton but would go one step further to say that S wins the competition...

As soon as S said "I know you don't", P (knowing his own number) now knows what both numbers were. Indeed this is the next thing he says. Without that information, S doesn't know what the numbers are. Instead of saying "Ah...now I know them" P should have just given the answer and won the competition.

So S should be considered the smartest person in the land because P must be really stupid to make such a silly mistake! By the way, there's another similar puzzle to this I posted a while back here. Nobody managed to get the right answer at the time and I ended up posting the solution after a while, so try having a crack at that one without reading the hints and solution first.

I really like this puzzle by the way - got the mind working again! ## Join the conversation

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