9
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The old rich Jeremiah has two sons, Timothy and Joseph. One day, Jeremiah decides to discuss about his heritage with his sons...

JEREMIAH:

Dear sons, I'm very old and exhausted, so I want to bequeath my collection of coins to you, if you prove me that you are smart enough.

You must know that I have a number $G$ of valuable gold coins and a number $S$ of precious silver coins, and I'm the only one in this world who knows $G$ and $S$. Also, $G$ and $S$ are two numbers between $2$ and $20$ (meaning that $2\leq G, S \leq 20$).

I'm going to tell you, Timothy, the sum of my gold and silver coins ($G+S$), and to you, Joseph, the product of them ($G\cdot S$). You have 24 hours from now to discover the number of your brother. If you both find the answer, I'll give my collection to both.

So, Jeremiah tells them the numbers, as he said. Of course, the two brothers are greedy, and each one doesn't want to share his information with his brother.

After six hours of reflecting...

TIMOTHY: Joseph, I don't think that you can determine my sum!

JOSEPH: Haha! You're a fool! Now I know your sum! The heritage is all mine!!!

TIMOTHY: I'm sorry for you, but if I'm a fool, then you are unwise! Now I know your product, and we'll have to share the legacy!


QUESTION: Can you determine the two numbers $G$ and $S$? Also, is the solution unique?

Note that your answer must be complete and detailed to be accepted!

EDIT: $2\leq G$ and $2\leq S$

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  • $\begingroup$ A similar puzzle: puzzling.stackexchange.com/questions/251/… $\endgroup$ – GOTO 0 Mar 20 '15 at 20:00
  • $\begingroup$ Yeah, this is slightly different because in this problem the "sum" guy is giving the first information. $\endgroup$ – leoll2 Mar 20 '15 at 20:09
10
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We are looking for two positive integers $G$ and $S$ with $2\le G,S\le20$ that fit with the conversation. Let us denote the sum $s=G+S$ and the product $p=GS$. Note that $4\le s\le40$ and $4\le p\le400$.

TIMOTHY: Joseph, I don't think that you can determine my sum!

(1) First of all, this statement implies that Joseph is not able to deduce the factors $S$ and $G$ from their product $p=GS$. This means that:

  • $p$ is not the product of (exactly) two primes $p=ab$; otherwise Joseph would immediately deduce that $\{G,S\}=\{a,b\}$.
  • $p$ does not have any prime factor $f$ with $11\le f\le19$; otherwise Joseph would deduce $\{G,S\}=\{f,p/f\}$.

(2) Secondly, this statement implies that for any possible way of writing $s$ as the sum of two summands $z$ and $s-z$, the corresponding product $z\cdot(s-z)$ does not allow Joseph to deduce $G$ and $S$. Equivalently, for any integer $z$ with $2\le z\le s/2$, the corresponding product $z\cdot(s-z)$ cannot be of the form as excluded in (1). This excludes most possible values for $s$:

  • $s\le12$ and $s\ne11$: Then $s$ can be written as the sum of two primes ($4=2+2$, $5=2+3$, $6=3+3$, $7=2+5$, $8=3+5$, $9=2+7$, $10=5+5$ and $12=5+7$).
  • $13\le s\le 31$: for the prime $z=11$, the product $z\cdot(s-z)$ has been excluded.
  • $32\le s\le 39$: for the prime $z=19$, the product $z\cdot(s-z)$ has been excluded.
  • $s=40$: then $G=S=20$, and Joseph easily deduces this from $p=400$.

(3) Summarizing we see that only the case $s=11$ remains. Indeed, $s=11$ agrees with Timothy's statement:

  • $\{G,S\}=\{2,9\}$: The product $2\cdot9=18$ allows the good factorizations $18=2\cdot9$ and $18=3\cdot6$.
  • $\{G,S\}=\{3,8\}$: The product $3\cdot8=24$ allows the good factorizations $24=2\cdot12$ and $24=3\cdot8$ and $24=4\cdot6$.
  • $\{G,S\}=\{4,7\}$: The product $4\cdot7=28$ allows the good factorizations $28=2\cdot14$ and $28=4\cdot7$.
  • $\{G,S\}=\{5,6\}$: The product $5\cdot6=30$ allows the good factorizations $30=2\cdot15$ and $30=3\cdot10$ and $30=5\cdot6$.

JOSEPH: Haha! You're a fool! Now I know your sum! The heritage is all mine!!!

(4) Joseph has thought through the matter as we did above, and hence is able to determine $s=11$.

TIMOTHY: I'm sorry for you, but if I'm a fool, then you are unwise! Now I know your product, and we'll have to share the legacy!

(5) This statement is impossible: Timothy does not learn anything new from Joseph's preceding statement.

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  • $\begingroup$ its true since numbers go till 20 there must be some mistake $\endgroup$ – Abr001am Mar 21 '15 at 17:45
  • $\begingroup$ the best way to bring that to work is extending upper limit to 30 or including 1 in the range $\endgroup$ – Abr001am Mar 21 '15 at 17:58
5
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the answer is variant about two eventual possibilities :


- If both sons and father know that numbers of coins are bigger than 1

the answer dosnt differ than the classic puzzle I don't know the two numbers... but now I do because there s noway the person with product can know the certainly the sum except the case S and G are pimes or $S*G = prime^{3}$ because:

lets assume for example $c$ divide S:

$S+G = \frac {S}{c}+Gc$

$Sc+Gc = S+Gc²$

$S = Gc$

So .. the only two couples sharing same product and sum are (S,G) = (cG,G) and (cG/c,Gc) and they are identical .

No hope for other solution except: 4,13 but this solution is expulsed since there doesnt exists any natural k>1 where 13*k <= 20


- If number 1 is included

TIMOTHY: Joseph, I don't think that you can determine my sum!

  • this means sum of numbers isnt prime+1 (considering 1 as prime)

JOSEPH: Haha! You're a fool! Now I know your sum! The heritage is all mine!!!

  • this means same product of whicheve two numbers , their sum is prime+1 except one unique couple

TIMOTHY: I'm sorry for you, but if I'm a fool, then you are unwise! Now I know your product, and we'll have to share the legacy!

  • this just meant that one unique couple among all possible couples which have common sum verifies the last condition

Such numbers are:

(1,4) , (1,8) , (1,9) , (1,15)

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  • $\begingroup$ This answer isn't clear, your deductions need argumenting. $\endgroup$ – leoll2 Mar 20 '15 at 20:19
  • $\begingroup$ so is it right answer or nt , because im not good at explanation $\endgroup$ – Abr001am Mar 20 '15 at 20:21
  • $\begingroup$ Try to figure it out on your own and let others understand your explanation. Why you deduce that sum of numbers isn't prime+1? What would happen if it was prime+1? $\endgroup$ – leoll2 Mar 20 '15 at 20:26
  • $\begingroup$ lol if its prime+1 summer would doubt that the person with product would have been given a prime number as product . so he wudnt be 100% sure $\endgroup$ – Abr001am Mar 20 '15 at 20:30
  • $\begingroup$ if u know the right solution , ushould hence identify and estimate mine so is it true ? $\endgroup$ – Abr001am Mar 20 '15 at 20:35
3
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I think the answers are

(1,4) or (3,4)

Reasoning:

The first statement means that there are no numbers that could sum together to be Timothy's such that their product would determine their sum. If any product determines either the sum or the numbers, then this cannot be the number. This eliminates:

Any sum greater than or equal to 12. If the sum is equal to 12, then the product 11*1=11 is uniquely defined as those digits. If the sum is greater than 12 but less than 23, then taking one of the digits to be 11 and the other digit to be sum-11 causes a situation where Joseph knows the number. As you cannot multiply 11 by 2 and get a possible number (and since 11 is prime), the factoring of this number is unique given the constraints. For sum >23, do the same thing with 19 (also prim) and the factoring idea is the same. This leaves only 20 and 20, which I think is obvious that the product could only be determined by those two numbers.

Further cases:

For 10, consider the digits 3 and 7. The product is 21, which is uniquely determined by these digits and Joseph knows. For 8,6,4,3,2 use the property noted by Agawa where any sum that is a prime has a possible product that is uniquely determined.

We are left with possible sums of 11, 9, 7, and 5.

For these sums, we need to consider each case. A possibility will be any sum that creates only one product where the sum of digits is a number that has not been eliminated. This leaves the choices above.

For example, if the sum were 7, then the possibilities are (1,6), (2,5) and (3,4). If the numbers were (1,6) then the product is 6. This could be obtained with either (1,6) or (2,3). The sum of (2,3) is 5. As Timothy would have claimed that Joseph couldnt' determine the numbers for either of these sums, Joseph cannot determine what the numbers are.

If the numbers were (2,5) then the product is 10. This could be obtained with either (1,10) or (2,5). Timothy would have similarly claimed that Joseph couldn't determine the numbers for either of these sums, and so Joseph cannot determine what the numbers are.

If the numbers were (3,4) then the product is 12. This could be obtained with either (1,12) [sum of 13 -> Timothy believes (11,2) is a possibility which means he would not claim Joseph couldn't determine numbers], (2,6) [which again has a sum of 8, which could be obtained by (7,1) and Timothy would not claim Joseph couldn't determine the numbers] or (3,4) [sum of 7. Timothy WOULD claim Joseph couldn't determine the numbers]. As (3,4) is the only possibility for which Timothy would claim Joseph couldn't determine the numbers out of this product, then Joseph knows this is the answer. As none of the other options for 7 lead to Joseph knowing this is the answer, now Timothy knows this is the answer too.

For 9, it breaks down because for both 1,8 and 2,7 Joseph would be able to determine the answer, which means that Timothy couldn't figure out which it was based on Joseph knowing.

For 11, it similarly breaks down for (3,8), (4,7) and (5,6).

For 5, only the choice (1,4) leads to Joseph determining the answer, so Timothy can figure this out too.

EDIT: changed first elimination to state SUM instead of numbers per Duncan's comment. Also updated first algorithm to change 20 to 19 and add note about primality of 11 and 19. Also note that it's not possible to determine which number is S and which is G as both addition and multiplication are associative. You can only figure out solutions where either number could be either number of coins.

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  • $\begingroup$ I'm not quite following your logic for the first one. You say that any number greater than or equal to (geq) 12 is eliminated. Is that for any SUM geq 12? Or that neither G nor S can be geq 12? $\endgroup$ – Duncan Mar 21 '15 at 4:09
  • $\begingroup$ It's any SUM geq 12. I'll edit. $\endgroup$ – OpiesDad Mar 21 '15 at 5:20
  • $\begingroup$ I'm not convinced that's right. For example, if the sum is 12, then it could be 11+1 (actually, the edit to the problem says that G and S are both greater than 1, so it can't be that, but following your original logic...). It could also be 10 and 2, or 9 and 3, or 8 and 4, etc. If it's 10 and 2, then the multiple would be 20, which means that Joseph wouldn't know if the numbers were 10 and 2 or 4 and 5. There are similar multiple values for 8 and 4 (product 32 = 8,4 or 2,16) and 6,6 (product 36 = 6,6 or 4, 9 or 2, 18) $\endgroup$ – Duncan Mar 21 '15 at 5:26
  • $\begingroup$ TImothy stated that "He doesn't think Joseph can determine the sum" Therefore if there is a possibility for his sum such that Joseph WOULD be able to determine the numbers, then he couldn't say this. Producing such a product of Joseph's numbers such that for Timothy's sum Joseph knows the numbers eliminates this sum as Timothy's. $\endgroup$ – OpiesDad Mar 21 '15 at 5:29
  • $\begingroup$ I think I see now. You're interpreting Timothy's first statement as "Based on this sum, there are NO possible G, S numbers such that the product is unique." I'm interpreting it as "Based on this sum, there's at least 1 possible G,S combination where the product is NOT unique." At least I think that's the difference. Does that seem right to you? $\endgroup$ – Duncan Mar 21 '15 at 5:59

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