The Ages of Mathematician´s sons
$begingroup$
Two mathematicians meet and talk:
"Do you have a son?" asked the first mathematician.
"Yes I actually have three sons, and none of them are twins." answered the second mathematician.
"How old are they?" asked the first mathematician.
"The product of their age is equal to the month number at this moment." answered the second mathematician.
"It is not sufficient!" said the first mathematician.
"True, if you sum their ages next year it will again be equal to the month number at this moment." said the second mathematician.
How old are his sons? (I was not able to evaluate this mathematically!)
puzzle
asked Dec 23 '18 at 9:41
Adam PáltikAdam Páltik
1149
$endgroup$
add a comment |
$begingroup$
Two mathematicians meet and talk:
"Do you have a son?" asked the first mathematician.
"Yes I actually have three sons, and none of them are twins." answered the second mathematician.
"How old are they?" asked the first mathematician.
"The product of their age is equal to the month number at this moment." answered the second mathematician.
"It is not sufficient!" said the first mathematician.
"True, if you sum their ages next year it will again be equal to the month number at this moment." said the second mathematician.
How old are his sons? (I was not able to evaluate this mathematically!)
puzzle
asked Dec 23 '18 at 9:41
Adam PáltikAdam Páltik
1149
$endgroup$
$begingroup$
Hint 1 : The ages must be distinct (because there are no twins)
$endgroup$
– Peter
Dec 23 '18 at 9:46
$begingroup$
Hint 2 : No son can be older than $6$ because then the product would exceed $12$
$endgroup$
– Peter
Dec 23 '18 at 9:47
2
$begingroup$
I would like to point out that "there are no twins" does not necessarily guarantee that there are no two sons of the same age. However I posted my solution assuming there are no sons of the same age.
$endgroup$
– Levent
Dec 23 '18 at 9:48
$begingroup$
@Levent Strictly speaking, you are right.
$endgroup$
– Peter
Dec 23 '18 at 9:49
$begingroup$
Hint 3 : The solution is $(1,2,6)$
$endgroup$
– Peter
Dec 23 '18 at 9:58
add a comment |
$begingroup$
Two mathematicians meet and talk:
"Do you have a son?" asked the first mathematician.
"Yes I actually have three sons, and none of them are twins." answered the second mathematician.
"How old are they?" asked the first mathematician.
"The product of their age is equal to the month number at this moment." answered the second mathematician.
"It is not sufficient!" said the first mathematician.
"True, if you sum their ages next year it will again be equal to the month number at this moment." said the second mathematician.
How old are his sons? (I was not able to evaluate this mathematically!)
puzzle
asked Dec 23 '18 at 9:41
Adam PáltikAdam Páltik
1149
$endgroup$
Two mathematicians meet and talk:
"Do you have a son?" asked the first mathematician.
"Yes I actually have three sons, and none of them are twins." answered the second mathematician.
"How old are they?" asked the first mathematician.
"The product of their age is equal to the month number at this moment." answered the second mathematician.
"It is not sufficient!" said the first mathematician.
"True, if you sum their ages next year it will again be equal to the month number at this moment." said the second mathematician.
How old are his sons? (I was not able to evaluate this mathematically!)
puzzle
puzzle
asked Dec 23 '18 at 9:41
Adam PáltikAdam Páltik
1149
asked Dec 23 '18 at 9:41
Adam PáltikAdam Páltik
1149
asked Dec 23 '18 at 9:41
Adam PáltikAdam Páltik
1149
asked Dec 23 '18 at 9:41
Adam PáltikAdam Páltik
1149
asked Dec 23 '18 at 9:41
Adam PáltikAdam Páltik
1149
1149
$begingroup$
Hint 1 : The ages must be distinct (because there are no twins)
$endgroup$
– Peter
Dec 23 '18 at 9:46
$begingroup$
Hint 2 : No son can be older than $6$ because then the product would exceed $12$
$endgroup$
– Peter
Dec 23 '18 at 9:47
2
$begingroup$
I would like to point out that "there are no twins" does not necessarily guarantee that there are no two sons of the same age. However I posted my solution assuming there are no sons of the same age.
$endgroup$
– Levent
Dec 23 '18 at 9:48
$begingroup$
@Levent Strictly speaking, you are right.
$endgroup$
– Peter
Dec 23 '18 at 9:49
$begingroup$
Hint 3 : The solution is $(1,2,6)$
$endgroup$
– Peter
Dec 23 '18 at 9:58
add a comment |
$begingroup$
Hint 1 : The ages must be distinct (because there are no twins)
$endgroup$
– Peter
Dec 23 '18 at 9:46
$begingroup$
Hint 2 : No son can be older than $6$ because then the product would exceed $12$
$endgroup$
– Peter
Dec 23 '18 at 9:47
2
$begingroup$
I would like to point out that "there are no twins" does not necessarily guarantee that there are no two sons of the same age. However I posted my solution assuming there are no sons of the same age.
$endgroup$
– Levent
Dec 23 '18 at 9:48
$begingroup$
@Levent Strictly speaking, you are right.
$endgroup$
– Peter
Dec 23 '18 at 9:49
$begingroup$
Hint 3 : The solution is $(1,2,6)$
$endgroup$
– Peter
Dec 23 '18 at 9:58
$begingroup$
Hint 1 : The ages must be distinct (because there are no twins)
$endgroup$
– Peter
Dec 23 '18 at 9:46
$begingroup$
Hint 1 : The ages must be distinct (because there are no twins)
$endgroup$
– Peter
Dec 23 '18 at 9:46
$begingroup$
Hint 2 : No son can be older than $6$ because then the product would exceed $12$
$endgroup$
– Peter
Dec 23 '18 at 9:47
$begingroup$
Hint 2 : No son can be older than $6$ because then the product would exceed $12$
$endgroup$
– Peter
Dec 23 '18 at 9:47
2
2
$begingroup$
I would like to point out that "there are no twins" does not necessarily guarantee that there are no two sons of the same age. However I posted my solution assuming there are no sons of the same age.
$endgroup$
– Levent
Dec 23 '18 at 9:48
$begingroup$
I would like to point out that "there are no twins" does not necessarily guarantee that there are no two sons of the same age. However I posted my solution assuming there are no sons of the same age.
$endgroup$
– Levent
Dec 23 '18 at 9:48
$begingroup$
@Levent Strictly speaking, you are right.
$endgroup$
– Peter
Dec 23 '18 at 9:49
$begingroup$
@Levent Strictly speaking, you are right.
$endgroup$
– Peter
Dec 23 '18 at 9:49
$begingroup$
Hint 3 : The solution is $(1,2,6)$
$endgroup$
– Peter
Dec 23 '18 at 9:58
$begingroup$
Hint 3 : The solution is $(1,2,6)$
$endgroup$
– Peter
Dec 23 '18 at 9:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $A_1,A_2,A_3$ be the ages of the sons respectively. Observe that if the month is $1,2,3,4,5,7,9,11$ then there are no solutions using the fact that there are no twins. If the month is $6,8$ or $10$ then there is a unique solutions so just by the first information it would be possible to determine the ages. Hence the month must be December.
$12$ has two decompositions : $(1,2,6),(1,3,4)$ and the sum of their ages next year is $12$ in the former case and $11$ in the latter case.
Thus the solution is $(1,2,6)$.
answered Dec 23 '18 at 9:47
LeventLevent
2,729925
$endgroup$
$begingroup$
Thank you very much for your mathematically based solution!
$endgroup$
– Adam Páltik
Dec 23 '18 at 9:53
$begingroup$
You are welcome. You can accept the solution if you are satisfied with it.
$endgroup$
– Levent
Dec 23 '18 at 9:54
$begingroup$
@Levent You have not considered that in a year all the three sons are one year older.
$endgroup$
– Peter
Dec 23 '18 at 9:56
$begingroup$
Oh god, I just added $1$ in the sums instead of $3$. Thank you very much for pointing out this stupid mistake. I'll edit.
$endgroup$
– Levent
Dec 23 '18 at 9:58
1
$begingroup$
@AdamPáltik You do not need mathematics of considerable level, if you apply some basic facts. You only have to look for duplicate products and check whether the claim with the sum fits. In fact, only $5$ triples are possible, considering the distinct numbers and that the product is not greater than $12$
$endgroup$
– Peter
Dec 23 '18 at 10:02
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
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votes
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oldest
votes
$begingroup$
Let $A_1,A_2,A_3$ be the ages of the sons respectively. Observe that if the month is $1,2,3,4,5,7,9,11$ then there are no solutions using the fact that there are no twins. If the month is $6,8$ or $10$ then there is a unique solutions so just by the first information it would be possible to determine the ages. Hence the month must be December.
$12$ has two decompositions : $(1,2,6),(1,3,4)$ and the sum of their ages next year is $12$ in the former case and $11$ in the latter case.
Thus the solution is $(1,2,6)$.
answered Dec 23 '18 at 9:47
LeventLevent
2,729925
$endgroup$
$begingroup$
Thank you very much for your mathematically based solution!
$endgroup$
– Adam Páltik
Dec 23 '18 at 9:53
$begingroup$
You are welcome. You can accept the solution if you are satisfied with it.
$endgroup$
– Levent
Dec 23 '18 at 9:54
$begingroup$
@Levent You have not considered that in a year all the three sons are one year older.
$endgroup$
– Peter
Dec 23 '18 at 9:56
$begingroup$
Oh god, I just added $1$ in the sums instead of $3$. Thank you very much for pointing out this stupid mistake. I'll edit.
$endgroup$
– Levent
Dec 23 '18 at 9:58
1
$begingroup$
@AdamPáltik You do not need mathematics of considerable level, if you apply some basic facts. You only have to look for duplicate products and check whether the claim with the sum fits. In fact, only $5$ triples are possible, considering the distinct numbers and that the product is not greater than $12$
$endgroup$
– Peter
Dec 23 '18 at 10:02
add a comment |
$begingroup$
Let $A_1,A_2,A_3$ be the ages of the sons respectively. Observe that if the month is $1,2,3,4,5,7,9,11$ then there are no solutions using the fact that there are no twins. If the month is $6,8$ or $10$ then there is a unique solutions so just by the first information it would be possible to determine the ages. Hence the month must be December.
$12$ has two decompositions : $(1,2,6),(1,3,4)$ and the sum of their ages next year is $12$ in the former case and $11$ in the latter case.
Thus the solution is $(1,2,6)$.
answered Dec 23 '18 at 9:47
LeventLevent
2,729925
$endgroup$
$begingroup$
Thank you very much for your mathematically based solution!
$endgroup$
– Adam Páltik
Dec 23 '18 at 9:53
$begingroup$
You are welcome. You can accept the solution if you are satisfied with it.
$endgroup$
– Levent
Dec 23 '18 at 9:54
$begingroup$
@Levent You have not considered that in a year all the three sons are one year older.
$endgroup$
– Peter
Dec 23 '18 at 9:56
$begingroup$
Oh god, I just added $1$ in the sums instead of $3$. Thank you very much for pointing out this stupid mistake. I'll edit.
$endgroup$
– Levent
Dec 23 '18 at 9:58
1
$begingroup$
@AdamPáltik You do not need mathematics of considerable level, if you apply some basic facts. You only have to look for duplicate products and check whether the claim with the sum fits. In fact, only $5$ triples are possible, considering the distinct numbers and that the product is not greater than $12$
$endgroup$
– Peter
Dec 23 '18 at 10:02
add a comment |
$begingroup$
Let $A_1,A_2,A_3$ be the ages of the sons respectively. Observe that if the month is $1,2,3,4,5,7,9,11$ then there are no solutions using the fact that there are no twins. If the month is $6,8$ or $10$ then there is a unique solutions so just by the first information it would be possible to determine the ages. Hence the month must be December.
$12$ has two decompositions : $(1,2,6),(1,3,4)$ and the sum of their ages next year is $12$ in the former case and $11$ in the latter case.
Thus the solution is $(1,2,6)$.
answered Dec 23 '18 at 9:47
LeventLevent
2,729925
$endgroup$
Let $A_1,A_2,A_3$ be the ages of the sons respectively. Observe that if the month is $1,2,3,4,5,7,9,11$ then there are no solutions using the fact that there are no twins. If the month is $6,8$ or $10$ then there is a unique solutions so just by the first information it would be possible to determine the ages. Hence the month must be December.
$12$ has two decompositions : $(1,2,6),(1,3,4)$ and the sum of their ages next year is $12$ in the former case and $11$ in the latter case.
Thus the solution is $(1,2,6)$.
answered Dec 23 '18 at 9:47
LeventLevent
2,729925
edited Dec 23 '18 at 9:59
answered Dec 23 '18 at 9:47
LeventLevent
2,729925
answered Dec 23 '18 at 9:47
LeventLevent
2,729925
answered Dec 23 '18 at 9:47
LeventLevent
2,729925
2,729925
$begingroup$
Thank you very much for your mathematically based solution!
$endgroup$
– Adam Páltik
Dec 23 '18 at 9:53
$begingroup$
You are welcome. You can accept the solution if you are satisfied with it.
$endgroup$
– Levent
Dec 23 '18 at 9:54
$begingroup$
@Levent You have not considered that in a year all the three sons are one year older.
$endgroup$
– Peter
Dec 23 '18 at 9:56
$begingroup$
Oh god, I just added $1$ in the sums instead of $3$. Thank you very much for pointing out this stupid mistake. I'll edit.
$endgroup$
– Levent
Dec 23 '18 at 9:58
1
$begingroup$
@AdamPáltik You do not need mathematics of considerable level, if you apply some basic facts. You only have to look for duplicate products and check whether the claim with the sum fits. In fact, only $5$ triples are possible, considering the distinct numbers and that the product is not greater than $12$
$endgroup$
– Peter
Dec 23 '18 at 10:02
add a comment |
$begingroup$
Thank you very much for your mathematically based solution!
$endgroup$
– Adam Páltik
Dec 23 '18 at 9:53
$begingroup$
You are welcome. You can accept the solution if you are satisfied with it.
$endgroup$
– Levent
Dec 23 '18 at 9:54
$begingroup$
@Levent You have not considered that in a year all the three sons are one year older.
$endgroup$
– Peter
Dec 23 '18 at 9:56
$begingroup$
Oh god, I just added $1$ in the sums instead of $3$. Thank you very much for pointing out this stupid mistake. I'll edit.
$endgroup$
– Levent
Dec 23 '18 at 9:58
1
$begingroup$
@AdamPáltik You do not need mathematics of considerable level, if you apply some basic facts. You only have to look for duplicate products and check whether the claim with the sum fits. In fact, only $5$ triples are possible, considering the distinct numbers and that the product is not greater than $12$
$endgroup$
– Peter
Dec 23 '18 at 10:02
$begingroup$
Thank you very much for your mathematically based solution!
$endgroup$
– Adam Páltik
Dec 23 '18 at 9:53
$begingroup$
Thank you very much for your mathematically based solution!
$endgroup$
– Adam Páltik
Dec 23 '18 at 9:53
$begingroup$
You are welcome. You can accept the solution if you are satisfied with it.
$endgroup$
– Levent
Dec 23 '18 at 9:54
$begingroup$
You are welcome. You can accept the solution if you are satisfied with it.
$endgroup$
– Levent
Dec 23 '18 at 9:54
$begingroup$
@Levent You have not considered that in a year all the three sons are one year older.
$endgroup$
– Peter
Dec 23 '18 at 9:56
$begingroup$
@Levent You have not considered that in a year all the three sons are one year older.
$endgroup$
– Peter
Dec 23 '18 at 9:56
$begingroup$
Oh god, I just added $1$ in the sums instead of $3$. Thank you very much for pointing out this stupid mistake. I'll edit.
$endgroup$
– Levent
Dec 23 '18 at 9:58
$begingroup$
Oh god, I just added $1$ in the sums instead of $3$. Thank you very much for pointing out this stupid mistake. I'll edit.
$endgroup$
– Levent
Dec 23 '18 at 9:58
1
1
$begingroup$
@AdamPáltik You do not need mathematics of considerable level, if you apply some basic facts. You only have to look for duplicate products and check whether the claim with the sum fits. In fact, only $5$ triples are possible, considering the distinct numbers and that the product is not greater than $12$
$endgroup$
– Peter
Dec 23 '18 at 10:02
$begingroup$
@AdamPáltik You do not need mathematics of considerable level, if you apply some basic facts. You only have to look for duplicate products and check whether the claim with the sum fits. In fact, only $5$ triples are possible, considering the distinct numbers and that the product is not greater than $12$
$endgroup$
– Peter
Dec 23 '18 at 10:02
add a comment |
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$begingroup$
Hint 1 : The ages must be distinct (because there are no twins)
$endgroup$
– Peter
Dec 23 '18 at 9:46
$begingroup$
Hint 2 : No son can be older than $6$ because then the product would exceed $12$
$endgroup$
– Peter
Dec 23 '18 at 9:47
2
$begingroup$
I would like to point out that "there are no twins" does not necessarily guarantee that there are no two sons of the same age. However I posted my solution assuming there are no sons of the same age.
$endgroup$
– Levent
Dec 23 '18 at 9:48
$begingroup$
@Levent Strictly speaking, you are right.
$endgroup$
– Peter
Dec 23 '18 at 9:49
$begingroup$
Hint 3 : The solution is $(1,2,6)$
$endgroup$
– Peter
Dec 23 '18 at 9:58