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"How can you guarantee that you won't change/quit job after just couple of months?" How to respond?

A king was born in a year that was a perfect square, lived a perfect square number of years, and also died in a year that was a perfect square

Riddle Man's ageWhy is every prime number (5 and higher) divisible by 3 when you square it and subtract 1?The Mendoza LineProfessor Halfbrain and the odd perfect numberEgyptian square pyramidWhat number am I ? (square number and triangle number)A Gathering of Number-TheoristsMother and DaughterA certain partition of 130Finni's tricky gameCreate the numbers 1 - 30 using the digits 2, 0, 1, 9 in this particular order!

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

4

$begingroup$

In which year could he have been born?
(A) 1936
(B) 1764
(C) 1600
(D) 1444

The answer's (C). Why?

number-theory

share|improve this question

edited Jun 6 at 15:01

Lawrence

6,59621449

asked Jun 5 at 11:04

Siddharth GargSiddharth Garg

516

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The answer's (C), but I guess the question kinda gives it away.
  $endgroup$
  – Annosz
  Jun 5 at 11:12
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  A related puzzle ... that said, did you find this in a book or somewhere and you need help with the explanation, or what?
  $endgroup$
  – Rand al'Thor
  Jun 5 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for not clarifying, I needed an explanation. And no, I did not find it in a book. My friend asked it from me.
  $endgroup$
  – Siddharth Garg
  Jun 6 at 8:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you've made a pretty big assumption about the lifespan of this king. For example, (A) could easily be the answer, provided the king lives for 1089 years.
  $endgroup$
  – Ian MacDonald
  Jun 6 at 15:14
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
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  @Randal'Thor Of course, the OP never said the King was Human. Could be an immortal jellyfish.
  $endgroup$
  – Tezra
  Jun 6 at 16:38
  

  
  
  
  

 | 
show 2 more comments

4

$begingroup$

In which year could he have been born?
(A) 1936
(B) 1764
(C) 1600
(D) 1444

The answer's (C). Why?

number-theory

share|improve this question

edited Jun 6 at 15:01

Lawrence

6,59621449

asked Jun 5 at 11:04

Siddharth GargSiddharth Garg

516

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The answer's (C), but I guess the question kinda gives it away.
  $endgroup$
  – Annosz
  Jun 5 at 11:12
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  A related puzzle ... that said, did you find this in a book or somewhere and you need help with the explanation, or what?
  $endgroup$
  – Rand al'Thor
  Jun 5 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry for not clarifying, I needed an explanation. And no, I did not find it in a book. My friend asked it from me.
  $endgroup$
  – Siddharth Garg
  Jun 6 at 8:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you've made a pretty big assumption about the lifespan of this king. For example, (A) could easily be the answer, provided the king lives for 1089 years.
  $endgroup$
  – Ian MacDonald
  Jun 6 at 15:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Randal'Thor Of course, the OP never said the King was Human. Could be an immortal jellyfish.
  $endgroup$
  – Tezra
  Jun 6 at 16:38
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

In which year could he have been born?
(A) 1936
(B) 1764
(C) 1600
(D) 1444

The answer's (C). Why?

number-theory

share|improve this question

edited Jun 6 at 15:01

Lawrence

6,59621449

asked Jun 5 at 11:04

Siddharth GargSiddharth Garg

516

$endgroup$

share|improve this question

edited Jun 6 at 15:01

Lawrence

6,59621449

asked Jun 5 at 11:04

Siddharth GargSiddharth Garg

516

share|improve this question

edited Jun 6 at 15:01

Lawrence

6,59621449

asked Jun 5 at 11:04

Siddharth GargSiddharth Garg

516

edited Jun 6 at 15:01

Lawrence

6,59621449

edited Jun 6 at 15:01

Lawrence

6,59621449

asked Jun 5 at 11:04

Siddharth GargSiddharth Garg

516

asked Jun 5 at 11:04

Siddharth GargSiddharth Garg

516

2 Answers
2

active

oldest

votes

20

$begingroup$

Given that the question contains its own answer, presumably what's wanted here is an explanation, so here is one.

Suppose the king is born in year $a^2$, lives $b^2$ years, and dies in year $c^2$ then we have $a^2+b^2=c^2$ (note: we'd better assume he lives an exact integer number of years, because otherwise this equation could be off by one).

There is an easy and stupid solution: we can take $b=0$ (this makes some assumptions about the rules of succession in the country whose king we're talking about!) in which case any perfect-square year will do. I assume we're not supposed to do this :-).

Otherwise, note that the next square after $a^2$ is $a^2+2a+1$, and the next is $a^2+4a+4$. Here we have $asimeq40$ and assuming our king is human his year of death is going to have to be $a^2+2a+1$ because $4a+4$ is too big to be anyone's age at death, which means that $2a+1$ is a square.

The values of $a$ are, in order, 44,42,40,38; so the values of $2a+1$ are 89,85,81,77. The third of these is a square and the others aren't.

(None of the other numbers there is within 1 of being a square, so actually we don't need to assume that the king lives an exact integer number of years after all.)

share|improve this answer

answered Jun 5 at 11:45

Gareth McCaughan♦Gareth McCaughan

74.7k3189289

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  About the eventuality of b=0, this guy was born 20 years to late : en.m.wikipedia.org/wiki/John_I_of_France
  $endgroup$
  – Evargalo
  Jun 5 at 15:10
  

  
  
  
  

add a comment | 

4

$begingroup$

So, the answer is C) 1600, which is $40^2$.

Ignoring the trivial case where he died the year he was born, $40+1=41$ so $41^2=1681$ (by definition a perfect square).

$1681-1600 = 81$ years, which is a reasonable life span and is $9^2$.

This is the easiest math puzzle ever! All I did was add 1 and it solved itself! :3

(Gareth was first, but this is a much simpler method. ^^;)

share|improve this answer

edited Jun 6 at 16:35

a stone arachnid

1717

answered Jun 5 at 19:40

TezraTezra

1414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this is a simpler method than Gareth's. It is the same method, but without the demonstration that this is the only solution.
  $endgroup$
  – Evargalo
  Jun 6 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Evargalo As Ian mentioned, living for 1089 years would make A valid (we were not told how old the king could live to be, or that he is human), so strictly there are other solutions. Also, Gareth solved for the years lived (I think, he lost me on "The values of A are"); At the very least, I think this is simpler for non-math people to follow. (it's easier to verify assertions of fact than a proof)
  $endgroup$
  – Tezra
  Jun 6 at 16:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  But the question wasn't "what's the solution". Indeed, the solution was given in the question. The question was "why is this the solution", indicating the OP wants an explanation of why that (and, one can safely infer, why only that) is a solution. You've answered the problem in the question, but not the question itself.
  $endgroup$
  – Rubio♦
  Jun 6 at 23:32
  

  
  
  
  

add a comment | 

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20

$begingroup$

Given that the question contains its own answer, presumably what's wanted here is an explanation, so here is one.

Suppose the king is born in year $a^2$, lives $b^2$ years, and dies in year $c^2$ then we have $a^2+b^2=c^2$ (note: we'd better assume he lives an exact integer number of years, because otherwise this equation could be off by one).

There is an easy and stupid solution: we can take $b=0$ (this makes some assumptions about the rules of succession in the country whose king we're talking about!) in which case any perfect-square year will do. I assume we're not supposed to do this :-).

Otherwise, note that the next square after $a^2$ is $a^2+2a+1$, and the next is $a^2+4a+4$. Here we have $asimeq40$ and assuming our king is human his year of death is going to have to be $a^2+2a+1$ because $4a+4$ is too big to be anyone's age at death, which means that $2a+1$ is a square.

The values of $a$ are, in order, 44,42,40,38; so the values of $2a+1$ are 89,85,81,77. The third of these is a square and the others aren't.

(None of the other numbers there is within 1 of being a square, so actually we don't need to assume that the king lives an exact integer number of years after all.)

share|improve this answer

answered Jun 5 at 11:45

Gareth McCaughan♦Gareth McCaughan

74.7k3189289

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  About the eventuality of b=0, this guy was born 20 years to late : en.m.wikipedia.org/wiki/John_I_of_France
  $endgroup$
  – Evargalo
  Jun 5 at 15:10
  

  
  
  
  

add a comment | 

20

$begingroup$

Given that the question contains its own answer, presumably what's wanted here is an explanation, so here is one.

Suppose the king is born in year $a^2$, lives $b^2$ years, and dies in year $c^2$ then we have $a^2+b^2=c^2$ (note: we'd better assume he lives an exact integer number of years, because otherwise this equation could be off by one).

There is an easy and stupid solution: we can take $b=0$ (this makes some assumptions about the rules of succession in the country whose king we're talking about!) in which case any perfect-square year will do. I assume we're not supposed to do this :-).

Otherwise, note that the next square after $a^2$ is $a^2+2a+1$, and the next is $a^2+4a+4$. Here we have $asimeq40$ and assuming our king is human his year of death is going to have to be $a^2+2a+1$ because $4a+4$ is too big to be anyone's age at death, which means that $2a+1$ is a square.

The values of $a$ are, in order, 44,42,40,38; so the values of $2a+1$ are 89,85,81,77. The third of these is a square and the others aren't.

(None of the other numbers there is within 1 of being a square, so actually we don't need to assume that the king lives an exact integer number of years after all.)

share|improve this answer

answered Jun 5 at 11:45

Gareth McCaughan♦Gareth McCaughan

74.7k3189289

$endgroup$

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  About the eventuality of b=0, this guy was born 20 years to late : en.m.wikipedia.org/wiki/John_I_of_France
  $endgroup$
  – Evargalo
  Jun 5 at 15:10
  

  
  
  
  

add a comment | 

$begingroup$

Given that the question contains its own answer, presumably what's wanted here is an explanation, so here is one.

Suppose the king is born in year $a^2$, lives $b^2$ years, and dies in year $c^2$ then we have $a^2+b^2=c^2$ (note: we'd better assume he lives an exact integer number of years, because otherwise this equation could be off by one).

There is an easy and stupid solution: we can take $b=0$ (this makes some assumptions about the rules of succession in the country whose king we're talking about!) in which case any perfect-square year will do. I assume we're not supposed to do this :-).

Otherwise, note that the next square after $a^2$ is $a^2+2a+1$, and the next is $a^2+4a+4$. Here we have $asimeq40$ and assuming our king is human his year of death is going to have to be $a^2+2a+1$ because $4a+4$ is too big to be anyone's age at death, which means that $2a+1$ is a square.

The values of $a$ are, in order, 44,42,40,38; so the values of $2a+1$ are 89,85,81,77. The third of these is a square and the others aren't.

(None of the other numbers there is within 1 of being a square, so actually we don't need to assume that the king lives an exact integer number of years after all.)

share|improve this answer

answered Jun 5 at 11:45

Gareth McCaughan♦Gareth McCaughan

74.7k3189289

$endgroup$

share|improve this answer

answered Jun 5 at 11:45

Gareth McCaughan♦Gareth McCaughan

74.7k3189289

answered Jun 5 at 11:45

Gareth McCaughan♦Gareth McCaughan

74.7k3189289

answered Jun 5 at 11:45

Gareth McCaughan♦Gareth McCaughan

74.7k3189289

4

$begingroup$

So, the answer is C) 1600, which is $40^2$.

Ignoring the trivial case where he died the year he was born, $40+1=41$ so $41^2=1681$ (by definition a perfect square).

$1681-1600 = 81$ years, which is a reasonable life span and is $9^2$.

This is the easiest math puzzle ever! All I did was add 1 and it solved itself! :3

(Gareth was first, but this is a much simpler method. ^^;)

share|improve this answer

edited Jun 6 at 16:35

a stone arachnid

1717

answered Jun 5 at 19:40

TezraTezra

1414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this is a simpler method than Gareth's. It is the same method, but without the demonstration that this is the only solution.
  $endgroup$
  – Evargalo
  Jun 6 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Evargalo As Ian mentioned, living for 1089 years would make A valid (we were not told how old the king could live to be, or that he is human), so strictly there are other solutions. Also, Gareth solved for the years lived (I think, he lost me on "The values of A are"); At the very least, I think this is simpler for non-math people to follow. (it's easier to verify assertions of fact than a proof)
  $endgroup$
  – Tezra
  Jun 6 at 16:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  But the question wasn't "what's the solution". Indeed, the solution was given in the question. The question was "why is this the solution", indicating the OP wants an explanation of why that (and, one can safely infer, why only that) is a solution. You've answered the problem in the question, but not the question itself.
  $endgroup$
  – Rubio♦
  Jun 6 at 23:32
  

  
  
  
  

add a comment | 

4

$begingroup$

So, the answer is C) 1600, which is $40^2$.

Ignoring the trivial case where he died the year he was born, $40+1=41$ so $41^2=1681$ (by definition a perfect square).

$1681-1600 = 81$ years, which is a reasonable life span and is $9^2$.

This is the easiest math puzzle ever! All I did was add 1 and it solved itself! :3

(Gareth was first, but this is a much simpler method. ^^;)

share|improve this answer

edited Jun 6 at 16:35

a stone arachnid

1717

answered Jun 5 at 19:40

TezraTezra

1414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think this is a simpler method than Gareth's. It is the same method, but without the demonstration that this is the only solution.
  $endgroup$
  – Evargalo
  Jun 6 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Evargalo As Ian mentioned, living for 1089 years would make A valid (we were not told how old the king could live to be, or that he is human), so strictly there are other solutions. Also, Gareth solved for the years lived (I think, he lost me on "The values of A are"); At the very least, I think this is simpler for non-math people to follow. (it's easier to verify assertions of fact than a proof)
  $endgroup$
  – Tezra
  Jun 6 at 16:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  But the question wasn't "what's the solution". Indeed, the solution was given in the question. The question was "why is this the solution", indicating the OP wants an explanation of why that (and, one can safely infer, why only that) is a solution. You've answered the problem in the question, but not the question itself.
  $endgroup$
  – Rubio♦
  Jun 6 at 23:32
  

  
  
  
  

add a comment | 

$begingroup$

So, the answer is C) 1600, which is $40^2$.

Ignoring the trivial case where he died the year he was born, $40+1=41$ so $41^2=1681$ (by definition a perfect square).

$1681-1600 = 81$ years, which is a reasonable life span and is $9^2$.

This is the easiest math puzzle ever! All I did was add 1 and it solved itself! :3

(Gareth was first, but this is a much simpler method. ^^;)

share|improve this answer

edited Jun 6 at 16:35

a stone arachnid

1717

answered Jun 5 at 19:40

TezraTezra

1414

$endgroup$

share|improve this answer

edited Jun 6 at 16:35

a stone arachnid

1717

answered Jun 5 at 19:40

TezraTezra

1414

edited Jun 6 at 16:35

a stone arachnid

1717

edited Jun 6 at 16:35

a stone arachnid

1717

answered Jun 5 at 19:40

TezraTezra

1414

answered Jun 5 at 19:40

TezraTezra

1414