Maximal sum of all numbers in the circle.
$begingroup$
$49$ integers are ordered around a circle. It is given that for any two adjacent numbers $x,y$ we have $x+y=(x-y)^2$. Find the maximal sum of all numbers in the circle.
Solution
Note that $x+y=(x-y)^2$ may be written as
$(2x-2y+1)^2=8y+1$ or also $(2y-2x+1)^2=8x+1$
And so all elements are such that $8x+1ge 0$
Replacing all elements $x$ on the circle by $X=sqrt{8x+1}$ the relation becomes
$X=pm Ypm 2$ (always with the constraint $X,Yge 0$)
So, applying $n$ times this process starting from $a$, we get $pm a+(overbrace{pm 1pm 1 ... pm 1}^{ntext{ times}})2$
In order to have a cycle after $49$ steps, we get $a=pm a+(overbrace{pm 1pm 1 ... pm 1}^{49text{ times}})2$
The second part is twice an odd number and so can never be zero and so we need the $pm a$ in RHS be $-a$
And so $a=(overbrace{pm 1pm 1 ... pm 1}^{49text{ times}})le 49$
Note that in order the $pm a$ becomes $-a$, at least one transformation needs to be $Xto 2-X$ and so $X=1$
It is easy to see that the optimum sequence for Xs then will be :
$(49,47,45,...,3,1,1,3,5,...,45,47),(49,47,45,...,3,1,1,3,5,...,45,47),...$
Which is, for the original sequence of $49$ numbers :
$frac{49^2-1}8,frac{47^2-1}8,...,frac{3^2-1}8,0,0,frac{3^2-1}8,...,frac{47^2-1}8$
And so the maximum sum is $sum_{k=0}^{24}frac{(2k+1)^2-1}8+sum_{k=0}^{23}frac{(2k+1)^2-1}8$
Which is $boxed{4900}$
My question: Could you please solve the generalize problem?
Thank you in advance!
combinatorics number-theory
asked Dec 19 '18 at 12:07
namnam
973
$endgroup$
add a comment |
$begingroup$
$49$ integers are ordered around a circle. It is given that for any two adjacent numbers $x,y$ we have $x+y=(x-y)^2$. Find the maximal sum of all numbers in the circle.
Solution
Note that $x+y=(x-y)^2$ may be written as
$(2x-2y+1)^2=8y+1$ or also $(2y-2x+1)^2=8x+1$
And so all elements are such that $8x+1ge 0$
Replacing all elements $x$ on the circle by $X=sqrt{8x+1}$ the relation becomes
$X=pm Ypm 2$ (always with the constraint $X,Yge 0$)
So, applying $n$ times this process starting from $a$, we get $pm a+(overbrace{pm 1pm 1 ... pm 1}^{ntext{ times}})2$
In order to have a cycle after $49$ steps, we get $a=pm a+(overbrace{pm 1pm 1 ... pm 1}^{49text{ times}})2$
The second part is twice an odd number and so can never be zero and so we need the $pm a$ in RHS be $-a$
And so $a=(overbrace{pm 1pm 1 ... pm 1}^{49text{ times}})le 49$
Note that in order the $pm a$ becomes $-a$, at least one transformation needs to be $Xto 2-X$ and so $X=1$
It is easy to see that the optimum sequence for Xs then will be :
$(49,47,45,...,3,1,1,3,5,...,45,47),(49,47,45,...,3,1,1,3,5,...,45,47),...$
Which is, for the original sequence of $49$ numbers :
$frac{49^2-1}8,frac{47^2-1}8,...,frac{3^2-1}8,0,0,frac{3^2-1}8,...,frac{47^2-1}8$
And so the maximum sum is $sum_{k=0}^{24}frac{(2k+1)^2-1}8+sum_{k=0}^{23}frac{(2k+1)^2-1}8$
Which is $boxed{4900}$
My question: Could you please solve the generalize problem?
Thank you in advance!
combinatorics number-theory
asked Dec 19 '18 at 12:07
namnam
973
$endgroup$
$begingroup$
By the generalized problem you mean the initial problem with 49 replaced by any integer $nge 2$, right?
$endgroup$
– Alex Ravsky
Dec 21 '18 at 3:32
1
$begingroup$
$ngeq 2$ integers are ordered around a circle. It is given that for any two adjacent numbers x,y we have $x+y=(x−y)^2$. Find the maximal sum of all numbers in the circle.
$endgroup$
– nam
Dec 23 '18 at 2:56
add a comment |
$begingroup$
$49$ integers are ordered around a circle. It is given that for any two adjacent numbers $x,y$ we have $x+y=(x-y)^2$. Find the maximal sum of all numbers in the circle.
Solution
Note that $x+y=(x-y)^2$ may be written as
$(2x-2y+1)^2=8y+1$ or also $(2y-2x+1)^2=8x+1$
And so all elements are such that $8x+1ge 0$
Replacing all elements $x$ on the circle by $X=sqrt{8x+1}$ the relation becomes
$X=pm Ypm 2$ (always with the constraint $X,Yge 0$)
So, applying $n$ times this process starting from $a$, we get $pm a+(overbrace{pm 1pm 1 ... pm 1}^{ntext{ times}})2$
In order to have a cycle after $49$ steps, we get $a=pm a+(overbrace{pm 1pm 1 ... pm 1}^{49text{ times}})2$
The second part is twice an odd number and so can never be zero and so we need the $pm a$ in RHS be $-a$
And so $a=(overbrace{pm 1pm 1 ... pm 1}^{49text{ times}})le 49$
Note that in order the $pm a$ becomes $-a$, at least one transformation needs to be $Xto 2-X$ and so $X=1$
It is easy to see that the optimum sequence for Xs then will be :
$(49,47,45,...,3,1,1,3,5,...,45,47),(49,47,45,...,3,1,1,3,5,...,45,47),...$
Which is, for the original sequence of $49$ numbers :
$frac{49^2-1}8,frac{47^2-1}8,...,frac{3^2-1}8,0,0,frac{3^2-1}8,...,frac{47^2-1}8$
And so the maximum sum is $sum_{k=0}^{24}frac{(2k+1)^2-1}8+sum_{k=0}^{23}frac{(2k+1)^2-1}8$
Which is $boxed{4900}$
My question: Could you please solve the generalize problem?
Thank you in advance!
combinatorics number-theory
asked Dec 19 '18 at 12:07
namnam
973
$endgroup$
$49$ integers are ordered around a circle. It is given that for any two adjacent numbers $x,y$ we have $x+y=(x-y)^2$. Find the maximal sum of all numbers in the circle.
Solution
Note that $x+y=(x-y)^2$ may be written as
$(2x-2y+1)^2=8y+1$ or also $(2y-2x+1)^2=8x+1$
And so all elements are such that $8x+1ge 0$
Replacing all elements $x$ on the circle by $X=sqrt{8x+1}$ the relation becomes
$X=pm Ypm 2$ (always with the constraint $X,Yge 0$)
So, applying $n$ times this process starting from $a$, we get $pm a+(overbrace{pm 1pm 1 ... pm 1}^{ntext{ times}})2$
In order to have a cycle after $49$ steps, we get $a=pm a+(overbrace{pm 1pm 1 ... pm 1}^{49text{ times}})2$
The second part is twice an odd number and so can never be zero and so we need the $pm a$ in RHS be $-a$
And so $a=(overbrace{pm 1pm 1 ... pm 1}^{49text{ times}})le 49$
Note that in order the $pm a$ becomes $-a$, at least one transformation needs to be $Xto 2-X$ and so $X=1$
It is easy to see that the optimum sequence for Xs then will be :
$(49,47,45,...,3,1,1,3,5,...,45,47),(49,47,45,...,3,1,1,3,5,...,45,47),...$
Which is, for the original sequence of $49$ numbers :
$frac{49^2-1}8,frac{47^2-1}8,...,frac{3^2-1}8,0,0,frac{3^2-1}8,...,frac{47^2-1}8$
And so the maximum sum is $sum_{k=0}^{24}frac{(2k+1)^2-1}8+sum_{k=0}^{23}frac{(2k+1)^2-1}8$
Which is $boxed{4900}$
My question: Could you please solve the generalize problem?
Thank you in advance!
combinatorics number-theory
combinatorics number-theory
asked Dec 19 '18 at 12:07
namnam
973
asked Dec 19 '18 at 12:07
namnam
973
asked Dec 19 '18 at 12:07
namnam
973
asked Dec 19 '18 at 12:07
namnam
973
asked Dec 19 '18 at 12:07
namnam
973
973
$begingroup$
By the generalized problem you mean the initial problem with 49 replaced by any integer $nge 2$, right?
$endgroup$
– Alex Ravsky
Dec 21 '18 at 3:32
1
$begingroup$
$ngeq 2$ integers are ordered around a circle. It is given that for any two adjacent numbers x,y we have $x+y=(x−y)^2$. Find the maximal sum of all numbers in the circle.
$endgroup$
– nam
Dec 23 '18 at 2:56
add a comment |
$begingroup$
By the generalized problem you mean the initial problem with 49 replaced by any integer $nge 2$, right?
$endgroup$
– Alex Ravsky
Dec 21 '18 at 3:32
1
$begingroup$
$ngeq 2$ integers are ordered around a circle. It is given that for any two adjacent numbers x,y we have $x+y=(x−y)^2$. Find the maximal sum of all numbers in the circle.
$endgroup$
– nam
Dec 23 '18 at 2:56
$begingroup$
By the generalized problem you mean the initial problem with 49 replaced by any integer $nge 2$, right?
$endgroup$
– Alex Ravsky
Dec 21 '18 at 3:32
$begingroup$
By the generalized problem you mean the initial problem with 49 replaced by any integer $nge 2$, right?
$endgroup$
– Alex Ravsky
Dec 21 '18 at 3:32
1
1
$begingroup$
$ngeq 2$ integers are ordered around a circle. It is given that for any two adjacent numbers x,y we have $x+y=(x−y)^2$. Find the maximal sum of all numbers in the circle.
$endgroup$
– nam
Dec 23 '18 at 2:56
$begingroup$
$ngeq 2$ integers are ordered around a circle. It is given that for any two adjacent numbers x,y we have $x+y=(x−y)^2$. Find the maximal sum of all numbers in the circle.
$endgroup$
– nam
Dec 23 '18 at 2:56
add a comment |
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$begingroup$
By the generalized problem you mean the initial problem with 49 replaced by any integer $nge 2$, right?
$endgroup$
– Alex Ravsky
Dec 21 '18 at 3:32
1
$begingroup$
$ngeq 2$ integers are ordered around a circle. It is given that for any two adjacent numbers x,y we have $x+y=(x−y)^2$. Find the maximal sum of all numbers in the circle.
$endgroup$
– nam
Dec 23 '18 at 2:56