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Number of circles required to reach a length of 100 million units in form of circle
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I need a formula to find out number of circles required to reach a length of 100 million units in form of circle. All circles are in offset from first circle with +1 unit in radius. all circles have a common centre.
N=X*3.14*2 (x is the value of radius of circle)(N=perimeter of circle)
Y=N+6.28 ((Y is the value of circumference of next circle's radius with x+1unit) (a circle with a radius of 100 units has a perimeter of 628 units,
a circle with 101 unit has a perimeter of 634.28 units, a circle with 102 unit has a perimeter of 640.56 units, the difference is always 6.28 units.))
so, addition of perimeter of first three circles is (628+634.28+640.56) 1902.84) units. what should be the formula to find out number of circles required to reach 100 million units.
number-theory
asked Jul 19 at 7:12
Sayan Mukherjee
11
 |Â
show 5 more comments
up vote
0
down vote
favorite
I need a formula to find out number of circles required to reach a length of 100 million units in form of circle. All circles are in offset from first circle with +1 unit in radius. all circles have a common centre.
N=X*3.14*2 (x is the value of radius of circle)(N=perimeter of circle)
Y=N+6.28 ((Y is the value of circumference of next circle's radius with x+1unit) (a circle with a radius of 100 units has a perimeter of 628 units,
a circle with 101 unit has a perimeter of 634.28 units, a circle with 102 unit has a perimeter of 640.56 units, the difference is always 6.28 units.))
so, addition of perimeter of first three circles is (628+634.28+640.56) 1902.84) units. what should be the formula to find out number of circles required to reach 100 million units.
number-theory
asked Jul 19 at 7:12
Sayan Mukherjee
11
If all the circles have a common center, then how can they be offset from the first circle, or reach much of any length at all?
– Gerry Myerson
Jul 19 at 7:21
@GerryMyerson the circles are in radiating pattern, like first circle is of radius 100mm, the second circle is of radius 101mm and expanding.
– Sayan Mukherjee
Jul 19 at 7:28
N= 100x3.14x2=628, N1=101x3.14x2=634.28, N2=102x3.14x2=640.56, N3=103x3.14x2=646.84, N4=104x3.14x2=653.12 and so on.....
– Sayan Mukherjee
Jul 19 at 7:29
I would recommend clarifying the question and making the terms consistent. It's a bit vague to say "circle that reaches $10^8$ units" - so what exactly reaches the value of $10^8$? The perimeter? or the radius? Also, if you want a circle with perimeter $10^8$, it's possible to simply solve for the radius and you're done.
– Matti P.
Jul 19 at 7:38
1
Read the wikipedia page: en.wikipedia.org/wiki/Arithmetic_progression
– Matti P.
Jul 19 at 8:37
 |Â
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need a formula to find out number of circles required to reach a length of 100 million units in form of circle. All circles are in offset from first circle with +1 unit in radius. all circles have a common centre.
N=X*3.14*2 (x is the value of radius of circle)(N=perimeter of circle)
Y=N+6.28 ((Y is the value of circumference of next circle's radius with x+1unit) (a circle with a radius of 100 units has a perimeter of 628 units,
a circle with 101 unit has a perimeter of 634.28 units, a circle with 102 unit has a perimeter of 640.56 units, the difference is always 6.28 units.))
so, addition of perimeter of first three circles is (628+634.28+640.56) 1902.84) units. what should be the formula to find out number of circles required to reach 100 million units.
number-theory
asked Jul 19 at 7:12
Sayan Mukherjee
11
I need a formula to find out number of circles required to reach a length of 100 million units in form of circle. All circles are in offset from first circle with +1 unit in radius. all circles have a common centre.
N=X*3.14*2 (x is the value of radius of circle)(N=perimeter of circle)
Y=N+6.28 ((Y is the value of circumference of next circle's radius with x+1unit) (a circle with a radius of 100 units has a perimeter of 628 units,
a circle with 101 unit has a perimeter of 634.28 units, a circle with 102 unit has a perimeter of 640.56 units, the difference is always 6.28 units.))
so, addition of perimeter of first three circles is (628+634.28+640.56) 1902.84) units. what should be the formula to find out number of circles required to reach 100 million units.
number-theory
asked Jul 19 at 7:12
Sayan Mukherjee
11
edited Jul 19 at 7:22
asked Jul 19 at 7:12
Sayan Mukherjee
11
asked Jul 19 at 7:12
Sayan Mukherjee
11
asked Jul 19 at 7:12
Sayan Mukherjee
11
11
If all the circles have a common center, then how can they be offset from the first circle, or reach much of any length at all?
– Gerry Myerson
Jul 19 at 7:21
@GerryMyerson the circles are in radiating pattern, like first circle is of radius 100mm, the second circle is of radius 101mm and expanding.
– Sayan Mukherjee
Jul 19 at 7:28
N= 100x3.14x2=628, N1=101x3.14x2=634.28, N2=102x3.14x2=640.56, N3=103x3.14x2=646.84, N4=104x3.14x2=653.12 and so on.....
– Sayan Mukherjee
Jul 19 at 7:29
I would recommend clarifying the question and making the terms consistent. It's a bit vague to say "circle that reaches $10^8$ units" - so what exactly reaches the value of $10^8$? The perimeter? or the radius? Also, if you want a circle with perimeter $10^8$, it's possible to simply solve for the radius and you're done.
– Matti P.
Jul 19 at 7:38
1
Read the wikipedia page: en.wikipedia.org/wiki/Arithmetic_progression
– Matti P.
Jul 19 at 8:37
 |Â
show 5 more comments
If all the circles have a common center, then how can they be offset from the first circle, or reach much of any length at all?
– Gerry Myerson
Jul 19 at 7:21
@GerryMyerson the circles are in radiating pattern, like first circle is of radius 100mm, the second circle is of radius 101mm and expanding.
– Sayan Mukherjee
Jul 19 at 7:28
N= 100x3.14x2=628, N1=101x3.14x2=634.28, N2=102x3.14x2=640.56, N3=103x3.14x2=646.84, N4=104x3.14x2=653.12 and so on.....
– Sayan Mukherjee
Jul 19 at 7:29
I would recommend clarifying the question and making the terms consistent. It's a bit vague to say "circle that reaches $10^8$ units" - so what exactly reaches the value of $10^8$? The perimeter? or the radius? Also, if you want a circle with perimeter $10^8$, it's possible to simply solve for the radius and you're done.
– Matti P.
Jul 19 at 7:38
1
Read the wikipedia page: en.wikipedia.org/wiki/Arithmetic_progression
– Matti P.
Jul 19 at 8:37
If all the circles have a common center, then how can they be offset from the first circle, or reach much of any length at all?
– Gerry Myerson
Jul 19 at 7:21
If all the circles have a common center, then how can they be offset from the first circle, or reach much of any length at all?
– Gerry Myerson
Jul 19 at 7:21
@GerryMyerson the circles are in radiating pattern, like first circle is of radius 100mm, the second circle is of radius 101mm and expanding.
– Sayan Mukherjee
Jul 19 at 7:28
@GerryMyerson the circles are in radiating pattern, like first circle is of radius 100mm, the second circle is of radius 101mm and expanding.
– Sayan Mukherjee
Jul 19 at 7:28
N= 100x3.14x2=628, N1=101x3.14x2=634.28, N2=102x3.14x2=640.56, N3=103x3.14x2=646.84, N4=104x3.14x2=653.12 and so on.....
– Sayan Mukherjee
Jul 19 at 7:29
N= 100x3.14x2=628, N1=101x3.14x2=634.28, N2=102x3.14x2=640.56, N3=103x3.14x2=646.84, N4=104x3.14x2=653.12 and so on.....
– Sayan Mukherjee
Jul 19 at 7:29
I would recommend clarifying the question and making the terms consistent. It's a bit vague to say "circle that reaches $10^8$ units" - so what exactly reaches the value of $10^8$? The perimeter? or the radius? Also, if you want a circle with perimeter $10^8$, it's possible to simply solve for the radius and you're done.
– Matti P.
Jul 19 at 7:38
I would recommend clarifying the question and making the terms consistent. It's a bit vague to say "circle that reaches $10^8$ units" - so what exactly reaches the value of $10^8$? The perimeter? or the radius? Also, if you want a circle with perimeter $10^8$, it's possible to simply solve for the radius and you're done.
– Matti P.
Jul 19 at 7:38
1
1
Read the wikipedia page: en.wikipedia.org/wiki/Arithmetic_progression
– Matti P.
Jul 19 at 8:37
Read the wikipedia page: en.wikipedia.org/wiki/Arithmetic_progression
– Matti P.
Jul 19 at 8:37
 |Â
show 5 more comments
1 Answer
1
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oldest
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up vote
0
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We consider $n$ circles, where the radius of each circle is equal to $99+n$. The perimeter of the $n$th circle is therefore
$$
p(n) = 2pi (99+n)
$$
When we add up the perimeters of the circles, we get the sum of an arithmetic progression
$$
beginsplit
S =& p(1) + p(2) +p(3) +ldots + p(n)\ =& 2pi left(100+101+102 + ldots + 99 +n right)\
=&2pi(underbrace99+99+99 +ldots+99_textn terms) + 2pi left(1+2+3+ldots+n right)
endsplit
$$
The summing the $99$'s together makes just $99n$, and for the sum $1+2+dots+n$ we can find the identity
$$
1+2+3+ldots+n = fracn(n+1)2
$$
and therefore
$$
S = 198pi n + 2pi fracn(n+1)2
$$
We want this to be at least $S= 10^8$, and therefore we get the equation
$$
2times 10^8 = 396pi n + 2pi n^2 + 2pi n qquad Rightarrow qquad 2pi n^2 + 398pi n - 2times 10^8 = 0
$$
Numerically, we can find $napprox 5~544$. This is the number of circles required so that the sum of perimeters is at least to 100 million.
answered Jul 19 at 8:52
Matti P.
1,308212
So the answer is 5544 circles. Many many thanks @Matti P.
– Sayan Mukherjee
Jul 19 at 9:13
You're welcome! If you think that the answer is correct, feel free to mark is as the accepted answer.
– Matti P.
Jul 19 at 9:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We consider $n$ circles, where the radius of each circle is equal to $99+n$. The perimeter of the $n$th circle is therefore
$$
p(n) = 2pi (99+n)
$$
When we add up the perimeters of the circles, we get the sum of an arithmetic progression
$$
beginsplit
S =& p(1) + p(2) +p(3) +ldots + p(n)\ =& 2pi left(100+101+102 + ldots + 99 +n right)\
=&2pi(underbrace99+99+99 +ldots+99_textn terms) + 2pi left(1+2+3+ldots+n right)
endsplit
$$
The summing the $99$'s together makes just $99n$, and for the sum $1+2+dots+n$ we can find the identity
$$
1+2+3+ldots+n = fracn(n+1)2
$$
and therefore
$$
S = 198pi n + 2pi fracn(n+1)2
$$
We want this to be at least $S= 10^8$, and therefore we get the equation
$$
2times 10^8 = 396pi n + 2pi n^2 + 2pi n qquad Rightarrow qquad 2pi n^2 + 398pi n - 2times 10^8 = 0
$$
Numerically, we can find $napprox 5~544$. This is the number of circles required so that the sum of perimeters is at least to 100 million.
answered Jul 19 at 8:52
Matti P.
1,308212
So the answer is 5544 circles. Many many thanks @Matti P.
– Sayan Mukherjee
Jul 19 at 9:13
You're welcome! If you think that the answer is correct, feel free to mark is as the accepted answer.
– Matti P.
Jul 19 at 9:36
add a comment |Â
up vote
0
down vote
We consider $n$ circles, where the radius of each circle is equal to $99+n$. The perimeter of the $n$th circle is therefore
$$
p(n) = 2pi (99+n)
$$
When we add up the perimeters of the circles, we get the sum of an arithmetic progression
$$
beginsplit
S =& p(1) + p(2) +p(3) +ldots + p(n)\ =& 2pi left(100+101+102 + ldots + 99 +n right)\
=&2pi(underbrace99+99+99 +ldots+99_textn terms) + 2pi left(1+2+3+ldots+n right)
endsplit
$$
The summing the $99$'s together makes just $99n$, and for the sum $1+2+dots+n$ we can find the identity
$$
1+2+3+ldots+n = fracn(n+1)2
$$
and therefore
$$
S = 198pi n + 2pi fracn(n+1)2
$$
We want this to be at least $S= 10^8$, and therefore we get the equation
$$
2times 10^8 = 396pi n + 2pi n^2 + 2pi n qquad Rightarrow qquad 2pi n^2 + 398pi n - 2times 10^8 = 0
$$
Numerically, we can find $napprox 5~544$. This is the number of circles required so that the sum of perimeters is at least to 100 million.
answered Jul 19 at 8:52
Matti P.
1,308212
So the answer is 5544 circles. Many many thanks @Matti P.
– Sayan Mukherjee
Jul 19 at 9:13
You're welcome! If you think that the answer is correct, feel free to mark is as the accepted answer.
– Matti P.
Jul 19 at 9:36
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We consider $n$ circles, where the radius of each circle is equal to $99+n$. The perimeter of the $n$th circle is therefore
$$
p(n) = 2pi (99+n)
$$
When we add up the perimeters of the circles, we get the sum of an arithmetic progression
$$
beginsplit
S =& p(1) + p(2) +p(3) +ldots + p(n)\ =& 2pi left(100+101+102 + ldots + 99 +n right)\
=&2pi(underbrace99+99+99 +ldots+99_textn terms) + 2pi left(1+2+3+ldots+n right)
endsplit
$$
The summing the $99$'s together makes just $99n$, and for the sum $1+2+dots+n$ we can find the identity
$$
1+2+3+ldots+n = fracn(n+1)2
$$
and therefore
$$
S = 198pi n + 2pi fracn(n+1)2
$$
We want this to be at least $S= 10^8$, and therefore we get the equation
$$
2times 10^8 = 396pi n + 2pi n^2 + 2pi n qquad Rightarrow qquad 2pi n^2 + 398pi n - 2times 10^8 = 0
$$
Numerically, we can find $napprox 5~544$. This is the number of circles required so that the sum of perimeters is at least to 100 million.
answered Jul 19 at 8:52
Matti P.
1,308212
We consider $n$ circles, where the radius of each circle is equal to $99+n$. The perimeter of the $n$th circle is therefore
$$
p(n) = 2pi (99+n)
$$
When we add up the perimeters of the circles, we get the sum of an arithmetic progression
$$
beginsplit
S =& p(1) + p(2) +p(3) +ldots + p(n)\ =& 2pi left(100+101+102 + ldots + 99 +n right)\
=&2pi(underbrace99+99+99 +ldots+99_textn terms) + 2pi left(1+2+3+ldots+n right)
endsplit
$$
The summing the $99$'s together makes just $99n$, and for the sum $1+2+dots+n$ we can find the identity
$$
1+2+3+ldots+n = fracn(n+1)2
$$
and therefore
$$
S = 198pi n + 2pi fracn(n+1)2
$$
We want this to be at least $S= 10^8$, and therefore we get the equation
$$
2times 10^8 = 396pi n + 2pi n^2 + 2pi n qquad Rightarrow qquad 2pi n^2 + 398pi n - 2times 10^8 = 0
$$
Numerically, we can find $napprox 5~544$. This is the number of circles required so that the sum of perimeters is at least to 100 million.
answered Jul 19 at 8:52
Matti P.
1,308212
answered Jul 19 at 8:52
Matti P.
1,308212
answered Jul 19 at 8:52
Matti P.
1,308212
answered Jul 19 at 8:52
Matti P.
1,308212
1,308212
So the answer is 5544 circles. Many many thanks @Matti P.
– Sayan Mukherjee
Jul 19 at 9:13
You're welcome! If you think that the answer is correct, feel free to mark is as the accepted answer.
– Matti P.
Jul 19 at 9:36
add a comment |Â
So the answer is 5544 circles. Many many thanks @Matti P.
– Sayan Mukherjee
Jul 19 at 9:13
You're welcome! If you think that the answer is correct, feel free to mark is as the accepted answer.
– Matti P.
Jul 19 at 9:36
So the answer is 5544 circles. Many many thanks @Matti P.
– Sayan Mukherjee
Jul 19 at 9:13
So the answer is 5544 circles. Many many thanks @Matti P.
– Sayan Mukherjee
Jul 19 at 9:13
You're welcome! If you think that the answer is correct, feel free to mark is as the accepted answer.
– Matti P.
Jul 19 at 9:36
You're welcome! If you think that the answer is correct, feel free to mark is as the accepted answer.
– Matti P.
Jul 19 at 9:36
add a comment |Â
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If all the circles have a common center, then how can they be offset from the first circle, or reach much of any length at all?
– Gerry Myerson
Jul 19 at 7:21
@GerryMyerson the circles are in radiating pattern, like first circle is of radius 100mm, the second circle is of radius 101mm and expanding.
– Sayan Mukherjee
Jul 19 at 7:28
N= 100x3.14x2=628, N1=101x3.14x2=634.28, N2=102x3.14x2=640.56, N3=103x3.14x2=646.84, N4=104x3.14x2=653.12 and so on.....
– Sayan Mukherjee
Jul 19 at 7:29
I would recommend clarifying the question and making the terms consistent. It's a bit vague to say "circle that reaches $10^8$ units" - so what exactly reaches the value of $10^8$? The perimeter? or the radius? Also, if you want a circle with perimeter $10^8$, it's possible to simply solve for the radius and you're done.
– Matti P.
Jul 19 at 7:38
1
Read the wikipedia page: en.wikipedia.org/wiki/Arithmetic_progression
– Matti P.
Jul 19 at 8:37