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2009-11-28 14:55:20
HaiBai
Your thoughts, insights, and musings on this matter intrigue me
+304|5901|Bolingbrook, Illinois
Basically: 216 = y + y + y + y + y + z + a + b + c

z - y = d

a - z = d

b - a = d

c - b = d

Basically, I want the sum of nine different integers to equal 216.  The first 5 integers are the same value.  The distance between y and z is also the distance between z and a, a and b, and b and c.  Can someone help me out?
2009-11-28 14:55:30
Gooners
Wiki Contributor
+2,700|7049

42
2009-11-28 14:58:20
Finray
Hup! Dos, Tres, Cuatro
+2,629|6205|Catherine Black
https://i.imgur.com/qwWEP9F.png
2009-11-28 14:59:08
GravyDan
Back from the Grave(y)
+768|6349|CA
He's a witch!
2009-11-28 14:59:27
Macbeth
Banned
+2,444|6003

I want the sum of nine different integers to equal 216.  The first 5 integers are the same value.
If the nine integers are different the first 5 cannot be the same.
2009-11-28 15:03:48
HaiBai
Your thoughts, insights, and musings on this matter intrigue me
+304|5901|Bolingbrook, Illinois

Macbeth wrote:

I want the sum of nine different integers to equal 216.  The first 5 integers are the same value.
If the nine integers are different the first 5 cannot be the same.
Sorry, 4 integers are different, 5 are the same, and the sum = 216.
2009-11-28 15:04:48
HaiBai
Your thoughts, insights, and musings on this matter intrigue me
+304|5901|Bolingbrook, Illinois
Or if it's simpler, all 9 integers can be different, as long as the distance between all of them is the same and it equals 216.
2009-11-28 15:08:24
Morpheus
This shit still going?
+508|6416|The Mitten
216? Isn't that the number in that movie, Π ?
EE (hats
2009-11-28 15:09:59
HaiBai
Your thoughts, insights, and musings on this matter intrigue me
+304|5901|Bolingbrook, Illinois
No, 216 / 9 = 24.  I'm trying to find the average of 9 different numbers to = 24.  And I need the distance between all of them to be the same.
2009-11-28 15:15:05
Morpheus
This shit still going?
+508|6416|The Mitten
i meant
http://en.wikipedia.org/wiki/Pi_%28film%29

and yea, it was a 216 digit long number.

Anyway, try some basic substitution? Such as, 5y + z + a etc, and replace z with y + d etc.... until you have less and less variables...

But then again, i failed out of math, so who knows
EE (hats
2009-11-28 15:53:18
nukchebi0
Пушкин, наше всё
+387|6741|New Haven, CT
216=5y+z+a+b+c

z=y+d
a=y+2d
b=y+3d
c=y+4d

216=9y+10d
24-10d/9=y

The nature of the problem is that you'll have a set of answers for y and d (from which you can determine corresponding z, a, b, and c values). The d is the independent variable, as the equation form suggests, and consequently you'll have to choose that yourself.
2009-11-28 15:57:56
(HUN)Rudebwoy
Member
+45|7172
^that

Usually if you have more variables than equations you cannot solve the problem.
2009-11-28 16:14:56
Flaming_Maniac
prince of insufficient light
+2,490|7124|67.222.138.85
If you have more variables than equations that usually means there is no answer or there is a range of answers.

 

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