Math Question...Bits Rewarded
Author 
Message 
Archi

Posted: Mon Jul 28, 2003 11:43 pm Post subject: Math Question...Bits Rewarded 


45 bits to the one who gets this question correct.
The problem is as follows:
ax+by/m=d
Where, when
a=40 & b=35 d=52
a=46, b=40, d=63
a=54, b=43, d=75
a=62, b=48, d=88
a=70, b=55, d=105
a=78, b=64, d=125
a=86, b=75, d=144
a=94, b=88, d=171
a=102, b=103, d=201
Find x,y,m.
Note: x,y,m must be constant throughout the entire solution. 





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bugzpodder

Posted: Tue Jul 29, 2003 4:11 pm Post subject: (No subject) 


no solution 





Archi

Posted: Tue Jul 29, 2003 4:19 pm Post subject: (No subject) 


hmm... Okey...well try this:
I have 9 weapons...I have designated a certian % chance to hit for each weapon. What I need though, is a formula which will work for each weapon and will take into account the players accuracy, and also the hit chance of the weapon.
Here are the number with the acc being that at the lowest lvl possible to equip the weapon.
Weap 1: a=40 & b=35 d=74%
Weap 3: a=46, b=40, d=77%
Weap 3: a=54, b=43, d=80%
Weap 4: a=62, b=48, d=83%
Weap 5: a=70, b=55, d=86%
Weap 6: a=78, b=64, d=89%
Weap 7: a=86, b=75, d=92%
Weap 8: a=94, b=88, d=95%
Weap 9: a=102, b=103, d=98%
This is for my hit/miss system where I will randomly pick a number between either 1(a+b) or 1100 depending on which has a valid solution.
The d value is what percent I want the player to hit. So for weapon 1, I want him to hit roughly 74% of the time. So the answer has to be greater than 74% of the numbers. 





bugzpodder

Posted: Thu Jul 31, 2003 12:22 pm Post subject: (No subject) 


take your ideal hit percentage of the weapon, multiply by the player's accuracy and some constant. so you need some kind of table that contains the information you gave me 





the_one

Posted: Wed Apr 21, 2004 5:45 pm Post subject: (No subject) 


i dun get it....confused 





the_one

Posted: Wed Apr 21, 2004 5:45 pm Post subject: (No subject) 


can ya teach? 





beard0

Posted: Fri May 28, 2004 9:21 am Post subject: (No subject) 


bugzpodder is right, no solution.
Evidence:
since y and m remain constant, make constant z=y/m
equation is now ax+bz=d
the first two equations read
1) 40x+35z=53
2) 46x+40z=63
mult 1) by 1.15
46x+40.25z=59.8
Subtract new equation from 2) to get
0x0.25z=3.2
0.25z=3.2
z=12.8
Substituting this into 1)
40x+35(12.8)=52
40x448=52
40x=500
x=12.5
These value in the third equation
d=54(12.5)+43(12.8)=675550.4=124.6
d is supposed to be 75.
Therefore, no solution. 






