Re: A tough mathematical problem
- From: Phil Carmody <thefatphil_demunged@xxxxxxxxxxx>
- Date: 23 Oct 2007 13:30:15 +0300
"mensanator@xxxxxxxxxxx" <mensanator@xxxxxxx> writes:
On Oct 21, 4:42 pm, DDP3...@xxxxxxxxx wrote:
Hello.
We have several sets of given numbers to which a spesific algorithm is
applied in order to get given numbers. We also have some clues about
the algorithm.
1)500000,5000,1000,300,100,10 gives 42000
2)500000,5000,300,100,10 gives 60000
3)500000,5000,300,100 gives 95000
4)500000,5000,300 gives 140000
5)500000,5000 gives 245000
Some more clues:
a) the result depends on the total value of the given numbers
b) the result depends on how many are the given numbers
c) the result is rounded to the nearest thousand or maybe? it is
rounded earlier by for example substituting numbers lower than 1000
with 0
The logical (n1+n2+....)*x=m gives
x1=0.0829
x2=0.1187
x3=0.1879
x4=0.277
x5=0.4851 with accuracy of 4 digits
So now we could just search for the algorithm giving the step among
these factors keeping in mind that it has to do with how many are the
given numbers.
But, it is not so simple because we are given one more clue as a
notice that:
6)200000,100,0.5 gives 52000 which is different from
(200000+100+0.5)*0.277=55427,8385~55000
the 6th clue, as i understand it, reminds us that the x has to do with
the values of the given set too.
But in the end it may not be so simple as a single *x solution...
What is the simplest possible solution?
The solution is:
(((((n[0]*1/2-n[1])*4/7(iif n[2]))*19/28(iif n[3]))*12/19(iif
n[4]))*7/10(iif n[5]))
# Python
import gmpy
t = [[500000,5000,1000,300,100,10], \
[500000,5000,300,100,10], \
[500000,5000,300,100], \
[500000,5000,300], \
[500000,5000], \
[200000,100,0.5]]
# in common denominator, not necessary just wanted to
# see relationship of 1/2, 4/7, 19/28, 12/19, 7/10
ratios = [gmpy.mpq(1330,2660), \
gmpy.mpq(1520,2660), \
gmpy.mpq(1805,2660), \
gmpy.mpq(1680,2660), \
gmpy.mpq(1862,2660)]
for i in t:
n = i[0]*ratios[0] - i[1]
if len(t)>2:
for j,k in enumerate(i[2:]):
n = n*ratios[j+1]
print '%8d' % (n),i
## 42000 [500000, 5000, 1000, 300, 100, 10]
## 60000 [500000, 5000, 300, 100, 10]
## 95000 [500000, 5000, 300, 100]
## 140000 [500000, 5000, 300]
## 245000 [500000, 5000]
## 57085 [200000, 100, 0.5]
## clue 6 must be a typo
Congratulations on even seeing a clue 6 buried in the gibber-jabber!
I can evolve my solution into yours by dividing everything everything
by 500000 and multiplying by n[0]. But would that make it simpler?
Clue to the OP:
What's the next number in this sequence -
0, 0, 0, 1, 1, 2, 3, 4, 6, 9, ...
Now multiply the number of missing parameters by an infinitude...
Phil
--
Dear aunt, let's set so double the killer delete select all.
-- Microsoft voice recognition live demonstration
.
- References:
- A tough mathematical problem
- From: DDP3000
- Re: A tough mathematical problem
- From: mensanator@xxxxxxxxxxx
- A tough mathematical problem
- Prev by Date: Re: Implementable Set Theory and Consistency of ZFC
- Next by Date: Re: Implementable Set Theory and Consistency of ZFC
- Previous by thread: Re: A tough mathematical problem
- Next by thread: Dense subgroup of R^n
- Index(es):
Relevant Pages
|
