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Topic: 2506=2*10^3+5*10^2+0*10+6^1*10^0
Replies: 8   Last Post: Oct 30, 2013 1:44 PM

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JT

Posts: 1,150
Registered: 4/7/12
Re: 2506=2*10^3+5*10^2+0*10+6^1*10^0
Posted: Oct 30, 2013 1:39 PM
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Den onsdagen den 30:e oktober 2013 kl. 18:09:46 UTC+1 skrev scattered:
> On Wednesday, October 30, 2013 10:40:55 AM UTC-4, jonas.t...@gmail.com wrote:
>

> > Den tisdagen den 29:e oktober 2013 kl. 21:00:44 UTC+1 skrev Virgil:
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> >
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> > > In article <19afd53e-b219-4480-a630-d534b39b5416@googlegroups.com>,
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> >
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> > >
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> >
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> > > jonas.thornvall@gmail.com wrote:
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> >
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> > >
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> >
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> > >
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> >
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> > >
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> > > > Den tisdagen den 29:e oktober 2013 kl. 20:16:53 UTC+1 skrev
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> >
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> > >
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> >
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> > > > jonas.t...@gmail.com:
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> >
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> > >
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> >
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> > > > > 2506=2*10^3+5*10^2+0*10+6^1*10^0
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> >
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> > >
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> >
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> > > > >
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> >
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> > >
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> >
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> > > The above is sometimes called a "base 10 expansion" or "decimal
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> >
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> > >
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> >
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> > > expansion".
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> > >
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> > >
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> > >
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> >
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> > > And while there are comparable binary, octal and hexadecimal expansions,
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> >
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> > >
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> >
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> > > particulatlry in computer technology, the expansions below are of quite
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> >
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> > >
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> >
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> > > a different type, and have no name that I am aware of.
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> >
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> > >
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> >
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> > > > >
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> >
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> > >
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> >
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> > > > > Is there a name for when you write out a number as the right side
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> >
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> > >
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> >
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> > > > > expression?
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> >
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> > >
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> > > >
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> > >
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> > > > If so there must be a name for when you write out expression like.
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> > >
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> > > > Exp=x^2
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> > >
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> > > > 7777777777= 88191 353 26 3 +2 = 88191^2+ 353^2+ 26^2+ 3^2 +2
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> >
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> > >
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> >
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> > > > Exp=x^3
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> >
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> > >
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> >
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> > > > 7777777777= 1981 153 33 10 4 3 3 +4=1981^3+ 153^3+ 33^3+ 10^3+ 4^3+ 3^3+ 3^3
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> >
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> > >
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> >
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> > > > +4
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> >
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> > >
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> >
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> > > > Exp=x^4
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> > >
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> > > > 7777777777= 296 100 33 12 9 5 4 3 3 2 2 2 +12 =296^4+ 100^4+ 33^4+ 12^4+ 9^4+
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> >
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> > >
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> > > > 5^4+ 4^4+ 3^4+ 3^4+ 2^4+ 2^4+ 2^4 +12
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> >
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> > >
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> >
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> > > >
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> > >
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> >
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> > > > Or maybe there simply isn't so what should i call these forms?
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> >
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> > >
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> >
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> > > --
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> > I managed to implement it in python also.
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> >
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> >
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> >
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> > #!/usr/bin/python
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> >
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> > import math
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> >
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> > # Function definition is here
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> >
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> > def sq(number):
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> >
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> >
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> >
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> > exp=1
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> >
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> > factor=2
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> >
>
> > multip=math.pow(2,exponent)
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> >
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> > print(x,"= ", end="")
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> >
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> > while number>=multip:
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> >
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> > while exp<=number:
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> >
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> > factor+=1
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> >
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> > exp=math.pow(factor,exponent)
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> >
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> > factor-=1
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> >
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> > print(factor,"^",exponent,"+",sep="",end="")
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> >
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> > exp=math.pow(factor,exponent)
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> >
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> > number=number-exp
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> >
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> > exp=1
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> >
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> > factor=1
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> >
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> > print(number)
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> >
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> >
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> >
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> > #Set exponent here
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> >
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> > exponent=2
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> >
>
> > print("Exp=x^",exponent,sep="")
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> >
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> > #Set range of numbers x
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> >
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> > for x in range (1,100):
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> >
>
> > sq(x)
>
>
>
> Here is a Python implementation of a fairly
>
> general function which can used to solve this and other, similar problems:
>
>
>
> #if f is an increasing function from positive
>
> #integers to positive integers with the
>
> #property that f(1) = 1
>
> #then every integer n can be written as
>
> #a sum of numbers of the form f(k) with k <=n
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> #the following implements a greedy algorithm to do so
>
>
>
> #first, a binary search algorithm
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> #to find largest k <= n with f(k) <= n
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>
>
> def largest(n,f):
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> if f(n) <= n: return n
>
> low = 1
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> high = n #f(low) <= n but f(high) > n
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> while low + 1 < high:
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> mid = (low + high) // 2
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> if f(mid) <= n:
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> low = mid
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> else:
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> high = mid
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> return low
>
>
>
> #as an application,
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> #we can get an integer square root function
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> #albeit not the most efficient one
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>
>
> def square(n): return n*n
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>
>
> def isqrt(n):
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> return largest(n,square)
>
>
>
> #now implement a greedy algorithm
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> #it returns a list of numbers in descending order
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>
>
> def find_terms(n,f):
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> k = largest(n,f)
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> terms = [k]
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> n = n - f(k)
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> while n > 0:
>
> k = largest(n,f)
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> terms.append(k)
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> n = n - f(k)
>
> return terms
>
>
>
> #for example, find_terms(78,square) returns [8,3,2,1]
>
> #since 78 = 64 + 9 + 4 + 1


Honestly my code seem more adept, and have you noticed that each number created is unique, so one actually do not need separators. A program could convert any exponential modulus formatted number as long it do know the exp^

#!/usr/bin/python
import math
# Function definition is here
def sq(number):

exp=1
factor=2
multip=math.pow(2,exponent)
print(x,"= ", end="")
while number>=multip:
while exp<=number:
factor+=1
exp=math.pow(factor,exponent)
factor-=1
print(factor,"^",exponent,"+",sep="",end="")
exp=math.pow(factor,exponent)
number=number-exp
exp=1
factor=1
print(number)

#Set exponent here
exponent=2
print("Exp=x^",exponent,sep="")
#Set range of numbers x
for x in range (1,100):
sq(x)



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