Basic

[Lasdem]

Now that WolframAlpha is available,
this chellange got very easy =)[/url]

Don't know what i mean?
click me[/url]

[megabreit]

It was easy for UNIX users long before that.
Check bc and the obase command:
$ bc
obase=7; 8
11

WolframAlpha is an amazing site, but can probably not solve most of the harder math challenges.
E.g. "Biggest Fib" runs into an timeout. You don't need a calculator anymore, but still have to use your own brain

[matter]

Yea, I used the BC libraries in PHP. Code was about 5 lines.

[Aghamemnon]

I solved it by hand
just for fun
These things amuse me

[chephy]

Yea, I used the BC libraries in PHP. Code was about 5 lines.

5 lines? Ruby: »28679718602997181072337614380936720482949.to_s 7«

[wolfamstart]

i know java is normaly realy gabby, but this was quit short:

Code: Select all

return new BigInteger("28679718602997181072337614380936720482949").toString(7); 

But the real challenge is to code the convertion by yourself. I try it soon.

[FreeFull]

I used dc

[laz0r]

Mathematica:

IntegerString[28679718602997181072337614380936720482949, 7]

[the_austria]

I wrote an own function in python (for fun)

Code: Select all

def base(number,base):
  newnumber=[]
  while number>0:
    newnumber.insert(0,number%base)
    number=number//base
  return newnumber

It returns a list of the digets.

[DHARMENDRA VERMA]

i solve this problem in java but i am wondering that how could I solve this in C ??
Is there any Header file which support integers with such large digits???

[AMindForeverVoyaging]

There are "big integer" libraries which can be used with C. One of them is the GNU Multiple Precision Arithmetic Library.

[Schnapphahn]

Thanks Mathematica

Code: Select all

BaseForm [28679718602997181072337614380936720482949, 7]

[Grusewolf]

I solved it in groovy in straight forward programing

Code: Select all

BigInteger number = 28679718602997181072337614380936720482949
def pow = 50
print "septal number: "
while(pow >= 0) {
    def digitValue = 7.power(pow)
    def currentDigit = 0
    while(number >= digitValue) {
        currentDigit++
        number -= digitValue
    }
    print currentDigit
    pow--
}

But after reading the solution of wolfamstart in this Forum I recognized, that groovy also provides a very quick solution:

Code: Select all

println 28679718602997181072337614380936720482949.toString(7)