Find the largest value of power.

Question

Some numbers can be represented as perfect powers of other numbers. A number x can be represented as x = base^power for some integer base and power.

Given an integer x you have to find the largest value of power, such that base is also an integer.

Sample Input:
9
Sample Output:
2

Answer 1

Golfscript - 26 chars

~:x,{:b;x,{b?x=b*}%+}*$-1>

Rough translation to Python

x=input()
acc = []
for b in range(x):
    for _ in range(x):
        acc.append((_**b==x)*b) # most of these are zeros
print max(acc)

So it loops way more times that necessary, but that often happens with golfed answers

Answer 2

APL (Dyalog Extended), 13 bytes

⌈/⍳×∘(⊢=⌊)⍳√⊢

Try it online!

⌈/⍳×∘(⊢=⌊)⍳√⊢     Monadic train taking an input n:
         ⍳√⊢     Get the 1st, ..., nth roots of n.
    (⊢=⌊)        Find whether each element equals its floor; i.e. is an integer.
                Returns a list of 0s and 1s.
  ⍳              List of 1, ..., n
   ×            Multiply that with the binary list.
                Now we have a list starting with 1 (since the 1st root of n is n, an integer)
                and larger nonzero values corresponding to the other roots that are integers.
⌈/              Find the maximum.

Answer 3

Haskell, 63 56 characters

Handles ℤ_>0 (56 characters)

main=do n<-readLn;print$last[p|p<-[0..n],b<-[0..n],b^p==n]

Handles ℤ_≥0 (58 characters)

main=do n<-readLn;print$last[p|p<-[0..n+1],b<-[0..n],b^p==n]

Handles ℤ (70 characters)

main=do n<-readLn;print$last[p|p<-[0..abs n+1],b<-[n..0]++[0..n],b^p==n]

Answer 4

Japt -h, 9 bytes

õ f@qX v1

Try it

õ f@qX v1     :Implicit input of integer U
õ             :Range [0,U]
  f@          :Filter as X
    qX        :  Xth root of U
       v1     :  Divisible by 1?
              :Implicit output of last element

Answer 5

05AB1E (legacy), 9 8 bytes

Lʒzm.ï}à

-1 byte thanks to @Don't be a x​‑triple dot.
Only works in the Python legacy version of 05AB1E, because .ï should be DïQ in the Elxiir rewrite due to a bug; and it's also way slower due to zm.

Try it online or verify all integers in the range [2,100].

Explanation:

\$b^{^\frac{1}a}\$ is another way to write \$\sqrt[^a]{b}\$

L           # Create a list in the range [1, (implicit) input]
 ʒ          # Filter this list of integers by:
  z         #  Do 1 divided by the integer
   m        #  And then take it to the power of the (implicit) input
    .ï      #  And check if it's an integer without decimal values after the comma
      }à    # After the filter: pop the list and push the maximum
            # (which is output implicitly)

Answer 6

Python, 72 chars

x=range(1,input()+1)
print max(p*sum(b**p==len(x)for b in x)for p in x)

Answer 7

Mathematica : 61 chars

. First try:

Max[k/.Table[Solve[Log@#/Log@b==k,k,Integers],{b,2,#}]][[1]]&

Answer 8

Buggy Golfscript - 59

~0\1{1{1$1$?3$={p;\)\.(.}{}if).3$=!}do;).2$=!}do;;{}{1p}if

This is my first attempt with golfscript, so it probably can be improved in length.

Now the buggy part comes from the fact that this runs perfectly on the interpreter written in perl, but in the original ruby version, I'm having a really weird bug. I spent a couple of hours trying to wrap my head around it but I can't figure it out. I decided to post anyway to maybe get some feedback.

Answer 9

CSharp - 130 chars

void Main(){var i = Int32.Parse(Console.ReadLine());Console.WriteLine(Enumerable.Range(2,i-2).Last(k => Math.Pow(i,1.0/k)%1==0));}

Answer 10

APL(NARS), 34 chars, 68 bytes

{⍵≤1:∞⋄⌈/i/⍨{0=1∣⍵√w}¨i←⍳n←⌊2⍟w←⍵}

The idea is that the max possible exponent is n←⌊2⍟argument, so i build one range on that: 1..n and to get the max possible exponent; test:

  f←{⍵≤1:∞⋄⌈/i/⍨{0=1∣⍵√w}¨i←⍳n←⌊2⍟w←⍵}
  f 25
2
  f 30
1
  f 8
3
  f  323*(2×3)
6

They want max exponent, right?

Answer 11

Stax, 4 bytes

|n|g

Run and debug it

|n gets the prime factor exponents of the input. |g calculates the GCD of the result.