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

Uiua, 11 10 bytes

Edit: -1 byte thanks to ovs

⊢⇌⍖◿1ⁿ÷⇡,1

Try it!

A different approach to Bubbler's Uiua answer, for the same number of bytes.

       ÷⇡,1   # reciprocals of 0..input
      ⁿ       # input to the power of those (so: roots)
    ◿1        # modulo 1 (so whole numbers become zero)         
   ⍖          # fall: indices if it was sorted descending 
              # (index of largest root that is a whole number is now last)
⊢⇌            # get the last item

Answer 4

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 5

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 6

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.

Answer 7

05AB1E (legacy), 9 8 4 bytes

Ó0K¿

-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.
-4 bytes porting @recursive's Stax answer.

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

Previous 9 8 bytes solution:

Lʒzm.ï}à

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

Explanation:

Ó      # Get the exponents of the (implicit) input's prime factorization
 0K    # Remove all 0s
   ¿   # Pop and push the Greatest Common Divisor (GCD) of this list
       # (which is output implicitly as result)

\$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    #  Pop the value and push 1/value instead
   m   #  Then take it to the power of the (implicit) input
    .ï #  And check if it's an integer without decimal values after the period
 }à    # After the filter: pop the list and push the maximum
       # (which is output implicitly as result)

Answer 8

Uiua, 10 bytes

⊢⍖=⊞ⁿ.⇡+1.

Try it online!

-1 thanks to Dominic van Essen.

⊢⍖=⊞ⁿ.⇡+1.   input: positive integer n >= 2
        ⇡+1    range from 0 to n inclusive
    ⊞ⁿ.       cartesian product with itself by power
               (the k-th row contains x^k for x = 0..n)
   =       .   for each element, 1 if it equals n, 0 otherwise
               (higher power gives 1 at an earlier position,
                so the highest power is the lexicographically highest row)
⊢⍖            index of the maximum row lexicographically

---

Uiua, 11 bytes

⊢⍏⊗∶⊞ⁿ.⇡+1.

Try it online!

⊢⍏⊗∶⊞ⁿ.⇡+1.    input: positive integer n >= 2
         ⇡+1    range from 0 to n inclusive
     ⊞ⁿ.       cartesian product with itself by power
                (the k-th row contains x^k for x = 0..n)
   ⊗∶       .   the index of n in each row of the matrix, n+1 if not present
                (higher power gives lower index)
⊢⍏             index of the minimum element

Answer 9

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 10

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 11

Vyxal, 23 bits^v2, 2.875 bytes

∆ǐġ

Try it Online!

Bitstring:

00010101110010111001101

Ports Recursive's answer in Stax

Answer 12

Mathematica : 61 chars

. First try:

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

Answer 13

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 14

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 15

Scala, 70 bytes

Golfed version. Try it online!

x=>(for{b<-0 to x-1;a<-0 to x-1;if Math.pow(a,b).toInt==x}yield b).max

Ungolfed version. Try it online!

object Main {
  def main(args: Array[String]): Unit = {
    def f(x: Int): Int = {
      val acc = for {
        b <- 0 until x
        a <- 0 until x
        if scala.math.pow(a, b).toInt == x
      } yield b 
      acc.max
    }

    println(f(9))
  }
}