How much does this Integer score?

Question

The Challenge

Given an integer, calculate its score as described below.

The score is calculated by raising each digit of the input Integer to the power n, where n denotes the index of the digit in Input (beginning with 1) when counted from left to right (the index of leftmost digit should be 1), and then summing them all together. If the original number happens to be prime, then add the largest digit of the input integer to the score, else subtract the smallest digit of the input integer from it.

Input

Your program should take an integer as input. You can take the input in whatever way you want to.

Output

Output the score of the input integer through whatever medium your programming language supports. Function return is also allowed.

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Test Cases

 5328   ->  (5^1)+(3^2)+(2^3)+(8^4)-2 = 4116
 3067   ->  (3^1)+(0^2)+(6^3)+(7^4)+7 = 2627
 7      ->  (7^1)+7                   = 14

Scoring

This is code-golf, so the shortest code in bytes wins!

Answer 1

Mathematica, 63 bytes

s+Max@If[PrimeQ[s=Tr[d^Range@Length[d=IntegerDigits@#]]],d,-d]&

Pure function taking a positive integer as input and returning an integer. Directly performs the indicated computation, with all the long Mathematica command names that requires. Perhaps the one innovation is noticing that, if d is the list of digits of the input, then the last addition/subtraction is the same as adding Max[d] if the initial sum was prime, and adding Max[-d] if not.

Answer 2

Mathematica, 68 bytes

#+If[PrimeQ@#,Max@x,-Min@x]&@Tr[(x=IntegerDigits@#)^Range[Tr[1^x]]]&

Explanation

Sets x equal to the list of digits of the input, then raises that to the power Range[Tr[1^x]], effectively raising the first digit to the power 1, the second digit to the power 2, and so on up to Tr[1^x], the length of x. Tr sums the resulting list, which is then passed into the function #+If[PrimeQ@#,Max@x,-Min@x]&.

Answer 3

05AB1E, 16 bytes

DpiZëW(}¹gL¹SmO+

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Explanation

D                  # duplicate input
 piZ               # if input is prime then max digit of input
    ëW(            # else -min digit of input
       }           # end if
        ¹gL        # range [1 ... len(input)]
           ¹S      # input split into digits
             m     # pow
              O    # sum
               +   # add

Answer 4

Python 3: 260 167 bytes

def s(n):
 t,l=0,list(str(n))
 for i,d in enumerate(l):t+=int(d)**(int(i)+1)
  for i in range(2,n-1):
   if n%i==0:t-=int(min(l))
  else:t+=int(max(l))
  break
 return t

93 bytes saved by Wheat Wizard

Answer 5

JavaScript, 173 bytes

Add an f= at the beginning and invoke like f(n).

n=>{k=(n+"").split``;t=k.map((c,i)=>c**(i+1)).reduce((c,p)=>c+p);j=1;for(i=2;i<n;i++){j=n%i||n==2?1:0;if(!j)break}return j?t+Math.max.apply(Math,k):t-Math.min.apply(Math,k)}

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Explanation

Takes in an Integer. Converts it into String. Then, converts it into Array. Uses map over the Array to make a new array having all its elements raised to their index+1^th (+1 because JavaScript used zero-based indexes). Then uses reduce to sum them all together. Checks whether the obtained number is prime. If it is then it returns the obtained number incremented by the largest digit of the Array otherwise returns the obtained number decremented by smallest digit of it.

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Test Cases

f=n=>{k=(n+"").split``;t=k.map((c,i)=>c**(i+1)).reduce((c,p)=>c+p);j=1;for(i=2;i<n;i++){j=n%i||n==2?1:0;if(!j)break}return j?t+Math.max.apply(Math,k):t-Math.min.apply(Math,k)}

console.log(f(5328));
console.log(f(3067));
console.log(f(7));
console.log(f(123));
console.log(f(1729));
console.log(f(2));
console.log(f(1458));
console.log(f(91));

Answer 6

Ohm, 17 16 bytes (CP437)

EDIT: Fixed so that it checks the input's primality.

▓_^≥ⁿ;Σ┼Dp?╧+¿╤-

One of these days I need to get around to implementing vectorization. Maybe then I'll finally beat 05AB1E!

Explanation:

▓_^≥ⁿ;Σ┼Dp?╧+¿╤-    Main wire: arguments: n

▓    ;              Map the following over all digits of n:
 _^≥ⁿ                 Raise the digit to the (index + 1)th power
      Σ             Push the sum of all that
       ┼Dp          Is n prime?
          ?         If so...
           ╧+         Add the maximum digit of n to the sum
             ¿      Else...
              ╤-      Subtract the minimum digit of n from the sum
                    (Implicit output)

Answer 7

Ruby, 98 bytes

->n{i= ~s=0
a=n.digits
a.map{|d|s+=d**(a.size-i+=1)}
s>1&&(2...s).all?{|i|s%i>0}?s+a.max: s-a.min}

Answer 8

PHP, 174 bytes

function s($i){$i=str_split($i);foreach($i as $k=>$x){$r+=pow($x,$k+1);}if(p($r)){$r+=max($i);}else{$r-=min($i);}print$r;}function p($n){for($i=$n;--$i&&$n%$i;);return$i==1;}

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Answer 9

Python 3, 118 bytes

*a,=map(int,input())
x=sum(n**-~i for i,n in enumerate(a))
print(x+(-min(a),max(a))[all(x%z for z in range(2,x))*x>1])

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Thanks to @math_junkie for pointing out a bug.

Answer 10

Python 2, 111 bytes

g=lambda n,z=1:n and(n%10)**len(`n`)+g(n/10,0)+int(['-'+min(`n`),max(`n`)][all(n%x for x in range(2,n))*n>1])*z

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Recursive function which calculates the sum starting with the last digit.

n and(n%10)**len(`n`)+g(n/10,1) is the recursive portion of the function

int(['-'+min(`n`),max(`n`)][...])*z returns 0 for all iterations except the first

all(n%x for x in range(2,n))*n>1] is the primality test

Alternate solution (115 bytes):

lambda n:sum(int(j)**(i+1)for i,j in enumerate(`n`))+int(['-'+min(`n`),max(`n`)][all(n%x for x in range(2,n))*n>1])

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