Write a program that takes two integers as an input; the first can be any integer and the second is less than or equal to the number of digits in the first number. Let these numbers be a and b respectively.

The program will do the following

  • Concatenate a minimal number of 1s to the end of a so the number of digits in a is divisible by b.
  • Split a along every b digits.
  • Multiply the digits in each section together.
  • Concatenate the products together (if one of the numbers is zero, then concatenate 0).
  • Repeat this process until a number with strictly fewer than b digits is formed. Print this as the output, as well as the number of the process is repeated. Units are not necessary, but some form of separation between the final number and number of iterations is.

In the following test cases, the individual steps are shown for the purpose of understanding. It is not necessary for your program to display the steps.

Test case 1

1883915502469, 3


1883915502469          //Iteration 1
188 391 550 246 911
64 27 0 48 9
64270489               //Iteration 2
642 704 891
48 0 72
48072                  //Iteration 3
480 721
0 14
014                    //Iteration 4

Sample Output: 0, 4

Test case 2

792624998126442, 4


792624998126442        //Iteration 1
7926 2499 8126 4421
756 648 96 32
7566489632             //Iteration 2
7566 4896 3211
1260 1728 6
126017286              //Iteration 3
1260 1728 6111
0 112 6
01126                  //Iteration 4
0112 6111
0 6

Sample Output: 06, 4

The program must return an error (or just not print anything) if b>len(a). Also, b cannot equal 1 or the program will result in an infinite loop.

This is code golf, so standard rules apply. Shortest code in bytes wins.

  • \$\begingroup\$ Does it need to be a complete program, or is a function enough? \$\endgroup\$
    – Ypnypn
    Nov 3, 2015 at 16:01
  • \$\begingroup\$ @Ypnypn A complete program. \$\endgroup\$
    – Arcturus
    Nov 3, 2015 at 16:02
  • \$\begingroup\$ So leading zeros count towards the length of a and are also included in the output? \$\endgroup\$
    – mbomb007
    Nov 3, 2015 at 22:44
  • \$\begingroup\$ @mbomb007 Yes, but only in the initial number. The chain of zeroes would be shortened to single zeroes in the concatenation of products. \$\endgroup\$
    – Arcturus
    Nov 4, 2015 at 3:04
  • \$\begingroup\$ @ypnypn you should say that explicitly in the question. The "standard rules"from the tag wiki say "the following defaults ... Answers may be either full programs or functions (or equivalent)." \$\endgroup\$ Nov 7, 2015 at 1:15

6 Answers 6


Perl 6, 116 bytes

my ($a,$b)=@*ARGS;for 0..* {if $b>$a.chars {$_&&say "$a,$_";last};$a=map({[*] @_},($a~1 x$b-1).comb.rotor($b)).join}
my ($a,$b)=@*ARGS;
for 0..* {
  if $b>$a.chars {$_&&say "$a,$_";last}; # you need a 「;」 if you remove the newline
    {[*] @_},
    ($a~1 x$b-1).comb.rotor($b)

Python 3 + functools + operator, 242 241 205 bytes

Shorter version (thanks!)

def g(a,b):
 while len(a)>=d:a=j(map(str,[reduce(mul,map(int,[*j(u)]))for u in zip(*[iter([a+"1"*n for n in range(d)if not len(a+"1"*n)%d][0//(d<=len(a))])]*d)]));c+=1
 return (a,c)

Original explained version:

def g(a,b):
 while len(a)>=int(b):a="".join(map(str,[reduce(mul,map(int,u))for u in map(list,map(''.join,zip(*[iter([a+"1"*n for n in range(int(b))if not len(a+"1"*n)%int(b)][0])]*int(b))))]));c+=1
 return None if int(b)>len(a)else(a,c)

This was a fun one! I wish I could've made it a one-liner but there was no real way to do that without massively increasing the length. Takes input as two strings, returns a 2-tuple of a string and an int (or None if b>len(a))


def challenge(a,b):
  while len(a) >= int(b):
        reduce(mul,map(int,u)) # Multiply the digits together. Uses reduce from functools and mul from operator, works similarly to sum() but for multiplication
        for u in map(list,map(''.join,
                                [a+"1"*n for n in range(int(b)) if not len(a+"1"*n)%int(b)][0] # Do the 1 concatenation
                             ) # Split a along every b digits
                    ) # Split those strings into lists
      ]) # Convert to a list of strings
    ) # Concatenate the products together
  return None if int(b)>len(a) else(a,c) # I don't really like the "return None if invalid input" requirement, but ah well

Python, 156 bytes

g=lambda a,d,c=0,k=-1,j=''.join:d>(l:=len(a))and(a,c/c*c)or g(j([str(-[k:=k*x for x in map(int,j((a+l%d*"1")[i:i+d]))][k:=-1])for i in range(0,l,d)]),d,-~c)

Attempt This Online!

Takes in a string and an integer, causes a DivideByZero error if the input is invalid.

Python 3 + functools + operator, 137 bytes

g=lambda a,d,c=0,j=''.join:d>(l:=len(a))and(a,c/c*c)or g(j([str(reduce(mul,map(int,j((a+l%d*"1")[i:i+d]))))for i in range(0,l,d)]),d,-~c)

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CJam, 42 bytes


Test it here.


Pyth, 32 bytes



Takes input on two lines, a followed by b. Gives output on two lines, operations followed by result.

Pad: +z*\1%_lzQ

Chop: c ... Q

Convert to list of ints: sMM

Take products: *M

Convert back to str: jk

Assign back: =z

Check for termination: <l ... Q

Print iterations taken: f ... )

Print result: z

Initial check of whether to print anything at all: IglzQ


Ruby, 139 bytes

c.size<b ?[c,i]:f[c,b,i+1]}
p f[*$*]

Attempt This Online!


f = ->n,b,i=1{                    # Recursive lambda definition; n is the number, b is the target length, i is the number of iterations
  c = [                             # Make an array of...
    *n.scan(/.{#{b=b.to_i}}/),        # the whole sections of length b
    *$'[0] && $'.ljust(b, ?1)         # plus, if there’s stray letters, the final section padded to length b
    _1.chars.map(&:to_i).reduce :*    # Multiply each section’s digits
  } * ""                            # Join back together

  c.size < b ?                      # If we’re under the target length...
    [c, i] :                          # return the remaining string and iteration count
    f[c, b, i+1]                      # otherwise call itself recursively, incrementing the iteration count

p f[*$*]                          # $* is ARGV; splat it into f's arguments

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