# Reduce a number by its largest digit

Given an integer number in decimal number system, reduce it to a single decimal digit as follows:

1. Convert the number to a list of decimal digits.
2. Find the largest digit, D
3. Remove D from the list. If there is more than one occurrence of D, choose the first from the left (at the most significant position), all others should remain intact.
4. Convert the resulting list to a decimal number and multiply it by D.
5. If the number is bigger than 9 (has more than 1 decimal digit), repeat the whole procedure, feeding the result into it. Stop when you get a single-digit result.
6. Display the result.

Example:

26364 ->
1. 2 6 3 6 4
2. The largest digit is 6, so D=6
3. There are two occurrences or 6: at positions 1 and 3 (0-based). We remove the left one,
at position 1 and get the list 2 3 6 4
4. we convert the list 2 3 6 4 to 2364 and multiply it by D:
2364 * 6 = 14184
5. 14184 is greater than 9 so we repeat the procedure, feeding 14184 into it.


We continue by repeating the procedure for 14184 and so on and we go through the following intermediate results, finally reaching 8:

11312
3336
1998
1782
1376
952
468
368
288
224
88
64
24
8


So the result for 26364 is 8.

Input: An integer / a string representing an integer

Output: A single digit, the result of the reduction applied to the number.

Test cases:

9 -> 9
27 -> 4
757 -> 5
1234 -> 8
26364 -> 8
432969 -> 0
1234584 -> 8
91273716 -> 6


This is , so the shortest answers in bytes in each language win.

• Which is it If the number is bigger than 10 or has more than 1 decimal digit. The number 10 has more than 1 decimal digit, but it isn't bigger than ten.
Nov 18 '17 at 22:00
• @Adám By coding logics, should then 10 -> 10? Nov 19 '17 at 0:16
• @Adám You are right, I should have written "bigger than 9". I'm going to edit the description. Thanks! Nov 19 '17 at 7:37
• Has someone examined the histogram of this function for sufficiently large regions? It seems to have a lot of zeroes; I also got many 8s while composing the test cases. Nov 19 '17 at 9:54
• Also, a random number divisible by 4 has 3/5 probability of the product of the last two digits being divisible by 8. Nov 21 '17 at 18:36

# 05AB1E, 6 bytes

### Code:

[Dg#à*


Uses the 05AB1E encoding. Try it online!

### Explanation

[Dg#     # While the length of the number is not 1
à    # Extract the largest element from the current number
*   # Multiply it with the leftover number


# JavaScript (ES6), 49 bytes

f=n=>n>9?f(""+n.replace(m=Math.max(...n),"")*m):n


Takes input as a string representation of an integer, like f("26364").

## Test Cases

f=n=>n>9?f(""+n.replace(m=Math.max(...n),"")*m):n

;["9", "27", "757", "1234", "26364", "432969", "1234584", "91273716"]
.forEach(num => console.log(num + " -> " + f(num)))
.as-console-wrapper{max-height:100%!important}

# Jelly, 13 bytes

œṡṀẎḌ×ṀD
DÇḊ¿


Try it online!

-1 thanks to a trick I found in Jonathan Allan's answer.

Full program.

# Pyth, 16 bytes

.WtH*s.-ZKeSZsK


Takes input as a String. Try it here! (Alternative: .WtH*s.-ZeSZseS)

# Pyth, 18 bytes

.WgHT*s.-ZKeSZsK


Takes input as an integer. Try it here!

### How it works

16-byter

.WtH*s.-ZKeSZsK ~ Full program.

.W               ~ Functional while. While A(value) is truthy, value = B(value).
~ The final value is returned.
tH             ~ A, condition: Is value[1:] truthy?  Is the length ≥ 2?
*s.-ZKeSZsK ~ B, setter.
.-        ~ Bagwise subtraction, used for removing the highest digit, with...
Z       ~ The current value Z, and...
KeSZ   ~ The highest digit of Z (as a String). Also assigns to a variable K.
s          ~ Casted to an integer.
*           ~ Multiplied by...
sK ~ The highest digit.
~ Convert to a String.


18-byter

.WgHT*s.-ZKeSZsK ~ Full program.

.W                 ~ Functional while. While A(value) is truthy, value = B(value).
~ The final value is returned.
gHT              ~ A, condition: is value (H) ≥ 10?
*s.-ZKeSZsK ~ B, setter.
.-          ~ Bagwise substraction (used for removing first occurrence).
Z        ~ The string representation of Z.
KeSZ   ~ And the highest (lexicographically) character of Z (highest digit).
It also assigns it to a variable called K.
s            ~ Cast to integer.
*             ~ Multiply by...
sK ~ K casted to int.


Being that close to Jelly at such type of challenge is very good for Pyth IMO :-)

# Husk, 1413 12 bytes

Thanks Zgarb for saving 1 byte.

Ω≤9oṠS*od-▲d


Try it online!

### Explanation:

Ω≤9            Repeat the following function until the result is ≤ 9
d     Convert to a list of digits
-▲      Remove the largest one
od        Convert back to an integer
oṠS*          Multiply by the maximum digit

• 12 bytes with some rearranging. Nov 18 '17 at 21:31
• @Zgarb Thanks, I was looking for something like that. Nov 18 '17 at 21:33

# R, 99 95 bytes

f=function(x)"if"(n<-nchar(x)-1,f(10^(n:1-1)%*%(d=x%/%10^(n:0)%%10)[-(M=which.max(d))]*d[M]),x)


Try it online!

A recursive function. Adding f(number) in the footer can be used to test for other values of number. Straightforward implementation, d is the list of digits, and 10^(n:2-2)%*%d[-M] computes the number with the largest digit removed.

# Python 2, 72 bytes

f=lambda n:n*(n<=9)or f(int(n.replace(max(n),'',1))*int(max(n)))


Try it online!

• ...I was debugging a dumb mistake in this. Darn, I got ninja'd. Nov 18 '17 at 20:21
• I'm getting an error on input 9 Nov 19 '17 at 7:27
• This seems to fail for the test case 432969. "ValueError: invalid literal for int() with base 10: ''" Nov 20 '17 at 13:25
• @JamesWebster should be fixed now. Nov 20 '17 at 18:15
• @recursive No, as then if n was 0 then n*(n<=9) would still evaluate to a falsy value, 0, making the recursion continue and causing an error, whereas the string '0' is a truthy value and therefore the recursion is halted. Nov 21 '17 at 22:16

# Ruby, 59 bytes

f=->n{m=n.digits.max;n>9?f[n.to_s.sub(m.to_s,"").to_i*m]:n}


Try it online!

Recursive lambda function called like f.

# Jelly, 15 bytes

D×Ṁ$œṡṀ$FḌµ>9µ¿


Try it online! or see the test-suite.

### How?

D×Ṁ$œṡṀ$FḌµ>9µ¿ - Link: number, n
¿ - while:
9   -    literal 9
>    -    loop value greater than (9)?
µ     - ...do (monadic):               e.g. 432969
D               -    convert to a decimal list        [4,3,2,9,6,9]
$- last two links as a monad: Ṁ - maximum 9 × - multiply (vectorises) [36,27,18,81,54,81]$        -    last two links as a monad:
Ṁ         -      maximum                         81
œṡ          -      split at first occurrence      [[36,27,18],[54,81]]
F       -    flatten                          [36,27,18,54,81]
Ḍ      -    convert from base 10              389421  (i.e. 360000 + 27000 + 1800 + 540 + 81)


Saved 3 4 bytes thanks to nimi!

f x|x<10=x|(a,b:c)<-span=<<(>).maximum$show x=f$read[b]*read(a++c)


Try it online!

• An explicit recursion is shorter than until. Try it online!.
– nimi
Nov 18 '17 at 21:52
• one more byte to save: ...span=<<(>).maximum....
– nimi
Nov 18 '17 at 22:06
• Ooh ... Nice :) Nov 18 '17 at 22:10

# C# (.NET Core), 126 bytes

int F(int n){var x=(n+"").ToList();var m=x.Max();x.RemoveAt(x.IndexOf(m));return n>9?F(int.Parse(string.Concat(x))*(m-48)):n;}


Try it online!

• Welcome to PPCG! You can remove a space. Nov 20 '17 at 11:25
• @EriktheOutgolfer Thank you, missed that one. Nov 20 '17 at 11:51
• @totallyhuman Thank you, down to 137 after some refactoring. Nov 20 '17 at 12:13
• You can change if(n<10)return n;...return F(...); to a single return with ternary-if, like this: int F(int n){var x=(n+"").ToList();var m=x.Max(d=>d);x.RemoveAt(x.IndexOf(m));return n<10?n:F(int.Parse(string.Concat(x))*(m-48));} (131 bytes) Nov 20 '17 at 13:38
• I think you need to include using System.Linq; (18 bytes) into the bytecount. Nov 21 '17 at 17:37

# APL (Dyalog), 3635 33 bytes

-1 due to updated OP specs. -2 thanks to ngn.

Anonymous tacit prefix function. Takes integer as argument.

{⍵>9:∇(⌈/×10⊥⊂⌷⍨¨⍳∘≢~⊢⍳⌈/)⍎¨⍕⍵⋄⍵}


Try it online!

{}a function where ⍵ is the argument:

⍵>9: if the argument is greater than 9, then:

⍕⍵ format (stringify) the argument

⍎¨ execute (evaluate) each (this gets us the digits as numbers)

() apply the following tacit function on those

⌈/ the largest digit

× times

10⊥ the base-10 decoding of (collects digits)

⊂ all the digits

⌷⍨¨ indexed by each of

⍳∘≢ the indices of the number of digits

≠ differs from

⊢⍳⌈/ the largest digit's index in the entire list of digits

∇ recurse (i.e. call self) on that

⋄ else

⍵ return the argument unmodified

• Shouldn't >10 be >9? Nov 18 '17 at 21:49
Nov 18 '17 at 21:59
• That's true, but >9 would save a byte. Nov 18 '17 at 22:10
• @EriktheOutgolfer Updated.
Nov 19 '17 at 9:33
• @Adám ∇ instead of ⍣= for -1 byte: {⍵>9:∇(⌈/ ... ⋄⍵}
– ngn
Nov 20 '17 at 18:26

# Perl 6,  45  41 bytes

{($_,{$/=.comb.max;S/"$/"//*$/}...10>*).tail}


Test it

{($_,{S/"{.comb.max}"//*$/}...10>*).tail}


Test it

## Expanded:

{  # bare block lambda with implicit parameter ｢$_｣ ( # generate the sequence$_,                      # start the sequence with the input

{                        # generate the rest of the values in the sequence

S/                   # find and replace (not in-place)
"{  .comb.max  }"  # find the max digit and match against it
//                   # replace it with nothing
*                    # multiply the result with
$/ # the digit that was removed } ... # keep generating values until 10 > * # the value is less than 10 ).tail # get the last value from the sequence }  # Retina, 67 bytes {1(..+)? 1$&;$& O\G\d .+((.);.*?)\2$1
\d+
$* 1(?=.*;(1+))|.$1
1


Try it online! Link includes the test cases fast enough not to hammer Dennis's server. Explanation:

{1(..+)?
1$&;$&


For two digit numbers, this duplicates the number with a ; separator, prefixing a 1 to the duplicate. For one digit numbers, this prefixes 1; to the number.

O\G\d


Sort the digits of the duplicate. (For one digit numbers, this has no effect.)

.+((.);.*?)\2
$1  Find the first occurrence of the largest digit, and delete it, and also the other digits in the duplicate, and the extra 1 that was added earlier. (For one digit numbers, the match fails so this does nothing.) \d+$*
1(?=.*;(1+))|.
$1 1  Multiply the number by the digit. For one digit numbers, this results in the original number, and the loop terminates. Otherwise, the program loops until a single digit is reached. # PowerShell, 123 bytes [Collections.ArrayList]$a=[char[]]"$args" while(9-lt-join$a){$a.remove(($b=($a|sort)[-1]));$a=[char[]]"$(+"$b"*-join$a)"}$a


Try it online!

Ooof. PowerShell arrays are immutable, so we need to use the lengthy [Collections.ArrayList] casting here so we can call .remove() later.

Takes input $args, converts it to a string, then a char-array, then an ArrayList. Stores that into $a. Then we while loop until we're at or below 9. Each iteration, we're calling .remove on the largest element of $a (done by sort and taking the last element [-1]), storing the largest element into $b at the same time. This happens to work because the ASCII values sort in the same fashion as the literal digits.

Next, we recompute $a, again as an char-array (and ArrayList implicitly), by casting our $b (which is currently a char) to a string, then an int with +, and multiplying that to $a -joined into a string (implicitly cast to int). This satisfies the "multiply by D" portion of the challenge. Finally, once we're out of the loop, we put $a onto the pipeline and output is implicit.

# C# (.NET Core), 177 164 + 18 bytes

Saved 13 bytes thanks to @raznagul!

int f(int n){string s=n+"",m=s.Max(d=>d)+"";if(n<10)return n;var x=s.ToList();x.RemoveAt(s.IndexOf(m));int y=int.Parse(string.Join("",x))*int.Parse(m);return f(y);}


Try it online!

• You can change s.Length<2 to n<10. Also, you can remove the ternary operator and just return f(y) at the end, as the case is handled by the if in the next step of the recursion. Nov 21 '17 at 13:04

# Java 8, 126 104 bytes

n->{for(;n>9;n=new Long((n+"").replaceFirst((n=(n+"").chars().max().getAsInt()-48)+"",""))*n);return n;}


-22 bytes thanks to @OlivierGrégoire.

Explanation:

Try it here.

n->{         // Method with long as both parameter and return-type
for(;n>9;  //  Loop as long as the number contains more than 1 digit
n=       //   Replace the current number with:
new Long((n+"").replaceFirst((n=(n+"").chars().max().getAsInt()-48)+"",""))
//    Remove the first largest digit from the number,
*n     //    and multiply this new number with the removed digit
);         //  End of loop
return n;  //  Return the result
}            // End of method

• 111 bytes Nov 22 '17 at 14:10
• 104 bytes (same as above, but iterative instead of recursive, also: n>9 and revert conditions instead of n<10). Nov 22 '17 at 14:14

# C 103 , 95 , 90 bytes

a,b;t,m;f(n){for(t=m=0,a=b=1e9;a/=10;)if((t=n/a%10)>m)m=t,b=a;n=n>9?f(m*=n/b/10*b+n%b):n;}


Try it online!

# Java 8: 115 bytes

-10 bytes thanks to Jo King

Unfortunately you can't call a lambda function recursively, so an extra 11 bytes is needed for the method header. I am aware there is a shorter Java answer that loops instead, but I decided to come up with this on my own.

long f(long n){int m=(n+"").chars().max().getAsInt()-48;return n>9?f(new Long((n+"").replaceFirst(""+m,""))*m):n;};


Try it online

• You can move the -48 from the map to the end of the m definition. Try it online! You also have some extra whitespace in your TIO link
– Jo King
Mar 8 '19 at 0:28
• @JoKing thanks. Mar 8 '19 at 0:43
• Technically you can have recursive lambda expressions, however, they require that you pass the lambda as a parameter when it is called Oct 13 '20 at 18:17
• @KrystosTheOverlord good to know, not sure I can use it here though due to the input rules (only 1 input which is the number). Oct 17 '20 at 23:30
• @BenjaminUrquhart ah my bad, well in the future it could be handy, as it makes recursive java solutions much smaller :) Oct 19 '20 at 17:32

# Jq 1.5, 86 bytes

until(.<10;"\(.)"|(./""|max)as$v|index($v)as$x|.[:$x]+.[1+$x:]|tonumber*($v|tonumber))


Expanded

until(
.<10                    # until number is below 10
; "\(.)"                  # convert to string
| (./""|max) as $v # find largest digit, call it$v
| index($v) as$x         # find index of digit
| .[:$x]+.[1+$x:]         # remove digit
| tonumber*($v|tonumber) # convert back to number and multiply by$v
)


Try it online!

# Perl 5, 41 + 1 (-p) = 42 bytes

$m=(sort/./g)[-1];s/$m//;($_*=$m)>9&&redo


Try it online!

• Fails on input of 9. :( but tinkered a little and managed to get 41 bytes with a fix: Try it online! Nov 20 '17 at 13:57

# Lua, 137 108 bytes

function f(n)while n>9 do b="0"g=b.gsub g(n,".",function(m)b=math.max(m,b)end)n=b*g(n,b,"",1)end print(n)end


Thanks to Jonathan S for golfing off 29 bytes.

Try it online!

• 108 bytes Nov 19 '17 at 12:28
• Thanks. That looks worthy of an answer of its own - will link to a post you make for it, otherwise will edit & credit. Nov 19 '17 at 12:34
• Just edit it in. It's still your code, I haven't written it from scratch. Nov 19 '17 at 12:46

# D, 188186 185 bytes

import std.conv,std.algorithm;T f(T,U=string)(T u){if(u<10)return u;T[]r;u.text.each!(n=>r~=n.to!T-48);T m=r.maxElement;U s;r.remove(r.maxIndex).each!(n=>s~=n.to!U);return f(m*s.to!T);}


Try it online!

I hate lazy evaluation, so much. Any tips are welcome!

## Lua, 154 Bytes

I should have some ways to golf this down, I'm experimenting right now.

n=...z=table
while n+0>9 do
t={}T={}n=n..''n:gsub(".",function(c)t[#t+1]=c T[#T+1]=c
end)z.sort(t)x=t[#t]z.remove(T,n:find(x))n=z.concat(T)*x
end
print(n)


Try it online!

### Explanations

n=...                    -- define n as a shorthand for the argument
z=table                  -- define z as a pointer to the object table
while n+0>9              -- iterate as long as n is greater than 9
do                       -- n+0 ensure that we're using a number to do the comparison
t={}                   -- intialise two tables, one is used to find the greatest digit
T={}                   -- the other one is used to remove it from the string
n=n..''                -- ensure that n is a string (mandatory after the first loop)
n:gsub(".",function(c) -- apply an anonymous function to each character in n
t[#t+1]=c -- fill our tables with the digits
T[#T+1]=c
end)
z.sort(t)              -- sort t to put the greatest digit in the last index
x=t[#t]                -- intialise x to the value of the greatest digit
z.remove(T,n:find(x))  -- remove the first occurence of x from the table T
-- based on its position in the input string
n=z.concat(T)*x        -- assign the new value to n
end                      -- if it still isn't a single digit, we're looping over again


# Pip, 22 21 bytes

Wa>9a:aRAa@?YMXax*:ya


Takes input as a command-line argument. Verify all test cases: Try it online!

### Explanation

                 a is 1st cmdline arg
W a>9 {          While a > 9:
Y MXa           Yank max(a) into y
a RA: a@?y ""   Find index of y in a; replace the character at that position with ""
a *: y          Multiply a by y
}
a                Autoprint a


In the golfed version, the loop body is condensed into a single expression:

a:aRAa@?YMXax*:y
YMXa      Yank max(a)
a@?          Find its index in a
aRA       x     Replace at that index with x (preinitialized to "")
*:y  Multiply that result by y (using : meta-operator to lower the precedence)
a:                Assign back to a


# Scala, 121 bytes

x=>def g(y:Int):Int={if(y>9){val m=y.toString.max.asDigit;g(y.toString.replaceFirst(m.toString,"").toInt*m)}else y}
g(x)}


Try it online!

# pure Bash, 116 112 bytes, or 101, or even 97?

Based on answers from iBug and Léa Gris and tips-for-golfing-in-bash (thanks, @AaronMiller and @BrowncatPrograms!):

#!/bin/bash

test112()
{
n=$1;shopt -s extglob;while((n>9));do m=0;for x in${n//?()/ };{ ((m<x))&&m=$x;};n=$[10#${n/$m/}*m];done;echo $n } for k in 9 27 757 1234 26364 432969 1234584 91273716; do printf '%d -> ' "$k"; test112 "$k"; done  Try it online! Also there are two shorter variants: #!/bin/bash f()(shopt -s extglob;(($1>9))||exit $1;m=0;for x in${1//?()/ };{ ((m<x))&&m=$x;};f$[10#${1/$m/}*m])

for k in 9 27 757 1234 26364 432969 1234584 91273716; do printf '%d -> ' "$k"; f "$k"; echo "$?"; done  (Try it online!) with 101 bytes and with 97 bytes (Try it online!): shopt -s extglob;(($1>9))||exit $1;m=0;for x in${1//?()/ };{ ((m<x))&&m=$x;};$0 $[10#${1/$m/}*m]  But I am not entirely sure the last two apply to the rules. The latter even uses $0 (I neither found a rule allowing that nor denying this) to recurse to itself, so it only works if put into a script (like golf.sh), made executable and then is called like golf.sh 91273716; echo $? Output of the scripts: 9 -> 9 27 -> 4 757 -> 5 1234 -> 8 26364 -> 8 432969 -> 0 1234584 -> 8 91273716 -> 6  Explained first variant (112 bytes): n=$1;                           # (*) get argument n
shopt -s extglob;               # enable ${n//?()/ } below while((n>9));do # loop until number is <=9 m=0; # preset maximum to 0 for x in${n//?()/ };{        # loop over the digits
((m<x))&&m=$x; # find max }; n=$[                          # calculate new number
10#                       # force base 10 as 0N becomes octal
${n/$m/}               # remove first occurance of max
*m];           # multiply m
done;
echo $n # output result  Notes: • Shebang #!/bin/bash+LF (12 byte) is not included in the count. • We are in bash, so this is already the default • AFAICS passing in the number directly as variable n (see marked (*) above) is not allowed. Hence these 5 bytes are needed. • Else ./golf.sh 1234 would look like n=1234 ./golf.sh • A read n; if number is passed in from STDIN is 2 byte longer • shopt -s extglob; can be left away, if this option is given on commandline • Like in n=1234 bash -Oextglob ./golf.sh • But extglob is not set by default, hence AFAICS it must be included in the count • 10# is needed as ${n/$m/} can leave a number with a leading 0 • Without the 10# input 91273716 wrongly returns 8. Explained shortest variant (97 bytes): shopt -s extglob; # see above (($1>9))||exit $1; # stop if goal reached m=0;for x in${1//?()/ };{ ((m<x))&&m=$x;}; # max calc, as above$0 $[10#${1/$m/}*m] # recurse to next step  Notes: • Actually the last line is a tail recursion, so bash uses exec and does not fork • Welcome to Code Golf, and nice first answer! Make sure to check out our tips for golfing in Bash to see if you can find ways to shorten your code. Sep 8 at 17:01 • Nice answer! For the input, you might want to look through the default I/O rules. Sep 8 at 17:06 • @AaronMiller Thanks for the link! 4 bytes less. – Tino Sep 8 at 17:26 # J, 40 bytes ((]*<^:3@i.{[)>./)&.(10&#.inv)^:(9&<)^:_  Try it online! ## explanation ( iterate )^:(9&<)^:_ NB. keep iterating while the number is > 9 ( stuff )&.(10&#.inv) NB. convert to digits, do stuff, convert back to number ( )>./) NB. stuff is a hook, with max digit >./ on the right (]*<^:3@i.{[) NB. so that in this phrase, ] means "max" and [ means "all digits" ] NB. the max digit... * NB. times... <^:3@ NB. triple box... i. NB. the first index of the max in the list of all digits { NB. "from" -- which because of the triple box means "take all indexes except..." [ NB. from all the digits of the number  • I learnt about the triple box selection from you today, thank you! Nov 19 '17 at 8:33 # PowerShell, 230 bytes $n="$args";do{$n=$n.ToString();$a=@();0..$n.Length|%{$a+=$n[$_]};$g=[convert]::ToInt32(($a|sort|select -last 1),10);[regex]$p=$g.ToString();[int]$s=$p.replace($n,'',1);if($n.Length-eq1){$n;exit}else{$r=$s*$g}$n=$r}until($r-lt10)$r


Try it online!

Wasted too much on all the type casting.

# PHP, 82 77+1 bytes

for($n=$argn;$n>9;)$n=join("",explode($d=max(str_split($n)),$n,2))*$d;echo\$n;


Run as pipe with -nR` or try it online.