# Smallest multiple being run of 9 followed by optional run of 0

Given a positive integer, find its smallest positive integer multiple which is a run of 9 followed by an optional run of 0. In other words, find its smallest positive integer multiple which is matched by the regex /^9+0*$/. For example, if the given positive integer is 2, then return 90, since 90 is a positive integer multiple of 2 and is the smallest which is matched by the regex /^9+0*$/.

Test cases:

n  f(n)
1  9
2  90
3  9
4  900
5  90
6  90
7  999999
8  9000
9  9
10 90
11 99
12 900
13 999999
14 9999990
15 90
16 90000


This is . Shortest answer in bytes wins. Standard loopholes apply.

• proof of well-definedness? Commented Jul 6, 2017 at 5:48
• @DestructibleLemon This proof suffices, since the result can be multiplied by 9.
– xnor
Commented Jul 6, 2017 at 5:50
• I think more test cases would be good to check that solutions require the 9's to come before the 0's.
– xnor
Commented Jul 6, 2017 at 5:55
• @LeakyNun maybe not, but 9900099 is, and shouldn't be allowed according to rules. Commented Jul 6, 2017 at 6:45
• @koita_pisw_sou the rule is that the program should "theoretically" work for any integer given arbitrary precision and memory and time. Commented Jul 6, 2017 at 13:02

# Jelly, 13 11 bytes

ṚḌ‘DS=ḍ@ð1#


Try it online!

### How it works

ṚḌ‘DS=ḍ@ð1#  Main link. Argument: n

ð    Start a dyadic chain with arguments n and n.
1#  Execute the chain to the left with left argument k = n, n+1, n+2, ...
and right argument n until 1 match has been found. Return the match.
Ṛ                Get the decimal digits of k, reversed.
Ḍ               Convert from base 10 to integer.
This essentially removes trailing zeroes. As a side effect, it
reverses the digits, which doesn't matter to us.
‘              Increment the resulting integer. If and only if it consisted
entirely of 9's, the result is a power of 10.
DS            Compute the sum of the digits. The sum is 1 if and only if the
integer is a power of 10. Note that the sum cannot be 0.
ḍ@         Test k for divisibility by n.
=           Compare the results.

• ಠ_ಠ how did you do it with neither 9 or 0 in your code Commented Jul 6, 2017 at 7:36
• I've added an explanation. Commented Jul 6, 2017 at 7:59

# Python 2, 55 54 bytes

n=r=input()
while int(10*r.lstrip('9')):r+=n
print r


Try it online!

• Not only do you have to outgolf us all, but you have to do it in Python... 3 times... :P Commented Jul 6, 2017 at 18:48

# Python 2, 51 bytes

f=lambda n,r=9:r%n and f(n,10*r-10**n*r%-n/n*9)or r


Try it online!

• Devious yet surprisingly simple! Commented Jul 6, 2017 at 18:50

# JavaScript (ES6), 4743 42 bytes

-4 bytes thanks to @Arnauld
-1 byte thanks to @Luke

n=>eval('for(i=0;!/^9+0*$/.test(i);)i+=n')  ## Tests let f= n=>eval('for(i=0;!/^9+0*$/.test(i);)i+=n')

for(let i=1;i<=16;i++)console.log(f(${i}) = +f(i)) ### Recursive solution (fails for 7, 13, and 14), 38 bytes n=>g=(i=0)=>/^9+0*$/.test(i+=n)?i:g(i)


Called like f(5)(). Reaches the max call stack size in Chrome and Firefox for n=7, n=13, and n=14.

f=
n=>g=(i=0)=>/^9+0*$/.test(i+=n)?i:g(i) for(let i=1;i<=16;i++){ try { console.log(f(${i}) = +f(i)()) }
catch(e) { console.log(f(${i}) = FAILED) } } • One byte shorter: n=>eval('for(i=0;!/^9+0*$/.test(i);)i+=n')
– Luke
Commented Jul 6, 2017 at 8:52

# Ruby, 36 bytes

->x{r=0;1until"#{r+=x}"=~/^9+0*$/;r}  Brute-forcing - takes forever for x=17. Try it online! • I came up with almost the same solution as you, but as a full program: codegolf.stackexchange.com/a/130106/60042. I borrowed the use of string interpolation from you, I hope that's ok. Commented Jul 6, 2017 at 7:26 # Java 8, 61 57 bytes n->{int r=0;for(;!(""+r).matches("9+0*");r+=n);return r;}  -4 bytes (and faster execution) thanks to @JollyJoker. Explanation: Try it here. n->{ // Method with integer as parameter and return-type int r=0; // Result-integer for(;!(""+r).matches("9+0*"); // Loop as long as r doesn't match the regex r+=n // And increase r by the input every iteration ); // End of loop return r; // Return the result-integer } // End of method  • Yeah for optimization! ^^ Commented Jul 6, 2017 at 12:02 • Incrementing by n avoids the r%n check, n->{int r=0;for(;!(""+(r+=n)).matches("9+0*"););return r;} Commented Jul 6, 2017 at 13:16 • for(;!(""+r).matches("9+0*");r+=n) Commented Jul 6, 2017 at 13:20 • I've tried, and tried to get on with integers and math, but I can't beat this! Congrats :) Commented Jul 7, 2017 at 12:53 # Python 3, 56 bytes f=lambda n,r=0:{*str(r).strip('0')}!={'9'}and n+f(n,n+r)  Try it online! # Brachylog, 16 bytes ;I×≜.ẹḅhᵐc~a₀90∧  Try it online! This is pretty slow ### Explanation ;I×≜. Output = Input × I .ẹḅ Deconcatenate into runs of consecutive equal digits hᵐ Take the head of each run c Concatenate into a number ~a₀90∧ That number is a prefix of 90 (i.e. it's 9 or 90)  # 05AB1E, 10 bytes 0[+D9Û0«_#  Try it online! It just keeps adding the input to 0, until the result minus leading 9's equals 0. # RProgN 2, 18 bytes x={x*'^9+0*$'E}éx*


## Explained

x={x*'^9+0*$'E}éx* x= # Set the value of "x" to the input. { }é # Find the first positive integer in which passing it to the defined function returns truthy. x* # Multiply the index by x, this essentially searches multiples now. '^9+0*$'       # A Regex defined by a literal string.
E      # Does the multiple match the regex?
x*  # Multiple the outputted index by x, giving the result.


Try it online!

# Ruby, 38 + 1 = 39 bytes

Uses -p flag.

$_=y=eval$_
1until"#{$_+=y}"=~/^9+0*$/


-p surrounds the program with:

while gets
...
end
puts $_  gets stores its result in $_. eval is used to convert it to a number, as it's shorter than .to_i, then brute force is used, incrementing $_ until it matches the regex. "#{}" is sttring interpolation, it's shorter than a .to_s call as that would require parantheses around $_+=y. Finally, $_ is printed. Try it online! Try all test cases! # Mathics, 71 bytes (x=#;While[!StringMatchQ[ToString@x,RegularExpression@"9+0*"],x+=#];x)&  Try it online! Not very intresting brute force solution, but it beats the other Mathematica answer, which uses some clever tricks. The one redeeming quality Mathematica has in regards to this challenge is the fact that StringMatchQ requires a full match, so I can do 9+0* rather than ^9+0*$.

• If you're willing to use Mathematica instead of Mathics, you can save a few bytes with "9"..~~"0"... instead of RegularExpression@"9+0*". Commented Jul 6, 2017 at 12:04
• @Notatree thanks, I'll keep it in mind for a later time, but I'll stick with Mathics. I prefer not to use syntax I don't understand, and that's the first time I've seen syntax like that. Commented Jul 6, 2017 at 19:51
• Fair enough. (Mathematica's pattern matching syntax is a powerful tool, but if you're familiar with regular expressions you probably already know that!) Commented Jul 6, 2017 at 21:45

## Batch, 175 bytes

@set/pn=
@set s=
:g
@set/ag=-~!(n%%2)*(!(n%%5)*4+1)
@if not %g%==1 set s=0%s%&set/an/=g&goto g
@set r=1
:r
@set s=9%s%
@set/ar=r*10%%n
@if %r% gtr 1 goto r
@echo %s%


Takes input on STDIN. Not a brute force solution but in fact based on my answer to Fraction to exact decimal so it will work for 17, 19, etc. which would otherwise exceed its integer limit anyway.

# Mathematica, 127 bytes

Select[FromDigits/@Select[Tuples[{0,9},c=#],Count[#,9]==1||Union@Differences@Flatten@Position[#,9]=={1}&],IntegerQ[#/c]&][[1]]&


Input

[17]

Output

9999999999999999

here are the first 20 terms

{9, 90, 9, 900, 90, 90, 999999, 9000, 9, 90, 99, 900, 999999, 9999990, 90, 90000, 9999999999999999, 90, 999999999999999999, 900}

• Clever, but the obvious solution seems to be the shortest: codegolf.stackexchange.com/a/130115/60042 Commented Jul 6, 2017 at 7:54
• your obvious solution can't do 17 ;-) Commented Jul 6, 2017 at 8:00
• What can I say, not fastest-code Commented Jul 6, 2017 at 8:01
• By the way, your solution works in Mathics, you could change it to that and add a TIO link. Commented Jul 6, 2017 at 8:02
• Yours can't do 17 either Commented Jul 6, 2017 at 8:04

f takes and returns an integer.

f n=filter(all(<'1').snd.span(>'8').show)[n,n+n..]!!0


Try it online!

This times out for 17, which conveniently is just beyond the test cases. A faster version in 56 bytes:

f n=[x|a<-[1..],b<-[0..a-1],x<-[10^a-10^b],mod x n<1]!!0


Try it online!

# How it works

• f generates all multiples of n, converts each to a string, filters out those with the right format, then takes the first one.

• The faster version instead uses that the required numbers are of the form 10^a-10^b, a>=1, a>b>=0. For golfing purposes it also uses the fact that for the minimal a, only one b can work, which allows it to generate the bs in the slightly shorter "wrong" order.

# Perl 5, 23 + 2 (-pa) = 25 bytes

Brute Force Method

$_+=$F[0]while!/^9+0*$/  Try it online! It's slow, but it's tiny. ## More Efficient Method: 41 + 2 (-pa) = 43 bytes $_=9;s/0/9/||($_=9 .y/9/0/r)while$_%$F[0]  Try it online! It works well for any input, but it's longer code. # Dyalog APL, 34 bytes {∧/'0'=('^9+'⎕R'')⊢⍕⍵:⍵⋄∇⍵+r}n←r←⎕  Recursive dfns, based on Dennis' Python solution. C++, 106 bytes int main(){long N,T=9,j=10,M;cin>>N;while(T%N){if(T/j){T+=(M/j);j*=10;}else{T=(T+1)*9;j=10;M=T;}}cout<<T;}  Detailed Form: int main() { long N,T=9,j=10,M; cin >> N; while (T%N) { if (T/j) { T += (M/j); j *= 10; } else { T = (T+1)*9; j = 10; M = T; } } cout << T; }  TRY it online! • Better golfed: [](int n){int T=9,j=10,m;while(t%n)if(t/j){t+=m/j;j*=10;}else{t=(t+1)*9;j=10;m=t;}return t;}}, takes 94 bytes. Essentially, treat it as a function task to save bytes, save up on unneeded parentheses, use lambda function to save on type naming and type. Commented Jul 6, 2017 at 14:41 • can't make it compile using lambda. could u give a hand? Commented Jul 7, 2017 at 6:25 • It may be the reason that I put too amuch parentheses on the end. Commented Jul 7, 2017 at 9:00 • In addition, lambda probably has not to exist in global scope, even though wrapping it into regular function takes 97 bytes. Commented Jul 7, 2017 at 12:07 # Python 2, 79 bytes x=input();n=10;y=9 while y%x: b=n while(b-1)*(y%x):b/=10;y=n-b n*=10 print y  Try it online! Some explanations It finds the smallest natural of form 10**n-10**b with n>b>=0 that divides the input. Some IO f(1) = 9 f(2) = 90 f(3) = 9 f(4) = 900 f(5) = 90 f(6) = 90 f(7) = 999999 f(8) = 9000 f(9) = 9 f(10) = 90 f(11) = 99 f(12) = 900 f(13) = 999999 f(14) = 9999990 f(15) = 90 f(16) = 90000 f(17) = 9999999999999999 f(18) = 90 f(19) = 999999999999999999  # PHP, 39 bytes for(;ltrim($r=$argn*++$i,9)>0;);echo$r;  Try it online! # PHP, 52 bytes for(;!preg_match("#^9+0*$#",$r=$argn*++$i););echo$r;


Try it online!

## Swift 3.0, Bytes:121

var i=2,m=1,n=""
while(i>0){n=String(i*m)
if let r=n.range(of:"^9+0*$",options:.regularExpression){print(n) break};m=m+1}  Try it online! • What does let r= do? I don't see r referred to anywhere else Commented Jul 11, 2017 at 2:49 • @Cyoce let r= checks whether the n.range returns value nil or not.You can use let _=.I am using optional binding here to reduce no of bytes. Commented Jul 11, 2017 at 7:16 # Python 3, 62 bytes This function takes an integer n and initializes m to zero. Then it removes all zeros from the ends of m and checks if the result only contains 9's, returning m if it does. If not, it adds n to m and checks again, etc. def f(n,m=0): while{*str(m).strip('0')}!={'9'}:m+=n return m  Try it online! # Java (OpenJDK 8), 66 bytes, doesn't choke on 17 n->{long a=10,b=1;for(;(a-b)%n>0;b=(b<10?a*=10:b)/10);return a-b;}  Try it online! Longer than @KevinCruijssen's solution but can handle slightly larger numbers. It calculates the candidate numbers like 10^6 - 10^3 = 999000. 64-bit longs are still the limit, breaking for n=23. Can probably be golfed a bit but already took too long to make it work... # ><>, 35 bytes &a:v ;n-< :,a/?(1:^!?%&:&-}:{ a*:\~  Try it online, or watch it at the fish playground! Assumes the input is already on the stack. Works by looking for numbers of the form 10a − 10b, with a < b (yes, that's a less than sign — it takes fewer bytes!) until that's divisible by the input, then printing 10b − 10a. This is much faster than the brute force method (which would be difficult in ><> anyway). # V, 19 14 bytes é0òÀ/[1-8]ü09  Try it online! ## Explanation é0 ' <m-i>nsert a 0 ò ' <m-r>ecursively À ' <m-@>rgument times  ' <C-A> increment the number (eventually gives all multiples) /[1-8]ü09 ' find ([1-8]|09) if this errors, the number is of the form ' (9+0*) (because there won't ever be just zeros) ' implicitly end the recursion which breaks on the above error  ## JavaScript (ES6), 51 49 bytes let f=(n,x=1,y=1)=>(x-y)%n?f(n,x,y*10):x-y||f(n,x*10) <input type=number value=1 step=1 min=1 oninput=O.value=f(value)> <input type=number value=9 id=O disabled> Not the shortest approach, but it is wicked fast. ## Mathematica, 82 bytes Using the pattern of submission from @Jenny_mathy 's answer... (d=x=1;y=0;f:=(10^x-1)10^y;n:=If[y>0,y--;x++,y=d;d++;x=1];While[Mod[f,#]!=0,n];f)&  Input: [17]  Output: 9999999999999999  And relative to the argument in comments at @Jenny_mathy's answer with @Phoenix ... RepeatedTiming[] of application to the input [17] gives {0.000518, 9999999999999999}  so half a millisecond. Going to a slightly larger input, [2003] : {3.78, 99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999}  a bit under 4 seconds. Test table: On the first 30 positive integers, the results are {9, 90, 9, 900, 90, 90, 999999, 9000, 9, 90, 99, 900, 999999, 9999990, 90, 90000, 9999999999999999, 90, 999999999999999999, 900, 999999, 990, 9999999999999999999999, 9000, 900, 9999990, 999, 99999900, 9999999999999999999999999999, 90}  Explanation: The only magic here is the custom iterator ("iterator" in the CS sense, not the M'ma sense) n := If[ y>0 , y-- ; x++ , y=d ; d++ ; x=1]  which acts on the global variables x, the number of leading "9"s, y, the number of trailing "0"s, and d, the total number of digits. We wish to iterate through the number of digits, and, for each choice of number of digits, start with the most "0"s and the least "9"s. Thus the first thing the code does is initialize d to 1, forcing x to 1 and y to 0. The custom iterator checks that the string of "0"s can be shortened. If so, it shortens the string of "0"s by one and increases the string of "1"s by one. If not, it increments the number of digits, sets the number of "0"s to one less than the number of digits, and sets the number of "9"s to 1. (This is actually done in a slightly different order to shave a few characters, since the old value of d is the desired value of y.) • And yet, still longer than brute force and regex. Commented Jul 6, 2017 at 19:52 • @Phoenix : So what's your timing on 2003? Commented Jul 20, 2017 at 23:42 # Ti-Basic (TI-84 Plus CE), 48 41 bytes Prompt X For(K,1,0 For(M,-K+1,0 10^(K)-10^(-M If 0=remainder(Ans,X Return End End  Input is Prompt-ed during the program; output is stored in Ans. Explanation: Tries numbers of the form (10n)(10m-1) = 10k-10m, where m+n=k starts at 1 and increases, and for each value of k, it tries m=1,n=k-1; m=2,n=k-2; ... m=k,n=0; until it finds a multiple of X. This works up to 16; 17 gives a domain error because remainder( can only accept dividends up to 9999999999999 (13 nines), and 17 should output 9999999999999999 (16 nines). Prompt X # 3 bytes, input number For(K,1,0 # 7 bytes, k in the description above; until a match is found For(M,-K+1,0 # 10 bytes, start with n=1, m=(k-n)=k-1; # then n=2, m=(k-n)=k-2, up to n=k, m=(k-n)=0 # (M=-m because it saved one byte) 10^(K)-10^(-M # 8 bytes, n=(k-m) nines followed by m zeroes → Ans If not(remainder(Ans,X # 8 bytes, If X is a factor of Ans (remainder = 0) Return # 2 bytes, End program, with Ans still there End # 2 bytes, End # 1 byte (no newline)  # QBIC, 53 bytes {p=p+1┘o=p*9_F!o$|┘n=!A!~(_l!n$|=_l!n+1$|)-(o%:)|\_Xo


## Explanation

{        infinitely DO
p=p+1    raise p (starts out as 0)
┘o=p*9   Get the mext multiple of 9 off of p
_F!o$| Flip a string representation of p*9 ┘n=!A! and set 'n' to be an int version of the flipped p*9 (this effectively drops trailing 0's) ~ This IF will subtract two values: the first is either 0 for n=x^10, or -1 and the second bit does (p*9) modulo 'a' (input number): also 0 for the numbers we want ( _l!n$|  the length of n's string representation
=        is equal to
_l!n+1\$| the length of (n+1)'s string rep (81 + 1 = 82, both are 2 long; 99 + 1 = 100, there's a difference)
)        The above yields -1 (Qbasic's TRUE value) for non-9 runs, 0 for n=x^10
-        Subtract from that
(o%:)    (p*9) modulo a     0 for p*9 = a*y
|       THEN (do nothing, since we want 0+0=0 in the conditionals above, execution of the right path jumps to ELSE
\_Xo    ELSE quit, printing (p*9)


# C (gcc), 126 bytes

#include<stdio.h>
main(x,n,y,b){n=10;y=9;scanf("%d",&x);while(y%x){b=n;while((b-1)*(y%x)){b/=10;y=n-b;}n*=10;}printf("%d",y);}


Try it online!

Some explanations It finds the smallest natural of form 10**n-10**b with n>b>=0 that divides the input.

Some IO

f(1) = 9
f(2) = 90
f(3) = 9
f(4) = 900
f(5) = 90
f(6) = 90
f(7) = 999999
f(8) = 9000
f(9) = 9
f(10) = 90
f(11) = 99
f(12) = 900
f(13) = 999999
f(14) = 9999990
f(15) = 90
f(16) = 90000