Largest number divisible by its digits: Difference between revisions
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=={{header|Prolog}}== |
=={{header|Prolog}}== |
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This will work with any radix, including base 10 and base 16. |
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<lang prolog>% Find the largest integer divisible by all it's digits, with no digit repeated. |
<lang prolog>% Find the largest integer divisible by all it's digits, with no digit repeated. |
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% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ |
% ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ |
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Revision as of 11:09, 19 October 2018
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Find the largest base 10 integer whose digits are all different, and is evenly divisible by each of its individual digits.
These numbers are also known as Lynch-Bell numbers, numbers n such that the
(base ten) digits are all different (and do not include zero) and n is divisible by each of its individual digits.
For example: 135 is evenly divisible by 1, 3 and 5.
Note that the digit zero (0) can not be in the number as integer division by zero is undefined. The digits must all be unique so a base 10 number will have at most 9 digits.
Feel free to use analytics and clever algorithms to reduce the search space your example needs to visit, but it must do an actual search. (Don't just feed it the answer and verify it is correct.)
- Stretch goal
Do the same thing for hexadecimal.
- Also see
- The OEIS sequence: A115569: Lynch-Bell numbers.
C
Base 10
The number can't contain 0 and 5, 0 is obvious, 5 because the number must end in 5 for it to be a multiple of that number and if that happens, all the even digits are ruled out which severely reduces the number's length since the other condition is that all digits must be unique. However, this means the number must be even and thus end only in 2,4,6,8. This speeds up the search by a factor of 2. The same approach when applied to hexadecimals takes a very long, long time. <lang C>
- include<stdio.h>
int main() { int num = 9876432,diff[] = {4,2,2,2},i,j,k=0; char str[10];
start:snprintf(str,10,"%d",num);
for(i=0;str[i+1]!=00;i++){ if(str[i]=='0'||str[i]=='5'||num%(str[i]-'0')!=0){ num -= diff[k]; k = (k+1)%4; goto start; } for(j=i+1;str[j]!=00;j++) if(str[i]==str[j]){ num -= diff[k]; k = (k+1)%4; goto start; } }
printf("Number found : %d",num); return 0; } </lang> Output:
Number found : 9867312
Base 16
<lang C>#include<stdio.h>
- include<string.h>
- define TRUE 1
- define FALSE 0
typedef char bool;
typedef unsigned long long uint64;
bool div_by_all(uint64 num, char digits[], int len) {
int i, d;
for (i = 0; i < len; ++i) {
d = digits[i];
d = (d <= '9') ? d - '0' : d - 'W';
if (num % d != 0) return FALSE;
}
return TRUE;
}
int main() {
uint64 i, magic = 15 * 14 * 13 * 12 * 11; uint64 high = 0xfedcba987654321 / magic * magic; int j, len; char c, *p, s[17], sd[16], found[16];
for (i = high; i >= magic; i -= magic) {
if (i % 16 == 0) continue; // can't end in '0'
snprintf(s, 17, "%llx", i); // always generates lower case a-f
if (strchr(s, '0') - s >= 0) continue; // can't contain '0'
for (j = 0; j < 16; ++j) found[j] = FALSE;
len = 0;
for (p = s; *p; ++p) {
if (*p <= '9') {
c = *p - '0';
} else {
c = *p - 87;
}
if (!found[c]) {
found[c] = TRUE;
sd[len++] = *p;
}
}
if (len != p - s) {
continue; // digits must be unique
}
if (div_by_all(i, sd, len)) {
printf("Largest hex number is %llx\n", i);
break;
}
}
return 0;
}</lang>
- Output:
Largest hex number is fedcb59726a1348
Factor
Base 10
This program works by filtering all the 8-digit permutations (of which there are only ~40,000) for all-digit-divisibility, and upon finding none, it will then generate the 7-digit combinations (of which there are 8) of the 8 possible digits, and then filter all permutations of the 8 combinations for all-digit-divisibility. Upon finding many, it will simply select the largest element which is our answer. If there hadn't been any 7-digit solutions, it would have gone down to six and then five, etc. <lang factor>USING: io kernel math math.combinatorics math.parser math.ranges sequences tools.time ; IN: rosetta-code.largest-divisible
- all-div? ( seq -- ? )
[ string>number ] [ string>digits ] bi [ mod ] with map sum 0 = ;
- n-digit-all-div ( n -- seq )
"12346789" swap <combinations> [ [ all-div? ] filter-permutations ] map concat ;
- largest-divisible ( -- str )
8 [ dup n-digit-all-div dup empty? ] [ drop 1 - ] while nip supremum ;
- largest-divisible-demo ( -- )
[ largest-divisible print ] time ;
MAIN: largest-divisible-demo</lang>
- Output:
9867312 Running time: 0.07224931499999999 seconds
Go
base 10
<lang go>package main
import (
"fmt" "strconv" "strings"
)
func divByAll(num int, digits []byte) bool {
for _, digit := range digits {
if num%int(digit-'0') != 0 {
return false
}
}
return true
}
func main() {
magic := 9 * 8 * 7
high := 9876432 / magic * magic
for i := high; i >= magic; i -= magic {
if i%10 == 0 {
continue // can't end in '0'
}
s := strconv.Itoa(i)
if strings.ContainsAny(s, "05") {
continue // can't contain '0'or '5'
}
var set = make(map[byte]bool)
var sd []byte // distinct digits
for _, b := range []byte(s) {
if !set[b] {
set[b] = true
sd = append(sd, b)
}
}
if len(sd) != len(s) {
continue // digits must be unique
}
if divByAll(i, sd) {
fmt.Println("Largest decimal number is", i)
return
}
}
}</lang>
- Output:
Largest decimal number is 9867312
base 16
<lang go>package main
import (
"fmt" "strconv" "strings"
)
func divByAll(num int64, digits []byte) bool {
for _, digit := range digits {
var d int64
if digit <= '9' {
d = int64(digit - '0')
} else {
d = int64(digit - 'W')
}
if num%d != 0 {
return false
}
}
return true
}
func main() {
var magic int64 = 15 * 14 * 13 * 12 * 11
high := 0xfedcba987654321 / magic * magic
for i := high; i >= magic; i -= magic {
if i%16 == 0 {
continue // can't end in '0'
}
s := strconv.FormatInt(i, 16) // always generates lower case a-f
if strings.IndexByte(s, '0') >= 0 {
continue // can't contain '0'
}
var set = make(map[byte]bool)
var sd []byte // distinct digits
for _, b := range []byte(s) {
if !set[b] {
set[b] = true
sd = append(sd, b)
}
}
if len(sd) != len(s) {
continue // digits must be unique
}
if divByAll(i, sd) {
fmt.Printf("Largest hex number is %x\n", i)
return
}
}
}</lang>
- Output:
Largest hex number is fedcb59726a1348
Haskell
base 10
Using the analysis provided in the Perl 6 (base 10) example:
<lang haskell>import Data.List (maximumBy, permutations, delete) import Data.Ord (comparing)
unDigits :: [Int] -> Int unDigits = foldl ((+) . (10 *)) 0
ds :: [Int] ds = [1, 2, 3, 4, 6, 7, 8, 9] -- 0 (and thus 5) are both unworkable
lcmDigits :: Int lcmDigits = foldr1 lcm ds -- 504
sevenDigits :: Int sevenDigits = (`delete` ds) <$> [1, 4, 7] -- Dropping any one of these three
main :: IO () main =
print $
maximumBy
(comparing
(\x ->
if rem x lcmDigits == 0 -- Checking for divisibility by all digits
then x
else 0))
(unDigits <$> concat (permutations <$> sevenDigits))</lang>
- Output:
Test run from inside the Atom editor:
9867312 [Finished in 0.395s]
base 16
First member of a descending sequence of multiples of 360360 that uses the full set of 15 digits when expressed in hex. <lang haskell>import Data.Set (fromList) import Numeric (showHex)
lcmDigits :: Int lcmDigits = foldr1 lcm [1 .. 15] -- 360360
upperLimit :: Int upperLimit =
let allDigits = 0xfedcba987654321 in allDigits - rem allDigits lcmDigits
main :: IO () main =
print $
head
(filter ((15 ==) . length . fromList) $
(`showHex` []) <$> [upperLimit,upperLimit - lcmDigits .. 1])</lang>
Test run from inside the Atom editor:
"fedcb59726a1348" [Finished in 2.319s]
Julia
Base 10
<lang julia>function main()
num = 9876432
dif = [4, 2, 2, 2]
local k = 1
@label start
local str = dec(num)
for (i, ch) in enumerate(str)
if ch in ('0', '5') || num % (ch - '0') != 0
num -= dif[k]
k = (k + 1) % 4 + 1
@goto start
end
for j in i+1:endof(str)
if str[i] == str[j]
num -= dif[k]
k = (k + 1) % 4 + 1
@goto start
end
end
end
return num
end
println("Number found: ", main())</lang>
- Output:
Number found: 9867312
Kotlin
Makes use of the Perl 6 entry's analysis:
base 10
<lang scala>// version 1.1.4-3
fun Int.divByAll(digits: List<Char>) = digits.all { this % (it - '0') == 0 }
fun main(args: Array<String>) {
val magic = 9 * 8 * 7
val high = 9876432 / magic * magic
for (i in high downTo magic step magic) {
if (i % 10 == 0) continue // can't end in '0'
val s = i.toString()
if ('0' in s || '5' in s) continue // can't contain '0' or '5'
val sd = s.toCharArray().distinct()
if (sd.size != s.length) continue // digits must be unique
if (i.divByAll(sd)) {
println("Largest decimal number is $i")
return
}
}
}</lang>
- Output:
Largest decimal number is 9867312
base 16
<lang scala>// version 1.1.4-3
fun Long.divByAll(digits: List<Char>) =
digits.all { this % (if (it <= '9') it - '0' else it - 'W') == 0L }
fun main(args: Array<String>) {
val magic = 15L * 14 * 13 * 12 * 11
val high = 0xfedcba987654321L / magic * magic
for (i in high downTo magic step magic) {
if (i % 16 == 0L) continue // can't end in '0'
val s = i.toString(16) // always generates lower case a-f
if ('0' in s) continue // can't contain '0'
val sd = s.toCharArray().distinct()
if (sd.size != s.length) continue // digits must be unique
if (i.divByAll(sd)) {
println("Largest hex number is ${i.toString(16)}")
return
}
}
}</lang>
- Output:
Largest hex number is fedcb59726a1348
Perl
Base 10
<lang perl>my $step = 9 * 8 * 7; # 504, interval between tests
my $initial = int(9876432 / $step) * $step; # largest 7 digit multiple of 504 < 9876432
for($test = $initial; $test > 0 ; $test -= $step) { # decrement by 504
next if $test =~ /[05]/; # skip numbers containing 0 or 5 next if $test =~ /(.).*\1/; # skip numbers with non unique digits
for (split , $test) { # skip numbers that don't divide evenly by all digits
next unless ($test / $_) % 1;
}
printf "Found $test after %d steps\n", ($initial-$test)/$step;
for (split , $test) {
printf "%s / %s = %s\n", $test, $_, $test / $_;
}
last
}</lang>
- Output:
Found 9867312 after 18 steps 9867312 / 9 = 1096368 9867312 / 8 = 1233414 9867312 / 6 = 1644552 9867312 / 7 = 1409616 9867312 / 3 = 3289104 9867312 / 1 = 9867312 9867312 / 2 = 4933656
Base 16
<lang perl>use bigint; use Math::Utils qw(lcm);
my $hex = 'FEDCBA987654321'; # largest possible hex number $step = lcm(1..15); $initial = int(hex($hex) / $step) * $step;
for($num = $initial; $num > 0 ; $num -= $step) { # decrement by lcm
my $test = sprintf '%x', $num; next if $test =~ /0/; # skip numbers containing 0 next if $test =~ /(.).*\1/; # skip numbers with non unique digits
push @res, sprintf "Found $test after %d steps\n", ($initial-$num)/$step;
push @res, ' 'x12 . 'In base 16' . ' 'x36 . 'In base 10';
for (split , $test) {
push @res, sprintf "%s / %s = %x | %d / %2d = %19d",
$test, $_, $num / hex($_),
$num, hex($_), $num / hex($_);
}
last
}
print join "\n", @res;</lang>
- Output:
Found fedcb59726a1348 after 954460 steps
In base 16 In base 10
fedcb59726a1348 / f = 10fda5b4be4f038 | 1147797065081426760 / 15 = 76519804338761784
fedcb59726a1348 / e = 1234561d150b83c | 1147797065081426760 / 14 = 81985504648673340
fedcb59726a1348 / d = 139ad2e43e0c668 | 1147797065081426760 / 13 = 88292081929340520
fedcb59726a1348 / c = 153d0f21ede2c46 | 1147797065081426760 / 12 = 95649755423452230
fedcb59726a1348 / b = 172b56538f25ed8 | 1147797065081426760 / 11 = 104345187734675160
fedcb59726a1348 / 5 = 32f8f11e3aed0a8 | 1147797065081426760 / 5 = 229559413016285352
fedcb59726a1348 / 9 = 1c5169829283b08 | 1147797065081426760 / 9 = 127533007231269640
fedcb59726a1348 / 7 = 2468ac3a2a17078 | 1147797065081426760 / 7 = 163971009297346680
fedcb59726a1348 / 2 = 7f6e5acb93509a4 | 1147797065081426760 / 2 = 573898532540713380
fedcb59726a1348 / 6 = 2a7a1e43dbc588c | 1147797065081426760 / 6 = 191299510846904460
fedcb59726a1348 / a = 197c788f1d76854 | 1147797065081426760 / 10 = 114779706508142676
fedcb59726a1348 / 1 = fedcb59726a1348 | 1147797065081426760 / 1 = 1147797065081426760
fedcb59726a1348 / 3 = 54f43c87b78b118 | 1147797065081426760 / 3 = 382599021693808920
fedcb59726a1348 / 4 = 3fb72d65c9a84d2 | 1147797065081426760 / 4 = 286949266270356690
fedcb59726a1348 / 8 = 1fdb96b2e4d4269 | 1147797065081426760 / 8 = 143474633135178345
Perl 6
Base 10
The number can not have a zero in it, that implies that it can not have a 5 either since if it has a 5, it must be divisible by 5, but the only numbers divisible by 5 end in 5 or 0. It can't be zero, and if it is odd, it can't be divisible by 2, 4, 6 or 8. So that leaves 98764321 as possible digits the number can contain. The sum of those 8 digits is not divisible by three so the largest possible integer must use no more than 7 of them (since 3, 6 and 9 would be eliminated). Strictly by removing possibilities that cannot possibly work we are down to at most 7 digits.
We can deduce that the digit that won't get used is one of 1, 4, or 7 since those are the only ones where the removal will yield a sum divisible by 3. It is extremely unlikely be 1, since EVERY number is divisible by 1. Removing it reduces the number of digits available but doesn't gain anything as far as divisibility. It is unlikely to be 7 since 7 is prime and can't be made up of multiples of other numbers. Practically though, the code to accommodate these observations is longer running and more complex than just brute-forcing it from here.
In order to accommodate the most possible digits, the number must be divisible by 7, 8 and 9. If that is true then it is automatically divisible by 2, 3, 4, & 6 as they can all be made from the combinations of multiples of 2 and 3 which are present in 8 & 9; so we'll only bother to check multiples of 9 * 8 * 7 or 504.
All these optimizations get the run time to well under 1 second.
<lang perl6>my $magic-number = 9 * 8 * 7; # 504
my $div = 9876432 div $magic-number * $magic-number; # largest 7 digit multiple of 504 < 9876432
for $div, { $_ - $magic-number } ... * -> $test { # only generate multiples of 504
next if $test ~~ / <[05]> /; # skip numbers containing 0 or 5 next if $test ~~ / (.).*$0 /; # skip numbers with non unique digits
say "Found $test"; # Found a solution, display it
for $test.comb {
printf "%s / %s = %s\n", $test, $_, $test / $_;
}
last
}</lang>
- Output:
Found 9867312 9867312 / 9 = 1096368 9867312 / 8 = 1233414 9867312 / 6 = 1644552 9867312 / 7 = 1409616 9867312 / 3 = 3289104 9867312 / 1 = 9867312 9867312 / 2 = 4933656
Base 16
There are fewer analytical optimizations available for base 16. Other than 0, no digits can be ruled out so a much larger space must be searched. We'll start at the largest possible permutation (FEDCBA987654321) and work down so as soon as we find a solution, we know it is the solution. The combination of .race.grep with .first lets us utilize concurrency and exit early when the single desired solution is found.
<lang perl6>sub find-match ($num) {
my $test = $num.base(16); return if index $test, 0; return if $test.comb.Bag.values.max > 1; $num;
} my $hex = 'FEDCBA987654321'; # largest possible hex number my $magic-number = [lcm] 1 .. 15; # find least common multiple my $div = :16($hex) div $magic-number * $magic-number;
- hunt for target stepping backwards in multiples of the lcm
my $target = ($div, { $_ - $magic-number } ... 0).race.grep(*.&find-match).first; my $hexnum = $target.base(16);
say "Found $hexnum"; # Found a solution, display it
say ' ' x 12, 'In base 16', ' ' x 36, 'In base 10'; for $hexnum.comb {
printf "%s / %s = %s | %d / %2d = %19d\n",
$hexnum, $_, ($target / :16($_)).base(16),
$target, :16($_), $target / :16($_);}</lang>
- Output:
Found FEDCB59726A1348
In base 16 In base 10
FEDCB59726A1348 / F = 10FDA5B4BE4F038 | 1147797065081426760 / 15 = 76519804338761784
FEDCB59726A1348 / E = 1234561D150B83C | 1147797065081426760 / 14 = 81985504648673340
FEDCB59726A1348 / D = 139AD2E43E0C668 | 1147797065081426760 / 13 = 88292081929340520
FEDCB59726A1348 / C = 153D0F21EDE2C46 | 1147797065081426760 / 12 = 95649755423452230
FEDCB59726A1348 / B = 172B56538F25ED8 | 1147797065081426760 / 11 = 104345187734675160
FEDCB59726A1348 / 5 = 32F8F11E3AED0A8 | 1147797065081426760 / 5 = 229559413016285352
FEDCB59726A1348 / 9 = 1C5169829283B08 | 1147797065081426760 / 9 = 127533007231269640
FEDCB59726A1348 / 7 = 2468AC3A2A17078 | 1147797065081426760 / 7 = 163971009297346680
FEDCB59726A1348 / 2 = 7F6E5ACB93509A4 | 1147797065081426760 / 2 = 573898532540713380
FEDCB59726A1348 / 6 = 2A7A1E43DBC588C | 1147797065081426760 / 6 = 191299510846904460
FEDCB59726A1348 / A = 197C788F1D76854 | 1147797065081426760 / 10 = 114779706508142676
FEDCB59726A1348 / 1 = FEDCB59726A1348 | 1147797065081426760 / 1 = 1147797065081426760
FEDCB59726A1348 / 3 = 54F43C87B78B118 | 1147797065081426760 / 3 = 382599021693808920
FEDCB59726A1348 / 4 = 3FB72D65C9A84D2 | 1147797065081426760 / 4 = 286949266270356690
FEDCB59726A1348 / 8 = 1FDB96B2E4D4269 | 1147797065081426760 / 8 = 143474633135178345
Prolog
This will work with any radix, including base 10 and base 16. <lang prolog>% Find the largest integer divisible by all it's digits, with no digit repeated. % ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ % We go for a generic algorithm here. Test for divisibility is done by % calculating the least common multiplier for all digits, and testing % whether a candidate can be divided by the LCM without remainder. % % Instead of iterating numbers and checking whether the number has % repeating digits, it is more efficient to generate permutations of % digits and then convert to a number. Doing it this way reduces search % space greatly. % % Notes: % For decimal numbers we could improve times by testing only numbers % of length 7 (since 5x2=10 and 0 is not one of our digits, and 9x2=18 % which needs 2 digits to store), but that sort of logic does not % hold for hexadecimal numbers. % We could also explicitly eliminate odd numbers, but the double validity % check actually slows us down very slightly instead of speeding us up.
- - dynamic
trial/1. % temporarily store digit combinations here.
gcd(X, X, X). % Calculate greatest common divisor gcd(M, N, X) :- N > M, B is N-M, gcd(M,B,X). gcd(M, N, X) :- N < M, A is M-N, gcd(A,N,X).
lcm(A, B, LCM) :- gcd(A,B,GCD), LCM is A * B / GCD.
lcm([H], H). % Calculate least common multiplier lcm([A|T], LCM) :- lcm(T, B), !, lcm(A,B,LCM).
mkint(_, Val, [], Val). % Result = Val where list is empty mkint(Radix, Val, [H|T], Int) :- % (((I0*10+I1)*10+I2)*10+In)... V0 is Val*Radix+H, !, mkint(Radix, V0, T, Int).
% Turn a list of digits into an integer number using Radix. mkint(Radix, [H|T], Int) :- mkint(Radix, H, T, Int).
domain(0, []). % For example, domain(5) is [1,2,3,4,5] domain(N, [N|Digits]) :-
succ(N0, N), !, domain(N0, Digits).
trial(0, Digits, Digits). % generates a combination of digits to test trial(N, D, Digits) :- % remove N digits, and find remaining combinations
append(L0,[_|L1],D), succ(N0, N), trial(N0, L1, Dx), append(L0, Dx, Digits). % trial(1, [3,2,1], D) -> D=[2,1]; D=[3,1]; D=[3,2].
make_trials(_,_) :- retractall(trial(_)), fail. make_trials(N,Domain) :- trial(N, Domain, Digits), asserta(trial(Digits)), fail. make_trials(_,_). % trials are stored highest values to lowest
combinations(Radix, NDigits) :- % Precalculate all possible digit combinations
succ(R0, Radix), domain(R0, Domain), Nskip is R0 - NDigits, make_trials(Nskip, Domain).
test(Radix, Digits, LCM, Number) :- % Make an integer and check for divisibility
mkint(Radix, Digits, Number), 0 is Number mod LCM.
bignum(Radix, Number) :-
succ(R0, Radix), between(1,R0,N), NDigits is Radix - N, % loop decreasing length combinations(Radix, NDigits), % precalc digit combos with length=NDigits trial(Digits), lcm(Digits, LCM), % for a combination, calculate LCM permutation(Digits, Trial), % generate a permutation test(Radix, Trial, LCM, Number). % test for divisibility
largest_decimal(N) :- bignum(10, N), !. largest_hex(N, H) :- bignum(16, N), !, sformat(H, '~16r', [N]).</lang>
?- time(largest_decimal(S)). % 20,043,250 inferences, 3.086 CPU in 3.089 seconds (100% CPU, 6493905 Lips) S = 9867312. ?- time(largest_hex(S,H)). % 73,332,059 inferences, 11.800 CPU in 11.803 seconds (100% CPU, 6214553 Lips) S = 1147797065081426760, H = "fedcb59726a1348".
Python
base 10
Using the insights presented in the preamble to the Perl 6 (base 10) example:
<lang python>from itertools import (chain, permutations) from functools import (reduce) from fractions import (gcd) from math import (floor)
- main :: IO ()
def main():
# (Division by zero is not an option, so we omit 0 and 5) digits = [1, 2, 3, 4, 6, 7, 8, 9]
# Least common multiple of the digits above lcmDigits = foldl1(lcm)(digits)
# Any set of 7 distinct digits obtained by deleting
# [1, 4 or 7] from the list of 8 above
sevenDigits = _map(
lambda x: delete(x)(digits)
)([1, 4, 7])
print (
maximumBy(lambda n: n if 0 == n % lcmDigits else 0)(
_map(unDigits)(
concat(_map(permutations)(sevenDigits))
)
)
)
- unDigits :: [Int] -> Int
def unDigits(xs):
return reduce(lambda a, x: a * 10 + x, xs, 0)
- GENERIC --------------------------------------------------
- concat :: a -> [a]
- concat :: [String] -> String
def concat(xxs):
xs = list(chain.from_iterable(xxs))
return [] if not xs else (
.join(xs) if type(xs[0]) is str else xs
)
- delete :: Eq a => a -> [a] -> [a]
def delete(x):
def go(xs):
ys = xs.copy()
ys.remove(x)
return ys
return lambda xs: go(xs)
- foldl1 :: (a -> a -> a) -> [a] -> a
def foldl1(f):
return lambda xs: reduce(
uncurry(f), xs[1:], xs[0]
) if (len(xs) > 0) else None
- lcm :: Int -> Int -> Int
def lcm(x):
return lambda y: (
0 if (0 == x or 0 == y) else abs(
y * floor(x / gcd(x, y))
)
)
- map :: (a -> b) -> [a] -> [b]
def _map(f):
return lambda xs: list(map(f, xs))
- maximumBy :: (a -> a -> a) -> [a] -> a
def maximumBy(f):
def go(x, y):
return y if f(y) > f(x) else x
return lambda xs: reduce(go, xs[1:], xs[0])
- uncurry :: (a -> b -> c) -> ((a, b) -> c)
def uncurry(f):
return lambda x, y: f(x)(y)
- MAIN ---
main()</lang>
- Output:
9867312
base 16
Descending from the upper limit, in steps of 360360 (least common multiple of the fifteen digit values), until the first number that uses all fifteen digits when expressed in hexadecimal.
<lang python>from functools import (reduce) from fractions import (gcd) from math import (floor)
- main :: IO ()
def main():
# Least common multiple of digits [1..15]
# ( -> 360360 )
lcmDigits = foldl1(lcm)(
enumFromTo(1)(15)
)
allDigits = 0xfedcba987654321
# ( -> 1147797409030632360 ) upperLimit = allDigits - (allDigits % lcmDigits)
# Possible numbers
xs = enumFromThenToGen(upperLimit)(
upperLimit - lcmDigits
)(1)
print (
hex(
until(lambda x: 15 == len(set(showHex(x))))(
lambda _: next(xs)
)(next(xs))
)
) # --> 0xfedcb59726a1348
- GENERIC ABSTRACTIONS -----------------------------------
- enumFromTo :: Int -> Int -> [Int]
def enumFromTo(m):
return lambda n: list(range(m, 1 + n))
- enumFromThenToGen :: Int -> Int -> Int -> Gen [Int]
def enumFromThenToGen(m):
def go(m, next, n):
d = next - m
v = m
while True:
yield v
v = d + v
return lambda next: lambda n: go(m, next, n)
- foldl1 :: (a -> a -> a) -> [a] -> a
def foldl1(f):
return lambda xs: reduce(
uncurry(f), xs[1:], xs[0]
) if xs else None
- lcm :: Int -> Int -> Int
def lcm(x):
return lambda y: (
0 if (0 == x or 0 == y) else abs(
y * floor(x / gcd(x, y))
)
)
- map :: (a -> b) -> [a] -> [b]
def _map(f):
return lambda xs: list(map(f, xs))
- showHex :: Int -> String
def showHex(n):
return hex(n)[2:]
- uncurry :: (a -> b -> c) -> ((a, b) -> c)
def uncurry(f):
return lambda x, y: f(x)(y)
- until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
def go(f, x):
v = x
while not p(v):
v = f(v)
return v
return lambda f: lambda x: go(f, x)
- MAIN --
main()</lang>
- Output:
0xfedcb59726a1348 [Finished in 1.39s]
REXX
base 10
This REXX version uses mostly the same logic and deductions that the Perl 6 example does, but it performs
the tests in a different order for maximum speed.
The inner do loop is only executed a score of times; the 1st if statement does the bulk of the eliminations. <lang rexx>/*REXX program finds the largest (decimal) integer divisible by all its decimal digits. */ $= 7 * 8 * 9 /*a number that it must be divisible by*/ start= 9876432 % $ * $ /*the number to start the sieving from.*/ t= 0 /*the number of divisibility trials. */
do #=start by -$ /*search from largest number downwards.*/
if # // $ \==0 then iterate /*Not divisible? Then keep searching.*/
if verify(50, #, 'M') \==0 then iterate /*does it contain a five or a zero? */
t= t+1 /*curiosity's sake, track # of trials. */
do j=1 for length(#) - 1 /*look for a possible duplicated digit.*/
if pos( substr( #, j, 1), #, j+1) \==0 then iterate #
end /*j*/ /* [↑] Not unique? Then keep searching*/
/* [↓] superfluous, but check anyways.*/
do v=1 for length(#) /*verify the # is divisible by all digs*/
if # // substr(#, v, 1) \==0 then iterate #
end /*v*/ /* [↑] ¬divisible? Then keep looking.*/
leave /*we found a number, so go display it. */
end /*#*/
say 'found ' # " (in " t ' trials)' /*stick a fork in it, we're all done. */</lang>
- output:
Timing note: execution time is under 1/2 millisecond (essentially not measurable in the granularity of the REXX timer under Microsoft Windows).
found 9867312 (in 11 trials)
base 16
The "magic" number was expanded to handle hexadecimal numbers.
Note that 15*14*13*12*11 is the same as 13*11*9*8*7*5. <lang rexx>/*REXX program finds the largest hexadecimal integer divisible by all its hex digits. */ numeric digits 20 /*be able to handle the large hex nums.*/ bigH= 'fedcba987654321' /*biggest hexadecimal number possible. */ bigN= x2d(bigH) /* " " " in decimal*/ $= 15 * 14 * 13 * 12 * 11 /*a number that it must be divisible by*/ start= bigN % $ * $ /*the number to start the sieving from.*/ t=0 /*the number of divisibility trials. */
do #=start by -$ /*search from largest poss. # downwards*/
if # // $ \==0 then iterate /*Not divisible? Then keep searching.*/
h= d2x(#) /*convert decimal number to hexadecimal*/
if pos(0, h) \==0 then iterate /*does hexadecimal number contain a 0? */
t= t+1 /*curiosity's sake, track # of trials. */
do j=1 for length(h) - 1 /*look for a possible duplicated digit.*/
if pos( substr(h, j, 1), h, j+1) \==0 then iterate #
end /*j*/ /* [↑] Not unique? Then keep searching*/
do v=1 for length(h) /*verify the # is divisible by all digs*/
if # // x2d(substr( h, v, 1) ) \==0 then iterate #
end /*v*/ /* [↑] ¬divisible? Then keep looking.*/
leave /*we found a number, so go display it. */
end /*#*/
say 'found ' h " (in " t ' trials)' /*stick a fork in it, we're all done. */</lang>
- output:
found FEDCB59726A1348 (in 287747 trials)
Ring
<lang ring>
- Project : Largest number divisible by its digits
for n = 9867000 to 9867400
numbers = list(9)
for t=1 to 9
numbers[t] = 0
next
flag = 1
flag2 = 1
flag3 = 1
str=string(n)
for m=1 to len(str)
if number(str[m]) > 0
numbers[number(str[m])] = numbers[number(str[m])] + 1
else
flag2 = 0
ok
next
if flag2 = 1
for p=1 to 9
if numbers[p] = 0 or numbers[p] = 1
else
flag = 0
ok
next
if flag = 1
for x=1 to len(str)
if n%(number(str[x])) != 0
flag3 = 0
ok
next
if flag3 = 1
see n + nl
ok
ok
ok
next </lang> Output:
9867312
Ruby
base 10
Following the reasoning of the Perl 6 sample. <lang ruby>magic_number = 9*8*7 div = 9876432.div(magic_number) * magic_number candidates = div.step(0, -magic_number)
res = candidates.find do |c|
digits = c.digits (digits & [0,5]).empty? && digits == digits.uniq
end
puts res # => 9867312</lang>
Sidef
base 10
<lang ruby>func largest_number(base) {
var digits = @(base ^.. 1)
digits.each {|k|
digits.variations(k, {|*a|
var n = Number(a.join, base)
if (a.all {|d| d.divides(n) }) {
return n
}
})
}
}
say largest_number(10) #=> 9867312</lang>
zkl
base 10
<lang zkl>const magic_number=9*8*7; # 504 const div=9876432 / magic_number * magic_number; #largest 7 digit multiple of 504 < 9876432
foreach test in ([div..0,-magic_number]){
text:=test.toString();
if(text.holds("0","5")) continue; # skip numbers containing 0 or 5
if(text.unique().len()!=text.len()) continue; # skip numbers with non unique digits
if(test.split().filter1('%.fp(test))) continue; # skip numbers that don't divide evenly by all digits
println("Found ",test); # Found a solution, display it
foreach d in (test.split()){
println("%s / %s = %s".fmt(test,d, test/d));
}
break;
}</lang>
- Output:
Found 9867312 9867312 / 9 = 1096368 9867312 / 8 = 1233414 9867312 / 6 = 1644552 9867312 / 7 = 1409616 9867312 / 3 = 3289104 9867312 / 1 = 9867312 9867312 / 2 = 4933656
base 16
<lang zkl>const bigN=0xfedcba987654321; // biggest hexadecimal number possible. lcm:=lcmNs([1..15]); // 360360, smallest # that will divide answer upperLimit:=bigN - bigN%lcm; // start at a mulitple of whatever the answer is
foreach test in ([upperLimit..1,-lcm]){
text:=test.toString(16); if(15!=text.unique().len()) continue; println(text); break;
}</lang> <lang zkl>fcn lcmNs(ns){ ns.reduce(fcn(m,n){ (m*n).abs()/m.gcd(n) }) }</lang>
- Output:
fedcb59726a1348