Largest number divisible by its digits
- Task
Find the largest base 10 integer whose digits are all different, and is evenly 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.
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]
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 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 next unless all $test.comb.map: $test %% *; # skip numbers that don't divide evenly by all 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.
<lang perl6>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;
for $div, { $_ - $magic-number } ... 0 -> $num { # Only generate multiples of the lcm
my $test = $num.base(16); next if $test ~~ / 0 /; # skip numbers containing 0 next if $test ~~ / (.).*$0 /; # skip numbers with non unique digits
say "Found $test"; # Found a solution, display it say ' ' x 12, 'In base 16', ' ' x 36, 'In base 10'; for $test.comb { printf "%s / %s = %s | %d / %2d = %19d\n", $test, $_, ($num / :16($_)).base(16), $num, :16($_), $num / :16($_); } last
}</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
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
- Date : 2017/09/22
- Author : Gal Zsolt (~ CalmoSoft ~)
- Email : <calmosoft@gmail.com>
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
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