Largest number divisible by its digits: Difference between revisions
m (→base 16: added wording to the REXX section header about equivalency of two forms of magic numbers.) |
m (→base 10: crunch) |
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foreach test in ([div..0,-magic_number]){ |
foreach test in ([div..0,-magic_number]){ |
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text,sz := test.toString(),text.len(); |
text,sz := test.toString(),text.len(); |
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if(text.holds("0" |
if(text.holds("0","5")) continue; # skip numbers containing 0 or 5 |
||
if(text.unique().len()!=sz) |
if(text.unique().len()!=sz) continue; # skip numbers with non unique digits |
||
if(test.split().filter1('%.fp(test))) |
if(test.split().filter1('%.fp(test))) continue; # skip numbers that don't divide evenly by all digits |
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println("Found ",test); # Found a solution, display it |
println("Found ",test); # Found a solution, display it |
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Revision as of 22:14, 2 September 2017
- Task
Find the largest base 10 integer whose digits are all different, that 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.
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, but practically, the code to accommodate that 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. Rather than using a magic number to try to filter the search space and then check for uniqueness, we'll just generate unique permutations and check them for divisibility. In practice this seems to be much faster. 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.
To verify that a number is divisible by 1 through 15, we only need to check 5, 7, 8, 9, 11, & 13.
<lang perl6>my $hex = 'FEDCBA987654321'; # largest possible hex number
for $hex.comb.permutations -> @test {
my $test = [~] @test; my $num = $test.parse-base(16);
next unless $num +& 8; # extremely cheap pre-check for divisibility by 8 next unless $num %% 45045; # check divisibility by other factors 5*7*9*11*13
say "Found $test"; # Found a solution, display it
say ' ' x 12, 'In base 16', ' ' x 36, 'In base 10';
for @test {
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.*/ L=length(start) /*the # of digits in the start number. */ i#=0 /*the number of divisibility trials. */
do #=9876432 by -2 /*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? */
i#=i#+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 L /*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 " i# ' 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).
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.*/ L=length( d2x(start) ) /*the length of the hexadecimal number.*/ i#=0 /*the number of divisibility trials. */
do #=start by -$ /*search from largest number 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? */
i#=i#+1 /*curiosity's sake, track # of trials. */
do j=1 for L /*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 L /*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 " i# ' trials)' /*stick a fork in it, we're all done. */</lang>
- output:
found FEDCB59726A1348 (in 287747 trials)
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,sz := test.toString(),text.len();
if(text.holds("0","5")) continue; # skip numbers containing 0 or 5
if(text.unique().len()!=sz) 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