Permuted multiples: Difference between revisions
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=={{header|REXX}}== |
=={{header|REXX}}== |
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<lang rexx>/*REXX program finds and displays the smallest positive integer n such that |
<lang rexx>/*REXX program finds and displays the smallest positive integer n such that ··· */ |
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/*───────────────────────── 2*n, 3*n, 4*5, 5*6, and 6*n contain the same decimal digits.*/ |
/*───────────────────────── 2*n, 3*n, 4*5, 5*6, and 6*n contain the same decimal digits.*/ |
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do n=1 /*increment N from unity 'til answer.*/ |
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b= 2*n /*calculate the product of: 2*n */ |
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t= 3*n /* " " " " 3*n */ |
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if verify(t, b)>0 then iterate /* T not have required digs? Skip*/ |
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q= 4*n /*calculate the product of: 4*n */ |
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if verify(q, b)>0 then iterate /* Q not have required digs? Skip*/ |
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if verify(q, t)>0 then iterate /* " " " " " " */ |
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v= 5*n /*calculate the product of: 5*n */ |
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if verify(v, b)<0 then iterate /* V not have required digs? Skip*/ |
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if verify(v, t)>0 then iterate /* " " " " " " */ |
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if verify(v, q)>0 then iterate /* " " " " " " */ |
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s= 6*n /*calculate the product of: 6*n */ |
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if verify(s, b)>0 then iterate /* S not have required digs? Skip*/ |
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if verify(s, t)>0 then iterate /* " " " " " " */ |
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if verify(s, q)>0 then iterate /* " " " " " " */ |
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if verify(s, v)>0 then iterate /* " " " " " " */ |
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say ' n =' commas(n) /*display the value of: n */ |
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say ' 2*n =' commas(b) /* " " " " 2*n */ |
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say ' 3*n =' commas(t) /* " " " " 3*n */ |
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say ' 4*n =' commas(q) /* " " " " 4*n */ |
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say ' 5*n =' commas(v) /* " " " " 5*n */ |
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say ' 6*n =' commas(s) /* " " " " 6*n */ |
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leave /*found the N number, time to leave.*/ |
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end /*n*/ |
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say _ '5*n =' commas(v) /* " " " 5*n. */ |
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say _ '6*n =' commas(s) /* " " " 6*n. */ |
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exit 0 /*stick a fork in it, we're all done. */ |
exit 0 /*stick a fork in it, we're all done. */ |
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/*──────────────────────────────────────────────────────────────────────────────────────*/ |
/*──────────────────────────────────────────────────────────────────────────────────────*/ |
Revision as of 01:03, 18 August 2021
- Attribution
The following task is taken from Project Euler.
- Task
Find the smallest positive integer n such that, when expressed in decimal, 2*n, 3*n, 4*n, 5*n, and 6*n contain exactly the same digits but in a different order.
AppleScript
— except that the 'steps' figure here is cumulative. Also, for six different numbers to have the same digits, each must have at least three digits, none of which can be 0. So the smallest possible value of n is 123. 6 times that is 738, which still doesn't fit the bill, but you have to start somewhere. 'Steps' are the number of 'n's actually tried when n * 6 doesn't exceed the next power of 10. While no power of 10 is exactly divisible by 6, it is technically the point at which the number of digits increases, so I've pedantically adjusted the logic to reflect this. :) The number of steps between 100,000 and the final arrival at 142,857 is 14,286, which is interesting.
<lang applescript>use AppleScript version "2.3.1" -- Mac OS X 10.9 (Mavericks) or later. use sorter : script "Insertion Sort" -- <https://www.rosettacode.org/wiki/Sorting_algorithms/Insertion_sort#AppleScript>
on decDigits(n)
set digits to {n mod 10 as integer} set n to n div 10 repeat until (n = 0) set beginning of digits to n mod 10 as integer set n to n div 10 end repeat return digits
end decDigits
on join(lst, delim)
set astid to AppleScript's text item delimiters set AppleScript's text item delimiters to delim set txt to lst as text set AppleScript's text item delimiters to astid return txt
end join
on task()
set {output, n, n10, steps} to {{}, 123, 1000, 0} repeat if (n * 6 < n10) then set steps to steps + 1 set nl to decDigits(n) tell sorter to sort(nl, 1, -1) set found to true repeat with i from 2 to 6 set inl to decDigits(n * i) tell sorter to sort(inl, 1, -1) if (inl ≠ nl) then set found to false exit repeat end if end repeat if (found) then exit repeat set n to n + 3 else set end of output to "Nothing below " & n10 & (" (" & steps & " steps)") set n to n10 + 2 set n10 to n10 * 10 -- set steps to 0 end if end repeat set end of output to " n = " & n & (" (" & steps & " steps altogether)") repeat with i from 2 to 6 set end of output to (i as text) & " * n = " & i * n end repeat return join(output, linefeed)
end task
task()</lang>
- Output:
<lang applescript>"Nothing below 1000 (15 steps) Nothing below 10000 (237 steps) Nothing below 100000 (2459 steps)
n = 142857 (16745 steps altogether)
2 * n = 285714 3 * n = 428571 4 * n = 571428 5 * n = 714285 6 * n = 857142"</lang>
Factor
<lang factor>USING: formatting io kernel lists lists.lazy math math.ranges math.vectors numspec present prettyprint sequences sets ;
- multiples ( n -- seq )
[ 2 * ] [ 6 * ] [ ] tri <range> [ present ] map ;
- all-set-eq? ( seq -- ? )
dup ?first [ set= ] curry all? ;
! Ordered lazy list of numbers that start with a '1' digit NUMSPEC: starting-with-one 1 1_ ... ;
- smallest-permuted-multiple ( -- n )
starting-with-one [ multiples all-set-eq? ] lfilter car ;
{ 2 3 4 5 6 } " n: " write smallest-permuted-multiple dup . over n*v [ "×%d: %d\n" printf ] 2each</lang>
- Output:
n: 142857 ×2: 285714 ×3: 428571 ×4: 571428 ×5: 714285 ×6: 857142
Go
<lang go>package main
import (
"fmt" "rcu" "sort"
)
// assumes l1 is sorted but l2 is not func areSame(l1, l2 []int) bool {
if len(l1) != len(l2) { return false } sort.Ints(l2) for i := 0; i < len(l1); i++ { if l1[i] != l2[i] { return false } } return true
}
func main() {
i := 100 // clearly a 1 or 2 digit number is impossible nextPow := 1000 for { digits := rcu.Digits(i, 10) if digits[0] != 1 { i = nextPow nextPow *= 10 continue } sort.Ints(digits) allSame := true for j := 2; j <= 6; j++ { digits2 := rcu.Digits(i*j, 10) if !areSame(digits, digits2) { allSame = false break } } if allSame { fmt.Println("The smallest positive integer n for which the following") fmt.Println("multiples contain exactly the same digits is:") fmt.Println(" n =", i) for k := 2; k <= 6; k++ { fmt.Printf("%d x n = %d\n", k, k*i) } return } i = i + 1 }
}</lang>
- Output:
The smallest positive integer n for which the following multiples contain exactly the same digits is: n = 142857 2 x n = 285714 3 x n = 428571 4 x n = 571428 5 x n = 714285 6 x n = 857142
Julia
<lang julia>n = minimum([n for n in 1:2000000 if sort(digits(2n)) == sort(digits(3n)) == sort(digits(4n)) == sort(digits(5n))== sort(digits(6n))]) println("n: $n, 2n: $(2n), 3n: $(3n), 4n: $(4n), 5n: $(5n), 6n: $(6n)")
</lang>
- Output:
n: 142857, 2n: 285714, 3n: 428571, 4n: 571428, 5n: 714285, 6n: 857142
Nim
Searching among multiples of 3 between 102 and 1_000 div 6, 1_002 and 10_000 div 6, 10_002 and 100_000 div 6, etc. (see discussion).
<lang Nim>from algorithm import sorted
func search(): int =
var start = 100 while true: for i in countup(start + 2, 10 * start div 6, 3): let digits = sorted($i) block check: for j in 2..6: if sorted($(i * j)) != digits: break check # Found. return i start *= 10
let n = search() echo " n = ", n for k in 2..6:
echo k, "n = ", k * n</lang>
- Output:
n = 142857 2n = 285714 3n = 428571 4n = 571428 5n = 714285 6n = 857142
Pascal
Create an array of the digits fixed 1 as first digit and 0 "1023456789"
Don't use the fact, that second digit must be < 6.Runtime negligible.
<lang pascal>program euler52;
{$IFDEF FPC}
{$MOde DElphi} {$Optimization On}
{$else}
{$Apptype console}
{$ENDIF} uses
sysutils;
const
Base = 10;
type
TUsedDigits = array[0..Base-1] of byte; tDigitsInUse = set of 0..Base-1;
var
UsedDigits :tUsedDigits; gblMaxDepth : NativeInt;
procedure InitUsed; Var
i : NativeInt;
Begin
For i := 2 to Base-1 do UsedDigits[i] := i; UsedDigits[0] := 1; UsedDigits[1] := 0;
end;
function GetUsedSet(const UsedDigits: tUsedDigits):tDigitsInUse; var
i : NativeInt;
begin
result := []; For i := 0 to gblMaxDepth do include(result,UsedDigits[i]);
end;
function CheckMultiples(const UsedDigits: tUsedDigits;OrgInUse:tDigitsInUse):NativeInt; var
SumDigits :tUsedDigits; i,c,s,j : integer;
begin
result := 0; SumDigits := UsedDigits; j := 2;// first doubled repeat c := 0; For i := gblMaxdepth downto 0 do Begin s := UsedDigits[i]+SumDigits[i]+c; c := ord(s >= base); SumDigits[i] := s-c*base; end; IF (c > 0) then break; if GetUsedSet(SumDigits) <> OrgInUse then break; inc(j); until j > 6; IF j > 6 then Begin result := 0; //Output in Base 10 For i := 0 to gblMaxdepth do result := result * Base +UsedDigits[i]; For i := 1 to 6 do writeln(i*result); writeln; end;
end;
procedure Check; Begin
CheckMultiples(UsedDigits,GetUsedSet(UsedDigits))
end;
procedure GetNextUsedDigit(StartIdx:NativeInt); var
i : NativeInt; DigitTaken: Byte;
Begin
For i := StartIDx to Base-1 do Begin //swap i with Startidx DigitTaken := UsedDigits[i]; UsedDigits[i]:= UsedDigits[StartIdx]; UsedDigits[StartIdx] := DigitTaken;
// write(StartIdx:3,i:3,DigitTaken:3,' ');
IF StartIdx <gblMaxDepth then GetNextUsedDigit(StartIdx+1) else check; //undo swap i with Startidx UsedDigits[StartIdx] := UsedDigits[i]; UsedDigits[i]:= DigitTaken; end;
end; var
T : INt64;
Begin
T := GetTickCount64; For gblMaxDepth := 2 to Base-1 do Begin InitUsed; writeln('With ',gblMaxdepth+1,' digits'); GetNextUsedDigit(1); end; T := GetTickCount64-T; write('Done in ',T/1000:0:3); {$IFDEF WINdows} readln; {$ENDIF}
end.</lang>
- Output:
With 3 digits With 4 digits With 5 digits With 6 digits 142857 285714 428571 571428 714285 857142 With 7 digits 1428570 2857140 4285710 5714280 7142850 8571420 1429857 2859714 4289571 5719428 7149285 8579142 With 8 digits 14298570 28597140 42895710 57194280 71492850 85791420 With 9 digits With 10 digits Done in 0.054
Phix
Maintain a limit (n10) and bump the iteration whenever *6 increases the number of digits, which (as [was] shown) cuts the number of iterations by a factor of nearly thirteen and a half times (as in eg [as was] 67 iterations instead of 900 to find nothing in 100..1,000). Also as noted on the talk page, since sum(digits(3n)) is a multiple of 3 and it uses the same digits as n, then sum(digits(n)) will also be the very same multiple of 3 and hence n must (also) be divisible by 3, so we can start each longer-digits iteration on 10^k+2 (since remainder(10^k,3) is always 1) and employ a step of 3, and enjoy a better than 40-fold overall reduction in iterations.
with javascript_semantics integer n = 3, n10 = 10, steps = 0 while true do if n*6>=n10 then printf(1,"Nothing less than %,d (%,d steps)\n",{n10,steps}) n = n10+2 n10 *= 10 steps = 0 else string ns = sort(sprintf("%d",n)) integer i -- (to test after loop) for i=2 to 6 do string ins = sort(sprintf("%d",n*i)) if ins!=ns then exit end if end for if i=7 then exit end if n += 3 steps += 1 end if end while constant fmt=""" Smallest positive integer n for which (2..6)*n contain the same digits: n = %d 2 x n = %d 3 x n = %d 4 x n = %d 5 x n = %d 6 x n = %d """ printf(1,fmt,sq_mul(n,tagset(6)))
- Output:
Nothing less than 10 (0 steps) Nothing less than 100 (2 steps) Nothing less than 1,000 (22 steps) Nothing less than 10,000 (222 steps) Nothing less than 100,000 (2,222 steps) Smallest positive integer n for which (2..6)*n contain the same digits: n = 142857 2 x n = 285714 3 x n = 428571 4 x n = 571428 5 x n = 714285 6 x n = 857142
Raku
<lang perl6>put display (^∞).map(1 ~ *).race.map( -> \n { next unless [eq] (2,3,4,5,6).map: { (n × $_).comb.sort.join }; n } ).first;
sub display ($n) { join "\n", " n: $n", (2..6).map: { "×$_: {$n×$_}" } }</lang>
- Output:
n: 142857 ×2: 285714 ×3: 428571 ×4: 571428 ×5: 714285 ×6: 857142
REXX
<lang rexx>/*REXX program finds and displays the smallest positive integer n such that ··· */ /*───────────────────────── 2*n, 3*n, 4*5, 5*6, and 6*n contain the same decimal digits.*/
do n=1 /*increment N from unity 'til answer.*/ b= 2*n /*calculate the product of: 2*n */ t= 3*n /* " " " " 3*n */ if verify(t, b)>0 then iterate /* T not have required digs? Skip*/ q= 4*n /*calculate the product of: 4*n */ if verify(q, b)>0 then iterate /* Q not have required digs? Skip*/ if verify(q, t)>0 then iterate /* " " " " " " */ v= 5*n /*calculate the product of: 5*n */ if verify(v, b)<0 then iterate /* V not have required digs? Skip*/ if verify(v, t)>0 then iterate /* " " " " " " */ if verify(v, q)>0 then iterate /* " " " " " " */ s= 6*n /*calculate the product of: 6*n */ if verify(s, b)>0 then iterate /* S not have required digs? Skip*/ if verify(s, t)>0 then iterate /* " " " " " " */ if verify(s, q)>0 then iterate /* " " " " " " */ if verify(s, v)>0 then iterate /* " " " " " " */ say ' n =' commas(n) /*display the value of: n */ say ' 2*n =' commas(b) /* " " " " 2*n */ say ' 3*n =' commas(t) /* " " " " 3*n */ say ' 4*n =' commas(q) /* " " " " 4*n */ say ' 5*n =' commas(v) /* " " " " 5*n */ say ' 6*n =' commas(s) /* " " " " 6*n */ leave /*found the N number, time to leave.*/ end /*n*/
exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?</lang>
- output when using the internal default input:
n = 142,857 2*n = 285,714 3*n = 428,571 4*n = 571,428 5*n = 714,285 6*n = 857,142
Ring
<lang ring> load "stdlib.ring"
see "working..." + nl see "Permuted multiples are:" + nl per = list(6) perm = list(6)
for n = 1 to 1000000
for x = 2 to 6 perm[x] = [] next perStr = list(6) for z = 2 to 6 per[z] = n*z perStr[z] = string(per[z]) for m = 1 to len(perStr[z]) add(perm[z],perStr[z][m]) next next for y = 2 to 6 perm[y] = sort(perm[y]) perStr[y] = list2str(perm[y]) perStr[y] = substr(perStr[y],nl,"") next if perStr[2] = perStr[3] and perStr[2] = perStr[4] and perStr[2] = perStr[5] and perStr[2] = perStr[6] see "n = " + n + nl see "2*n = " + (n*2) + nl see "3*n = " + (n*3) + nl see "4*n = " + (n*4) + nl see "5*n = " + (n*5) + nl see "6*n = " + (n*6) + nl exit ok
next
see "done..." + nl </lang>
- Output:
working... Permuted multiples are: n = 142857 2*n = 285714 3*n = 428571 4*n = 571428 5*n = 714285 6*n = 857142 done...
Wren
One thing that's immediately clear is that the number must begin with '1' otherwise the higher multiples will have more digits than it has. <lang ecmascript>import "/math" for Int
// assumes l1 is sorted but l2 is not var areSame = Fn.new { |l1, l2|
if (l1.count != l2.count) return false l2.sort() for (i in 0...l1.count) { if (l1[i] != l2[i]) return false } return true
}
var i = 100 // clearly a 1 or 2 digit number is impossible var nextPow = 1000 while (true) {
var digits = Int.digits(i) if (digits[0] != 1) { i = nextPow nextPow = nextPow * 10 continue } digits.sort() var allSame = true for (j in 2..6) { var digits2 = Int.digits(i * j) if (!areSame.call(digits, digits2)) { allSame = false break } } if (allSame) { System.print("The smallest positive integer n for which the following") System.print("multiples contain exactly the same digits is:") System.print(" n = %(i)") for (k in 2..6) System.print("%(k) x n = %(k * i)") return } i = i + 1
}</lang>
- Output:
The smallest positive integer n for which the following multiples contain exactly the same digits is: n = 142857 2 x n = 285714 3 x n = 428571 4 x n = 571428 5 x n = 714285 6 x n = 857142