Comments on: $7.11

By: Elvis Montero

Elvis Montero — Sat, 12 Jun 2010 21:47:47 +0000

Here's my solution. My Java code resembles some of the Python code posted by other commentators.

By: $7.11 in four prices and the Decimal « Wrong Side of Memphis

$7.11 in four prices and the Decimal « Wrong Side of Memphis — Fri, 22 Jan 2010 16:21:54 +0000

[...] 2009 de khelben There’s an fun and not very difficult programming exercise, presented on Programming Praxis . The problem is easy to [...]

By: Scott Haug

Scott Haug — Sun, 03 Jan 2010 03:59:11 +0000

Late to the game on this one, but I'm using this site to get some practice writing clojure. Here's my submission: [sourcecode lang="css"] (use '[clojure.contrib.combinatorics :only (combinations)]) (filter #(and (= (apply + %) 711) (= (apply * %) 711000000)) (combinations (filter #(zero? (mod 711000000 (int %))) (range 1 712)) 4)) [/sourcecode] Not the fastest by any means, but it gets the job done in about 1.4 secs on my laptop.

By: Dan Schoch

Dan Schoch — Fri, 04 Dec 2009 15:58:00 +0000

/* FWIW — Here’s my Oracle PL/SQL solution */
declare
done boolean := false ;
item4 pls_integer ;
begin
for item1 in 1..711
loop
for item2 in item1..(711-item1)
loop
for item3 in item2..(711-item1-item2)
loop
item4 := 711-(item1+item2+item3) ;
if item1*item2*item3*item4 = 711000000 then
dbms_output.put_line(‘Item 1 = $’||to_char(item1/100,’fm999.00′)) ;
dbms_output.put_line(‘Item 2 = $’||to_char(item2/100,’fm999.00′)) ;
dbms_output.put_line(‘Item 3 = $’||to_char(item3/100,’fm999.00′)) ;
dbms_output.put_line(‘Item 4 = $’||to_char(item4/100,’fm999.00′)) ;
done := true ;
end if ;
exit when done ;
end loop ;
exit when done ;
end loop ;
exit when done ;
end loop ;
end ;
/

By: PHW

PHW — Tue, 01 Dec 2009 22:20:49 +0000

Rejigged my earlier solution to take advantage of the idea that all 4 costs must be a factor of the overall product. Also (hopefully) I pretty-printed the code this time. The code now completes in less than a second. #!/usr/bin/perl use strict; my $PRODUCT = 711000000; for (my $a=708; $a>=1 ; $a-= 1) { if ( ($PRODUCT % $a) !=0 ) {next;} for (my $b=(709-$a); $b >=$a ; $b-=1) { if ( ($PRODUCT % $b) !=0 ) {next;} for (my $c=(710-$a-$b); $c>=$b; $c-=1) { if ( ($PRODUCT % $c) !=0 ) {next;} my $d = 711 - $a - $b - $c; if ($d>$c) { my $prd = $a*$b*$c*$d; if ($prd==$PRODUCT) { printf "%3.2f %3.2f %3.2f %3.2f\n",$a/100,$b/100,$c/100,$d/100; } } } } }

By: kernelbob

kernelbob — Mon, 30 Nov 2009 01:32:31 +0000

Oh. You dropped the requirement that c <= d. I had missed that.

Thanks for a fun problem.

By: programmingpraxis

programmingpraxis — Mon, 30 Nov 2009 00:31:53 +0000

The product 7.11 × 10⁸ factors as 2⁶ × 3² × 5⁶ × 79. The 79 factor has to be there someplace, and d is as good a place as any. And by the way, the multiplier can't be 7, because there is no 7 in the factorization, can't be 9, because 9 × 79 = 711 leaving no room for the other factors, and you've already demonstrated that it can't be 1 or 2.

By: kernelbob

kernelbob — Sun, 29 Nov 2009 22:56:05 +0000

677040, correct. I transcribed the number wrong.

It’s not clear to me that d has to be a multiple of 79. Can you explain your reasoning? It seems to me that it could be any of the 62 factors, though it’s at least the fourth root of 711000000 (i.e., greater than 163).

By: programmingpraxis

programmingpraxis — Sun, 29 Nov 2009 19:28:30 +0000

I get 677,040, which is close to your calculation. See http://programmingpraxis.codepad.org/tvysKxmF. Since the prime factors of 711 are 3, 3, and 79, you could do even better by having d iterate through 1×79, 2×79, ... 9×79. See http://programmingpraxis.codepad.org/B0W4qEU0. That reduces the size of the cross-product to 374,976.

By: kbob

kbob — Sun, 29 Nov 2009 15:42:40 +0000

Indentation weirded out, sorry. I guess WordPress assumes tab stops every four columns. I assumed eight.