1
$\begingroup$

A friend posed this question to me a few days ago, and I just haven't been able to get it off of my mind.

"There are ten numbers from 0 to 9. You can use any number you like in the box, but the number cannot repeat itself."

 [][][]
+[][][]

[][][][]

If you can provide the way in which arrived to the answer, that would be great. I would really like to learn.

$\endgroup$

migrated from math.stackexchange.com Feb 17 '17 at 23:06

This question came from our site for people studying math at any level and professionals in related fields.

  • $\begingroup$ Can the first digit of the answer be a zero? If that's the case, I've found a solution, $\endgroup$ – астон вілла олоф мэллбэрг Feb 17 '17 at 1:05
  • $\begingroup$ астон вілла олоф мэллбэрг: Normally not in puzzles like this, but there are exceptions. In fact, I start with the thousands digit of the sum being $1$ $\endgroup$ – Ross Millikan Feb 17 '17 at 1:06
  • 1
    $\begingroup$ A more usual way to write this is $$\ \ ABC\\ \underline {+DEF}\\GHJK$$ which gives a way to refer to the digits. My previous comment then could include $G=1$ because that is all you can carry from the addition of two numbers. $\endgroup$ – Ross Millikan Feb 17 '17 at 1:13
  • 1
    $\begingroup$ One example would be 537+489=1026 $\endgroup$ – Smurf Feb 17 '17 at 1:34
  • $\begingroup$ And $342+756=1098$, $742+356=1098$ ... my program counts exactly $96$ solutions... $\endgroup$ – ringø Feb 17 '17 at 1:56
2
$\begingroup$
1)    246 + 789 = 1035   
2)    249 + 786 = 1035   
3)    264 + 789 = 1053   
4)    269 + 784 = 1053   
5)    284 + 769 = 1053   
6)    286 + 749 = 1035   
7)    289 + 746 = 1035   
8)    289 + 764 = 1053   
9)    324 + 765 = 1089   
10)   325 + 764 = 1089    
11)   342 + 756 = 1098    
12)   346 + 752 = 1098    
13)   347 + 859 = 1206    
14)   349 + 857 = 1206    
15)   352 + 746 = 1098    
16)   356 + 742 = 1098    
17)   357 + 849 = 1206    
18)   359 + 847 = 1206    
19)   364 + 725 = 1089    
20)   365 + 724 = 1089    
21)   423 + 675 = 1098    
22)   425 + 673 = 1098    
23)   426 + 879 = 1305    
24)   429 + 876 = 1305    
25)   432 + 657 = 1089    
26)   437 + 589 = 1026    
27)   437 + 652 = 1089    
28)   439 + 587 = 1026    
29)   452 + 637 = 1089    
30)   457 + 632 = 1089    
31)   473 + 589 = 1062    
32)   473 + 625 = 1098    
33)   475 + 623 = 1098    
34)   476 + 829 = 1305    
35)   479 + 583 = 1062    
36)   479 + 826 = 1305    
37)   483 + 579 = 1062    
38)   487 + 539 = 1026    
39)   489 + 537 = 1026    
40)   489 + 573 = 1062    
41)   624 + 879 = 1503    
42)   629 + 874 = 1503    
43)   674 + 829 = 1503    
44)   679 + 824 = 1503    
45)   743 + 859 = 1602    
46)   749 + 853 = 1602    
47)   753 + 849 = 1602    
48)   759 + 843 = 1602    

Here's the brute-force Ruby program I used to enumerate these:

(2..10)
  .to_a
  .permutation
  .map { |a, b, c, d, e, f, g, h, i| ["#{a}#{b}#{c}".to_i, "#{d}#{e}#{f}".to_i, "1#{g}#{h}#{i}".to_i] }
  .select { |a, b, c| a < b && a + b == c }
  .map { |a, b, c| "#{a} + #{b} = #{c}" }
$\endgroup$
  • $\begingroup$ Hmm I count 96 lines (commutativity is not duplication) ... and that fits the output of my program :-) Btw I think OP looks for a mathematical resolution, rather than a raw output ... Or at least show your algorithm! (well, I could have provided mine as well, but this is not stackoverflow, so not sure it qualifies!) $\endgroup$ – ringø Feb 17 '17 at 2:00
  • $\begingroup$ With $48$ answers to cover, a complete solution based on purely analytic techniques seems difficult. $\endgroup$ – Oscar Lanzi Feb 17 '17 at 2:02
  • $\begingroup$ @OscarLanzi well, 96 out of 10 billions possibilities is not so much... $\endgroup$ – ringø Feb 17 '17 at 2:05
  • $\begingroup$ @ringo not $10^{10}$. All digits must be distinct which removes most of the tuples; digit G has to be $1$; the other initial digits cannot be zero; and in my book two solutions that differ only by switching the addends are not really different. Thus I am left with only about $130,000$ candidates. $\endgroup$ – Oscar Lanzi Feb 17 '17 at 2:38
  • 1
    $\begingroup$ Conversely, this could be cut down to 12 truly unique solutions, annotated with digit-wise commutativity (e.g., 246 + 789 = 249 + 786 = 286 + 749 = 289 + 746). $\endgroup$ – Peregrine Rook Feb 18 '17 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy