0
$\begingroup$

When I was in highschool I became obsessed with this strange combination I discovered. I don't really know a lot about math but I've recently re-discovered it and believe I have found an answer, but no real way to prove it. I'd like someone with a fair bit more knowledge than me to try at it.

You have 4 single digit number, 4 double digit numbers, and 2 triple digit numbers, demonstrated with "space filling" variables like so; X + X + X + X + XX + XX + XX + XX + XXX + XXX = SUM there are 18 X's above. You are permitted to use each digit from 1 to 9 exactly twice each, no more no less. No zeroes are allowed. They can be in any order, for example: 1 + 2 + 3 + 4 +56 +78 +91 +23 +456 +789 = 1503 Using this setup, how many different possible combinations of digits are there so that the SUM is equal to 1350? And can you prove there are no other possible solutions?

I believe there are 36 possible permutations that sum to 1350. If someone can mathematically prove that or disprove it by discovering additional solutions that I have not found I would be greatly appreciative.

$\endgroup$
8
  • 2
    $\begingroup$ This seems like a job for brute force via programming, not a job for pen&paper approaches. $\endgroup$
    – JMoravitz
    Commented Oct 20, 2018 at 1:31
  • 2
    $\begingroup$ Any particular reason for the choice of 1350? $\endgroup$
    – Sambo
    Commented Oct 20, 2018 at 1:35
  • 1
    $\begingroup$ There's definitely more than $36$ since it is possible to switch the digits of either the ones, tens, or hundredths columns around. $\endgroup$
    – Abraham Zhang
    Commented Oct 20, 2018 at 2:37
  • 1
    $\begingroup$ Do you mean that each set of numbers contains each digit from 1 to 9 exactly once? $\endgroup$
    – Fabio Somenzi
    Commented Oct 20, 2018 at 5:09
  • 1
    $\begingroup$ @Sagarmatha: You should put any clarifying remarks into the body of your question. Not everyone reads the comments. $\endgroup$
    – Blue
    Commented Oct 20, 2018 at 6:06

1 Answer 1

Reset to default
1
$\begingroup$

Here are the first 37 elements of a list of 456 that are obtained by imposing the constraint that ones, tens, and hundreds should appear in non-decreasing order from left to right. (As noted by Abraham Zhang in a comment, permuting digits of equal weight does not change the sum.) Many more combinations are possible if that constraint is removed or weakened.

Which elements (if any) of this list contradict your rules? If some do, why?

1 : 2 + 2 + 3 + 3 + 14 + 46 + 57 + 57 + 168 + 998 = 1350
2 : 1 + 1 + 2 + 3 + 43 + 54 + 65 + 86 + 297 + 798 = 1350
3 : 1 + 2 + 3 + 4 + 14 + 25 + 65 + 68 + 379 + 789 = 1350
4 : 2 + 2 + 3 + 4 + 14 + 15 + 65 + 67 + 389 + 789 = 1350
5 : 1 + 1 + 2 + 2 + 36 + 36 + 47 + 47 + 589 + 589 = 1350
6 : 1 + 1 + 2 + 2 + 36 + 36 + 47 + 48 + 578 + 599 = 1350
7 : 1 + 1 + 2 + 3 + 24 + 37 + 47 + 68 + 568 + 599 = 1350
8 : 1 + 1 + 2 + 3 + 24 + 36 + 47 + 68 + 579 + 589 = 1350
9 : 1 + 1 + 2 + 2 + 34 + 37 + 47 + 68 + 569 + 589 = 1350
10 : 1 + 1 + 2 + 4 + 25 + 35 + 47 + 68 + 368 + 799 = 1350
11 : 1 + 2 + 2 + 3 + 15 + 45 + 47 + 68 + 368 + 799 = 1350
12 : 1 + 1 + 2 + 3 + 25 + 45 + 47 + 68 + 369 + 789 = 1350
13 : 1 + 1 + 2 + 3 + 25 + 46 + 47 + 58 + 368 + 799 = 1350
14 : 1 + 2 + 2 + 4 + 15 + 35 + 46 + 68 + 378 + 799 = 1350
15 : 1 + 1 + 2 + 2 + 43 + 54 + 65 + 87 + 397 + 698 = 1350
16 : 1 + 1 + 2 + 2 + 33 + 64 + 65 + 87 + 497 + 598 = 1350
17 : 1 + 1 + 2 + 2 + 33 + 54 + 76 + 86 + 497 + 598 = 1350
18 : 1 + 1 + 2 + 2 + 34 + 54 + 75 + 86 + 397 + 698 = 1350
19 : 1 + 2 + 2 + 3 + 53 + 54 + 64 + 76 + 197 + 898 = 1350
20 : 1 + 2 + 4 + 4 + 25 + 35 + 36 + 67 + 178 + 998 = 1350
21 : 2 + 2 + 3 + 3 + 14 + 74 + 75 + 85 + 196 + 896 = 1350
22 : 1 + 3 + 3 + 4 + 14 + 26 + 26 + 77 + 598 + 598 = 1350
23 : 2 + 2 + 3 + 4 + 14 + 35 + 56 + 67 + 178 + 989 = 1350
24 : 1 + 2 + 3 + 4 + 24 + 35 + 57 + 67 + 168 + 989 = 1350
25 : 2 + 2 + 3 + 3 + 14 + 45 + 57 + 67 + 168 + 989 = 1350
26 : 1 + 1 + 2 + 2 + 53 + 64 + 74 + 75 + 389 + 689 = 1350
27 : 1 + 2 + 2 + 3 + 53 + 64 + 64 + 75 + 187 + 899 = 1350
28 : 1 + 1 + 2 + 3 + 53 + 64 + 64 + 75 + 288 + 799 = 1350
29 : 2 + 2 + 3 + 3 + 14 + 14 + 67 + 67 + 589 + 589 = 1350
30 : 1 + 1 + 3 + 4 + 24 + 35 + 57 + 67 + 269 + 889 = 1350
31 : 1 + 1 + 3 + 3 + 25 + 45 + 47 + 67 + 269 + 889 = 1350
32 : 1 + 1 + 3 + 4 + 25 + 35 + 46 + 67 + 279 + 889 = 1350
33 : 1 + 1 + 2 + 4 + 35 + 35 + 47 + 67 + 269 + 889 = 1350
34 : 1 + 3 + 3 + 4 + 25 + 25 + 46 + 76 + 178 + 989 = 1350
35 : 1 + 1 + 2 + 2 + 34 + 64 + 75 + 85 + 387 + 699 = 1350
36 : 1 + 1 + 3 + 3 + 24 + 26 + 47 + 67 + 589 + 589 = 1350
37 : 1 + 1 + 3 + 3 + 26 + 26 + 47 + 47 + 598 + 598 = 1350

EDIT: With the "original problem" described in a comment, in which

$$(a + bc + def + gh + i) + (i + hg + fed + cb + a) = 1350 \enspace,$$

the following are the first 37 of 480 solutions obtained by imposing the constraint that $a<i$, $c<h$, and $d<f$:

1 : 1 + 23 + 456 + 78 + 9 + 9 + 87 + 654 + 32 + 1 = 1350
2 : 6 + 14 + 397 + 25 + 8 + 8 + 52 + 793 + 41 + 6 = 1350
3 : 6 + 24 + 397 + 15 + 8 + 8 + 51 + 793 + 42 + 6 = 1350
4 : 7 + 64 + 318 + 25 + 9 + 9 + 52 + 813 + 46 + 7 = 1350
5 : 7 + 54 + 318 + 26 + 9 + 9 + 62 + 813 + 45 + 7 = 1350
6 : 7 + 24 + 318 + 65 + 9 + 9 + 56 + 813 + 42 + 7 = 1350
7 : 1 + 73 + 258 + 64 + 9 + 9 + 46 + 852 + 37 + 1 = 1350
8 : 2 + 84 + 367 + 15 + 9 + 9 + 51 + 763 + 48 + 2 = 1350
9 : 2 + 14 + 367 + 85 + 9 + 9 + 58 + 763 + 41 + 2 = 1350
10 : 8 + 63 + 427 + 15 + 9 + 9 + 51 + 724 + 36 + 8 = 1350
11 : 8 + 53 + 427 + 16 + 9 + 9 + 61 + 724 + 35 + 8 = 1350
12 : 8 + 35 + 427 + 16 + 9 + 9 + 61 + 724 + 53 + 8 = 1350
13 : 8 + 15 + 427 + 36 + 9 + 9 + 63 + 724 + 51 + 8 = 1350
14 : 8 + 61 + 427 + 53 + 9 + 9 + 35 + 724 + 16 + 8 = 1350
15 : 8 + 61 + 427 + 35 + 9 + 9 + 53 + 724 + 16 + 8 = 1350
16 : 8 + 51 + 427 + 63 + 9 + 9 + 36 + 724 + 15 + 8 = 1350
17 : 8 + 13 + 427 + 65 + 9 + 9 + 56 + 724 + 31 + 8 = 1350
18 : 8 + 31 + 427 + 65 + 9 + 9 + 56 + 724 + 13 + 8 = 1350
19 : 8 + 31 + 427 + 56 + 9 + 9 + 65 + 724 + 13 + 8 = 1350
20 : 8 + 51 + 427 + 36 + 9 + 9 + 63 + 724 + 15 + 8 = 1350
21 : 8 + 13 + 427 + 56 + 9 + 9 + 65 + 724 + 31 + 8 = 1350
22 : 6 + 12 + 397 + 54 + 8 + 8 + 45 + 793 + 21 + 6 = 1350
23 : 6 + 52 + 397 + 14 + 8 + 8 + 41 + 793 + 25 + 6 = 1350
24 : 6 + 42 + 397 + 15 + 8 + 8 + 51 + 793 + 24 + 6 = 1350
25 : 4 + 82 + 357 + 19 + 6 + 6 + 91 + 753 + 28 + 4 = 1350
26 : 4 + 92 + 357 + 18 + 6 + 6 + 81 + 753 + 29 + 4 = 1350
27 : 4 + 28 + 357 + 19 + 6 + 6 + 91 + 753 + 82 + 4 = 1350
28 : 4 + 25 + 387 + 16 + 9 + 9 + 61 + 783 + 52 + 4 = 1350
29 : 8 + 43 + 526 + 17 + 9 + 9 + 71 + 625 + 34 + 8 = 1350
30 : 8 + 34 + 526 + 17 + 9 + 9 + 71 + 625 + 43 + 8 = 1350
31 : 8 + 73 + 526 + 14 + 9 + 9 + 41 + 625 + 37 + 8 = 1350
32 : 7 + 42 + 516 + 38 + 9 + 9 + 83 + 615 + 24 + 7 = 1350
33 : 7 + 32 + 516 + 48 + 9 + 9 + 84 + 615 + 23 + 7 = 1350
34 : 7 + 23 + 516 + 48 + 9 + 9 + 84 + 615 + 32 + 7 = 1350
35 : 8 + 13 + 526 + 47 + 9 + 9 + 74 + 625 + 31 + 8 = 1350
36 : 8 + 14 + 526 + 37 + 9 + 9 + 73 + 625 + 41 + 8 = 1350
37 : 8 + 31 + 526 + 47 + 9 + 9 + 74 + 625 + 13 + 8 = 1350
$\endgroup$
6
  • $\begingroup$ I believe all of these permutations are within the guidelines, thank you! I'm going to take these answers and feed them back in to my method of solving and see what they look like. $\endgroup$
    – Sagarmatha
    Commented Oct 20, 2018 at 16:17
  • $\begingroup$ And for the record, my method of achieving the 36 permutations was from a more specific set. If you assign all nine digits a variable from a-i, and writing (a + bc + def + gh + i) + (i + hg + fed + cb + a), I figured there are only 36 ways of achieving 1350. I came here wondering if this series also applies to (X + X + X + X + XX + XX + XX + XX + XXX + XXX) in general, which is now clearly disproven. $\endgroup$
    – Sagarmatha
    Commented Oct 20, 2018 at 16:21
  • $\begingroup$ To follow up to your new list, I've noticed some permutations are simply re-arrangements of other permutations that have the same addition. For example, #29 in your new list is simply #28 but with the triple digits flipped around. The order of these additions really doesn't matter, only their actual value and their sum. Regardless, #28 and obviously many other permutations on that list are permutations I had not found, so that list still disproves my 36 answers theory. Thank you for bringing closure to this problem for me! $\endgroup$
    – Sagarmatha
    Commented Oct 21, 2018 at 17:01
  • $\begingroup$ You're welcome. Imposing the constraint $d < f$ eliminates those "duplicates" and--predictably--cuts the number of solutions in half, just like the other two constraints. I'll update my answer accordingly. $\endgroup$
    – Fabio Somenzi
    Commented Oct 21, 2018 at 19:34
  • $\begingroup$ So are the 480 permutations all possible methods given the current constraints of the problem or is it only a sample? $\endgroup$
    – Sagarmatha
    Commented Oct 22, 2018 at 3:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .