4
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Sudoku friends, this one's for you. This maths puzzle was set for eight-year-old students in Bao Loc, Vietnam, and they have actually been able to solve the puzzle. Even some people with doctorate in mathematics have been stumped. Nobody in France nor Belgium has been able to solve this to date. Let us try.

You just need to fill in each of the digits from 1 to 9 so that the grid makes sense. It is unknown if we can use the digits once only or as much as necessary.

Here it is.

Puzzle

Can you solve it?

I spent several minutes trying to find a common or hidden pattern, and eventually noticed that in the third row of the grid the numbers are in descending order: 13, 12, 11, 10. Now, their sum is 46, and the only number written in the first row is 66. Would a similar pattern to 66, 46, 26... help in something?

Additionally, someone offered this comment:

$\bigcirc+13\times\bigcirc:\bigcirc+\bigcirc+12\times\bigcirc-\bigcirc-11+\bigcirc\times\bigcirc:\bigcirc-10=66$

$X+(13\times1:1)+Y+(12\times1)-Z-11+(1\times1:1)-10=66$

$X+Y-Z+13+12-11+1-10=66$

$X+Y-Z-5=66$

$X+Y-Z=71$

(P.S. Please do not consider the $\bigcirc$ as a binary operation symbol in the equation above, but as a blank space to represent the empty squares in the grid.)

It looks like it can help with something, but this would perhaps work if there were not the 1 to 9 number limit, as $X+Y-Z$ cannot result to $71$ (being replace with digits from 1 to 9).

Thank you.

Source

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  • 1
    $\begingroup$ Just clarifying, the colons ":" are division? Also, do we do each action in order, or do we still follow order of operations? $\endgroup$ – JonTheMon May 21 '15 at 13:39
  • $\begingroup$ Yes, they represent division as $\div$ does. As instructions concerning this are not given, I think we may choose which suits us best. $\endgroup$ – Veo May 21 '15 at 13:40
  • $\begingroup$ I would consider removing the first source, as it appears to have an answer in the comments... $\endgroup$ – Bailey M May 21 '15 at 13:41
  • 3
    $\begingroup$ Um, why the downvotes? Isn't this a valid puzzle? +1 from me - looks fun! $\endgroup$ – Rand al'Thor May 21 '15 at 17:32
  • 3
    $\begingroup$ The downvotes must have come from France or Belgium, probably from people who solved it many years ago and are sick of hearing the lies. $\endgroup$ – Kingrames Sep 23 '15 at 18:44
5
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My answer is

6 9 3 5 2 1 7 8 4
enter image description here

And I follow the regular order of operations (first multiplications, then sums).

$6+13\times9/3+5+12\times2-1-11+7\times8/4-10=$
$6+39+5+24-1-11+14-10=$
$66$

Few considerations:

Assuming that the result of any operation is integer (8 year old children usually don't know reals, do they?), A+13*B/C implies that C divides (13*B), so C divides B. This excludes many solutions, leaving us with (9,1),(9,3),(8,4),(8,2),(8,1),(7,1),(6,1),(6,2),(6,3),(5,1),(4,1),(4,2),(3,1),(2,1) instead of 72 couples for B and C!
Anyway, there are many different valid combinations that fit this problem, I could write a program that does the heavy work for us, but what would the point of it? 8 year old children don't have C++! Or maybe they have, since they're asians (joking, of course)?

Additional answers:

5,2,1,3,4,7,9,8,6
5,9,3,6,2,1,7,8,4

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  • 3
    $\begingroup$ You beat me for time but feel free to use this: i.stack.imgur.com/U9WmB.png $\endgroup$ – Engineer Toast May 21 '15 at 13:57
  • $\begingroup$ Great. However, may I ask how you got the numbers at the first place? Thank you. $\endgroup$ – Veo May 21 '15 at 18:02
  • $\begingroup$ Just semi-random attempts, I tried to think as children since they were able to solve it! $\endgroup$ – leoll2 May 21 '15 at 18:04
5
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Time for the brute force computing brigade to pick up the remaining solutions.

Assuming:

  • each number must be used exactly once
  • division must result in a whole number

If we drop either of these assumptions, then this answer would be about 100 times longer.

If order of operations is respected, according to my script the possible answers are

3, 2, 1, 5, 4, 7, 8, 9, 6
3, 2, 1, 5, 4, 7, 9, 8, 6
5, 2, 1, 3, 4, 7, 8, 9, 6
5, 2, 1, 3, 4, 7, 9, 8, 6
5, 3, 1, 7, 2, 6, 8, 9, 4
5, 3, 1, 7, 2, 6, 9, 8, 4
5, 4, 1, 9, 2, 7, 3, 8, 6
5, 4, 1, 9, 2, 7, 8, 3, 6
5, 9, 3, 6, 2, 1, 7, 8, 4
5, 9, 3, 6, 2, 1, 8, 7, 4
6, 3, 1, 9, 2, 5, 7, 8, 4
6, 3, 1, 9, 2, 5, 8, 7, 4
6, 9, 3, 5, 2, 1, 7, 8, 4
6, 9, 3, 5, 2, 1, 8, 7, 4
7, 3, 1, 5, 2, 6, 8, 9, 4
7, 3, 1, 5, 2, 6, 9, 8, 4
9, 3, 1, 6, 2, 5, 7, 8, 4
9, 3, 1, 6, 2, 5, 8, 7, 4
9, 4, 1, 5, 2, 7, 3, 8, 6
9, 4, 1, 5, 2, 7, 8, 3, 6

If the equation is evaluated left-to-right, ignoring order of operations, the possible answers are

1, 2, 4, 7, 5, 8, 3, 6, 9
1, 2, 7, 5, 3, 4, 9, 8, 6
1, 2, 7, 5, 8, 9, 4, 3, 6
1, 3, 7, 5, 2, 6, 9, 8, 4
1, 4, 2, 8, 5, 7, 6, 3, 9
1, 4, 7, 5, 2, 9, 8, 6, 3
1, 4, 7, 5, 3, 9, 2, 8, 6
1, 5, 7, 4, 9, 3, 8, 2, 6
1, 6, 7, 8, 2, 5, 9, 4, 3
1, 8, 4, 7, 5, 2, 6, 3, 9
1, 8, 7, 2, 6, 3, 5, 4, 9
1, 9, 3, 7, 2, 5, 8, 4, 6
1, 9, 6, 2, 3, 7, 8, 4, 5
1, 9, 7, 2, 5, 3, 6, 4, 8
1, 9, 7, 3, 5, 8, 6, 2, 4
1, 9, 7, 5, 2, 8, 6, 4, 3
2, 1, 3, 6, 8, 7, 5, 4, 9
2, 1, 5, 3, 7, 9, 8, 4, 6
2, 1, 5, 3, 9, 6, 7, 4, 8
2, 1, 5, 8, 6, 3, 9, 4, 7
2, 3, 5, 6, 7, 8, 1, 4, 9
2, 6, 3, 5, 1, 7, 9, 8, 4
2, 6, 3, 9, 1, 7, 5, 8, 4
2, 6, 5, 4, 7, 8, 9, 1, 3
2, 6, 9, 1, 7, 3, 5, 4, 8
2, 7, 3, 5, 1, 9, 6, 8, 4
2, 7, 3, 6, 1, 9, 5, 8, 4
2, 8, 4, 5, 1, 7, 9, 6, 3
2, 8, 4, 9, 1, 7, 5, 6, 3
2, 9, 5, 6, 1, 3, 7, 8, 4
2, 9, 5, 7, 1, 3, 6, 8, 4
3, 1, 4, 8, 5, 2, 7, 6, 9
3, 1, 4, 9, 8, 6, 7, 2, 5
3, 4, 8, 7, 9, 5, 1, 2, 6
3, 5, 2, 7, 8, 9, 4, 1, 6
3, 5, 4, 1, 2, 7, 9, 8, 6
3, 6, 4, 8, 9, 7, 2, 1, 5
3, 6, 8, 7, 5, 1, 9, 2, 4
3, 7, 2, 4, 1, 9, 5, 8, 6
3, 7, 2, 5, 1, 9, 4, 8, 6
3, 7, 8, 5, 4, 1, 2, 6, 9
3, 9, 4, 6, 5, 1, 8, 2, 7
3, 9, 6, 5, 4, 8, 7, 1, 2
4, 5, 1, 7, 3, 6, 9, 2, 8
4, 6, 1, 5, 2, 7, 8, 3, 9
4, 6, 1, 7, 2, 8, 5, 3, 9
4, 6, 3, 9, 7, 2, 8, 1, 5
4, 9, 3, 2, 6, 7, 8, 1, 5
5, 1, 3, 9, 6, 7, 8, 2, 4
5, 1, 6, 2, 3, 9, 7, 8, 4
5, 1, 6, 3, 9, 7, 8, 2, 4
5, 1, 6, 9, 7, 8, 3, 2, 4
5, 1, 9, 2, 3, 6, 7, 8, 4
5, 1, 9, 2, 4, 3, 7, 8, 6
5, 1, 9, 3, 7, 2, 8, 4, 6
5, 2, 4, 1, 3, 7, 9, 8, 6
5, 2, 4, 9, 8, 7, 6, 1, 3
5, 3, 2, 7, 1, 6, 9, 8, 4
5, 3, 2, 9, 1, 6, 7, 8, 4
5, 4, 1, 9, 3, 8, 6, 2, 7
5, 4, 3, 9, 6, 1, 8, 2, 7
5, 4, 9, 1, 3, 2, 7, 8, 6
5, 4, 9, 1, 6, 8, 7, 2, 3
5, 4, 9, 1, 8, 7, 2, 3, 6
5, 4, 9, 3, 7, 6, 8, 1, 2
5, 4, 9, 6, 1, 3, 7, 8, 2
5, 4, 9, 7, 1, 3, 6, 8, 2
5, 6, 9, 3, 1, 4, 7, 8, 2
5, 6, 9, 7, 1, 4, 3, 8, 2
5, 7, 6, 9, 4, 8, 3, 1, 2
5, 7, 9, 1, 6, 2, 3, 4, 8
5, 8, 3, 1, 2, 4, 7, 6, 9
5, 8, 6, 7, 1, 3, 9, 4, 2
5, 8, 6, 9, 1, 3, 7, 4, 2
6, 2, 1, 5, 3, 9, 7, 4, 8
6, 8, 4, 5, 3, 9, 7, 1, 2
6, 9, 3, 1, 4, 8, 5, 2, 7
7, 1, 2, 4, 5, 8, 3, 6, 9
7, 1, 2, 8, 6, 3, 5, 4, 9
7, 1, 4, 9, 2, 8, 5, 6, 3
7, 2, 1, 8, 4, 6, 5, 3, 9
7, 2, 5, 1, 3, 4, 9, 8, 6
7, 2, 5, 1, 8, 9, 4, 3, 6
7, 2, 8, 1, 9, 4, 5, 3, 6
7, 3, 1, 9, 2, 5, 6, 4, 8
7, 3, 4, 9, 2, 5, 1, 8, 6
7, 3, 6, 4, 1, 5, 9, 8, 2
7, 3, 6, 9, 1, 5, 4, 8, 2
7, 4, 8, 5, 9, 6, 2, 1, 3
7, 6, 4, 2, 9, 8, 3, 1, 5
7, 6, 8, 9, 2, 5, 1, 4, 3
7, 9, 4, 8, 6, 2, 3, 1, 5
7, 9, 5, 3, 1, 8, 6, 4, 2
7, 9, 5, 6, 1, 8, 3, 4, 2
7, 9, 5, 6, 3, 1, 2, 4, 8
7, 9, 5, 8, 3, 6, 1, 2, 4
8, 1, 3, 9, 5, 2, 6, 4, 7
8, 1, 7, 3, 9, 5, 6, 2, 4
8, 1, 7, 5, 6, 4, 9, 2, 3
8, 1, 7, 9, 2, 4, 5, 6, 3
8, 2, 1, 6, 3, 5, 7, 4, 9
8, 2, 1, 7, 3, 6, 5, 4, 9
8, 3, 9, 5, 7, 6, 1, 2, 4
8, 4, 3, 9, 5, 7, 1, 2, 6
8, 4, 6, 5, 9, 3, 1, 2, 7
8, 4, 7, 3, 9, 5, 1, 2, 6
8, 6, 9, 3, 5, 2, 1, 4, 7
8, 6, 9, 5, 2, 1, 7, 4, 3
8, 6, 9, 7, 5, 3, 1, 2, 4
8, 7, 3, 1, 2, 4, 5, 6, 9
8, 9, 7, 1, 4, 2, 5, 3, 6
9, 2, 4, 7, 8, 6, 5, 1, 3
9, 3, 2, 4, 8, 7, 6, 1, 5
9, 3, 2, 5, 1, 7, 6, 8, 4
9, 3, 2, 6, 1, 7, 5, 8, 4
9, 3, 2, 7, 6, 5, 8, 1, 4
9, 4, 8, 6, 7, 3, 1, 2, 5
9, 6, 3, 8, 5, 7, 2, 1, 4
9, 7, 2, 8, 4, 3, 6, 1, 5
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  • $\begingroup$ I wrote a python script to solve this and I get a lot more answers than what you have and I'm following order of operations. This is my calculation x = a+((13*b)/c)+d+(12*e)-f-11+((g*h)/i)-10 which gives me 128 unique solutions. I don't see a problem in mine but I might be missing something $\endgroup$ – SirParselot Sep 23 '15 at 18:03
  • $\begingroup$ Oh i figured it out. It has to do with the fact that g*h is the same as h*g but my code thinks they are different since they are in a different order $\endgroup$ – SirParselot Sep 23 '15 at 18:19
  • $\begingroup$ I get 272 solutions (in about 5 milliseconds) in Java on an i7-7500U, without reusing digits, counting duplicates that would otherwise be eliminated due to commutation. (Running on just one core, no attempt to parallelize it.) $\endgroup$ – David Conrad Apr 14 '17 at 17:09
2
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1 2 6 4 7 8 3 5 9

An article with this problem in was passed around my office yesterday

solution

Image thanks to Engineer Toast

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  • 1
    $\begingroup$ And here's a picture for you: i.stack.imgur.com/udmzE.png $\endgroup$ – Engineer Toast May 21 '15 at 13:59
  • 1
    $\begingroup$ This one is interesting because 13 * 2 / 6 isn't an integer, but you end up with the right answer anyway. $\endgroup$ – Kevin May 21 '15 at 14:10
  • $\begingroup$ @Kevin I find 16 solutions where the intermediate terms are not always integral, in addition to the 272 solutions where they are. $\endgroup$ – David Conrad Apr 14 '17 at 17:23
1
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Chucking it into a computer, it seems that there are 119 solutions if you evaluate each sum in order as you go round the snake. The first few are:

7 3 4 9 2 5 1 8 6

7 6 8 9 2 5 1 4 3

1 4 7 5 3 9 2 8 6

5 6 9 7 1 4 3 8 2

7 9 5 6 1 8 3 4 2

7 3 6 9 1 5 4 8 2

3 7 2 5 1 9 4 8 6

8 1 7 9 2 4 5 6 3

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0
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Here are my answers:

$5 + 13 \times 4/9 + 6 + 12 \times 1 - 3 - 11 + 7 \times 8/2 - 10 = 66$

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