5
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The rules:

  1. each row must have the digits from 0 to 9.
  2. The sum of the digits in the column is shown at the bottom of the column.
  3. The same number can appear more than once in a column.
  4. Adjacent boxes cannot contain the same number, including diagonally adjacent.

Have fun!

enter image description here

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4
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I put some row/column codes onto the grid:

enter image description here

Column 9 is easy to solve:

we have two numbers summing to 7; it can't be $2,5$ or $3,4$ or $0,7$, so it must be $1,6$, which means B9 is $6$ and C9 is $1$.

Now consider column 6.

We have four numbers summing to 8. None of them can be $1$ or $4$, at most one of them (D6) can be $2$, and also that one must be at least $2$.

Assume D6 is $3$; then C6 must be $0$ and we have two numbers summing to 5 which can't be $2,3$ or $1,4$ or $0,5$. Contradiction, so D6 is $2$. We're left with two numbers summing to 6, and we can't use any of $1,2,4$; the only possibilities are $6,0,0$ or $3,3,0$. We don't know which one of these, but if there's a $6$ it must be in C6.

In column 7,

we have two numbers summing to 13; A7 must be one of $6,8$ and E7 must be one of $7,5$.

Now consider column 0.

The bottom cell F0 must be one of $1,5$. If it's $1$, then A0 and D0 are both $9$. If it's $5$, then the other two sum to 14, and D0 can't be $6,7,8$, so they must be $9$ and $5$ in some order.

So far we have the following (numbers on the right of a cell are certain, numbers to the left are a list of possibilities for that cell):

enter image description here

Consider column 8:

four numbers summing to 21. We cannot have any of $1,2,4,6$, but there must be at least one even number, so it must be either $0$ or $8$.
If there is $0$, the others are three odd numbers summing to 21, which must be $9,9,3$ (in which case A8 to D8 are $0,9,3,9$ in order) or $9,7,5$ (in which case A8 to D8 are $5,9,0,7$ in order).
If there is $8$, the others are three odd numbers (not $1$) summing to 13, which must be $7,3,3$ (impossible) or or $5,5,3$ (impossible).
In either case, B8 is $9$ and A0 is also $9$.

Column 1 has

three numbers summing to 18, none of which can be $2$. That's $9,9,0$ (impossible) or $9,8,1$ or $9,6,3$ or $9,5,4$ or $8,7,3$ or $8,6,4$ or $8,5,5$ or $7,7,4$ (impossible) or $7,6,5$. Too many possibilities!

Everywhere I look now seems to have too many possibilities - even why I try to do a What-If and assume something hoping to get a contradiction. So I'm going to post my progress so far and leave this as a partial solution for now:

enter image description here

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  • $\begingroup$ Why can't 3,4 come in column 9? 3 in C9 and 4 in B9 will also satisfy right? $\endgroup$ – Chief A Apr 1 at 15:08
  • $\begingroup$ @ChiefA Nope, there's already a 4 in row B. $\endgroup$ – Rand al'Thor Apr 1 at 15:11
  • $\begingroup$ Ok my bad. I misread the conditions (・_・;) $\endgroup$ – Chief A Apr 1 at 15:15
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I could not solve this with deduction. So I wrote a computer program,

Here is the solution. 7 2 4 0 6 1 8 5 3 9 8 1 7 3 5 0 4 9 6 2 2 3 4 8 9 6 5 0 1 7 0 8 5 1 3 2 4 7 6 9 3 9 7 6 4 0 5 1 2 8 2 0 4 8 5 9 6 3 7 1

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  • 1
    $\begingroup$ You have 2 5s diagonally adjacent (Row2-Column6 and Row3-Column7) and 2 6s diagonally adjacent (Row1-Column8 and Row2-Column9), which is prohibited $\endgroup$ – Chronocidal Apr 8 at 16:50
  • $\begingroup$ @Chronocidal thanks for noting! fixed the program and edited the answer. $\endgroup$ – daw Apr 8 at 18:19
  • $\begingroup$ Congratulations ! That was really impossible without a cmputer program... $\endgroup$ – perayu Apr 9 at 18:45

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