# Another puzzle for who has long boring days

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.

Have fun!

I put some row/column codes onto the grid:

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):

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:

• Why can't 3,4 come in column 9? 3 in C9 and 4 in B9 will also satisfy right? – Chief A Apr 1 at 15:08
• @ChiefA Nope, there's already a 4 in row B. – Rand al'Thor Apr 1 at 15:11
• Ok my bad. I misread the conditions (・_・;) – Chief A Apr 1 at 15:15

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 

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