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Can someone solve this? Please tell me how you did it.enter image description here

Source: Puzzle Page app

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3 Answers 3

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Without guessing:

Take a look at all the cells in the bottom two rows. Their sum must be $39 + 12 + 17 = 68$. The sum of the eight cells in the corners is $17 + 9 + 17 + 15 = 58$. This makes the sum of the two middle cells $68 - 58 = 10$ which only leaves $4$ and $6$ as possibilities there which quickly solves both corners.

Then

The middle $17$ has to have the $9$ on the left because the downwards $20$ would otherwise have to be $8 + 8 + 4$ or $6 + 8 + 6$.

Finally

The first cell of the across $24$ has to be a $7$ because otherwise the downwards $27$ would have to be $2 + 9 + 4 + 8 + 4$ or $3 + 9 + 3 + 8 + 4$.

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If

We put $6$ below $15$ clue

Then

Below $6$ is $9$, it's left will be $8$, it's up will be $9$

Also note that

We must put $5$ below $9$ clue as $6$ is used on the row

Thus

Below $5$ is $4$, it's left will be $8$, it's up will be $9$

Finally

There will be double $9$ on the row, hence don't put $6$ below $15$ clue, put $7$ instead

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I play this game too! I think athin's answer gets you the bottom-right 2x2 and the bottom-left 2x2. I did a similar guess-and-solve to figure the rest. (The undo feature helps this, since it preserves your pencil marks.)

The 6-long row at the bottom has 4 and 6 remaining in the center two squares. I guessed 4 in the left one (in the 20 column) and solved from there until I found a contradiction. So I put a 6 there instead (so the row becomes 9 5 6 4 8 7) and solved the 20 column.

Then

The 27 column becomes ? ? ? 8 4. That middle blank should have 7 or 9 penciled in. 8 + 4 is 12, leaving 15 for the three squares. If you fill in a 9, 15 - 9 = 6 and there's no way to make 6 with the remaining pencil marks, so that one must be a 7. The rest solves accordingly.

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  • $\begingroup$ Thank you for the answers! But isn’t there a way to do it without guessing? I have tried for hours to add numbers together to figure out if a certain row or column must/can’t contain a certain number, but I couldn’t narrow it down more than I have 😅 $\endgroup$
    – Olga
    Commented Apr 30, 2019 at 18:25
  • $\begingroup$ Typically there is, this puzzle as pretty unusual. Perhaps there's an advanced technique to eliminate one of the numbers but guessing as the only way I saw it. $\endgroup$
    – Somebody
    Commented May 1, 2019 at 16:21

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