1
$\begingroup$

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

Source: Puzzle Page app

$\endgroup$
3
$\begingroup$

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 $3 + 9 + 3 + 8 + 4$.

$\endgroup$
  • $\begingroup$ Wow, good job, thank you! $\endgroup$ – Olga May 9 at 17:32
5
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Apr 30 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 May 1 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.