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Very Nice! I did indeed find a fully logical path. No guesses needed.
>!I solved it using a combination of methods. With the total of all 5 rows set at 75 and primes known you can come up with a list of primes, then strike 23 because the c-shape doesn’t allow 23 even though a 5-square could get to 23 in a different shape. Then sum the primes and strike 2. The two-cell box is then 3, with a 1 and 2. Now figure out where the 5 box can be. Must be the 7-shaped one in the middle. Only one way to place the 2,1,2 to get 5. And you’re on your way…There is a neat bit of reasoning on a sum to 8 placing a 4 somewhere else.!<
Thank you!
>!Yep, you can deduce that the middle 3-cage is 5! Maybe you were being brief with your comment and already dealt with this, but 5 could also be 1+3+1 instead of 2+1+2. In my "intended solution path" one first needs to figure out that the bottom left 4-cage sums to 7, and use that to figure out the placements of 1s and 2s in the 2-cage and central 3-cage. Did you "disambiguate" the positions of the 1 and 2 in the 2-cage some other way?!<
That's how I did it and I must admit I had too much fun tonight. Took me a good 20 minutes but I was watching hockey.
Thank you for designing such an intriguing puzzle.
Superb! The way I came to that same conclusion >!was by way of figuring out that the 3 cell box with the 113/212 cage made it impossible to have 1s anywhere along the lines that cross it, making the bottom right box disambiguate from its possibilities of being both 7 or 17 as it would definitely need a repeating 1 to sum to 7 (both my two bottom boxes were able to sum to both 7 and 17 in the way I solved it) so that made the bottom left a 7 sum box!<
Right after that, everything unfolded beautifully.
Yeah - I realise that now.
>!But you can use the sum of a column to tie down the bottom-left box. Then the sum to 7 in the z-shaped box to prove where the 1s and 2s go. Which is what I did next, so the solution still worked out. But if luck on my part, but it can be reached by logic.!<
38 minutes, 15+ of which were me trying to think of any possible way to get into the puzzle after figuring out what the left side had to be.
But I eventually figured out >!all the possible primes were on the board, 2 is impossible, so there were exactly 7 in total and 7 boxes. Just needed to find which box was which prime, and everything fell into place from there.!<
Nice puzzle! I was surprised it was indeed solvable logically!
First off, let's name the pieces to make it easier to refer to them: the domino (2 squares), the inner and outer hooks (the 3-square pieces), and the C, S, T and I (like in Tetris, except the C).
>!The total value of all the numbers is (1+2+3+4+5)×5 = 75. If each cage sums to a distinct prime number, then it can't include 2 (since an even plus six odds is even, and 75 is odd), and the seven smallest other primes are 3+5+7+11+13+17+19 = 75 exactly (since they're the smallest, anything else would be larger), so those must be the sums of the cages.!<
>!Looking at what sums can go in what cages, remember that squares in the same row or column must have distinct numbers. This means the domino can have anything from 1+2=3 to 4+5=9, the hooks 1+2+1 = 4 to 5+4+5 = 14, the I 1+2+3+4 = 10 to 2+3+4+5 = 14, the T 1+2+3+1 = 7 to 5+4+3+5 = 17, the S 1+2+1+2 = 6 to 5+4+5+4 = 18, and the C 1+2+3+1+2 = 9 to 5+4+3+5+4 = 21.!<
>!Some of the primes from before immediately fit into certain ranges; 3 can only be the domino and 19 the C, 17 must be either the T or S, and 5 can only be a hook from what's left.!<
>!Then we can start filling in the puzzle; the domino with 3 can only be filled with 1 and 2. This stops the outer hook from being the one with 5 (since two of its squares can't have a 1 or 2), so it must be the inner hook (filled with either 212 or 131). The other hook has at minimum a 3 and 4 in the right column and a 2 in the remaining square (the inner hook has taken the 1), meaning it has a minimum of 9. That means it can't have the 7, which can only otherwise go in T or S; since one has 7 and the other 17, the I and outer hook must have the remaining two of 13 and 11.!<
>!Since the T must have at least a 3 in the right, and can't have a 1 in the middle due to the inner hook having the 1, it can't total 7; that means it must be 17, and the S is 7. The only way the S can have 7 is 1-2-1-3, which means it has the 1 in the bottom two columns; that lets us collapse several choices in the domino and inner hook. The domino has 2 on bottom, 1 top, and the inner hook is 212. We also must have 13 on the upper half of the S (since the 2 in that row is in the domino) and 12 in the bottom.!<
>!Since 12 is taken in the bottom row, the bottom of the T must have 345, meaning the top bit must have 5 to total 17, and the remaining square in that column on the left must be 4. In order for the I to total 11 or 13, we must leave the 2 or 4 out of 15 in the remaining cell; 4 is not available in that cell, so we can have 13 in the I and 11 in the outer hook, and collapse the rest of the S.!<
>!Since we have 1 and 3 in the S, the left side of the C must have 245 in some order, but the top row is the only one that has a 2 free, leaving 45 in the other two. This totals 11, so the other two cells in the C must total 8 to make 19, which can only be 35; this eliminates some possibilities in the T, letting you narrow down its bottom row to 435 left to right.!<
>!With 125 established in the right column, the other two cells in the outer hook must have 34; these sum up to 7, so making 11 requires a 4 in the last cell, forcing the other 4 into the lower column. This causes a chain reaction that collapses the rest of the C, and we can fill in the rest of the I as only one is left in each row.!<
So the final solution is left to right, top to bottom >!12543 35214 54321 43152 21435!<.
Nice, fun one. Was weird using duplicate digits in a cage.
I kind of half and halfed figuring which cages go where while putting in digits. The final slotting in of everything was very satisfying.
>! First is identifying the primes that can be made in these cages. 3, 5,7,11,13,17 and 19. 19 has to go in the five cage as there are no four cages that could support three fives. 3 has to go in the two cage as none of the three cages can support three ones.!<
>!Top left four cage has to be 11 or 13 due to its shape, bottom left corner cell thus has to be a two or a four. !<
>!The only way for the bottom left four cell to be the 17 cage would be to have two fives, which can't be done now as the bottom left cell needs to be two or four, and the 19 cage needs at least one five to sum to 19. This means the bottom right cage must be the 17 cage. This can again only be made with two fives, one of which must be placed in the top cell of the cage- first placed digit!!<
>!This then places two and one in the first two cells of row five, disambiguating the top left four cage as the 13 cage. !<
>!The central three cage must now be the 5 cage as the top right cage needs to have two cells be three four or five. Now the bottom left four cage must be the 7 cage as it cannot contain a five or two fours to make it the 11. Top right three cage becomes the 11 cage, and then it's just a little sudoku to fill out the digits!!<
>!!<
Very cute, thanks for the fun after work puzzle OP
Oh I had fun with this, and even though I’m late to the party I need to share my logic because it was very unnecessarily complicated but also very fun.
>!I made a crucial mistake right away that made solving the puzzle much harder: I counted six cages instead of seven, so I assumed one of the primes between 3-19 was missing. Obviously I could have noticed that six odd numbers sum to an even number, and the grid’s total of 75 is odd, but I didn’t think to check that either for some reason.!<
>!Consider which primes any 4-cell cage can sum to: it must be 7, 11, 13, or 17 (notice that 19 is impossible because you cannot fit three 5s in 4 connected cells). But also notice that the Z-cage and the T-cage together sum to 15 (the entire bottom row) + an odd number, either 1/3/5 or 2/4/(1,3,5). The possible sums here are 22 (15+1/2/4), 24 (15+1/3/5 or 2/3/4), or 26 (15+2/4/5). Comparing this with the possible sums of two of (7,11,13,17) makes 24 the only possible sum for these two cages: 7+17 or 11+13.!<
>!But now notice that the I-cage on the left must sum to 11 or 13! You can’t have three different cages summing to only two primes, so the T-cage and Z-cage must sum to some combination of 7 and 17.!<
>!Next, pay attention to the two 3-cell cages: now that they can no longer sum to 7, they must sum to 5, 11, or 13. That forms a triple with the I-cage we just discussed; those three prime sums are taken by those three cages, in some combination. At long last, I have shown that 5 and 7 must be used elsewhere, and so I have in the most roundabout way possible proven that the 2-cell cage must sum to 3.!<
>!Finally, to start putting numbers in the grid, pay attention to the parity of various cells in the I-, T-, Z-, and 2-cages. A cursory glance at the I-cage shows that the bottom left corner must be even, and therefore the remaining 3 cells in the Z-cage must be odd (since three evens and one odd cannot sum to 7 or 17 without three 2s or three 4s, which are both impossible). We also know by the same logic that the T-cage contains three odds and one even, and we furthermore know that the even number in that cage must be in the bottom row, so we have now successfully placed all three odds numbers in row 4. Put the 2 in column 5 since we know that cell sums to 3, the 4 in column 1 by process of elimination, and the 2 in the bottom-left corner again by process of elimination, and now you know the Z-cage sums to 7; the remaining three digits are 1/1/3. Fill those in, put the 5 in the T-cage in row 4 and the 1 in the 2-cage, and you’re off to the races.!<
Discussion: OP, I LOVED this concept! You should [submit this to CtC!](https://crackingthecryptic.com/#submissions)
[Here's a video of it being solved, which I hope you enjoy.](https://www.twitch.tv/videos/2132681582?t=0h30m13s) (I suggested it to him, and he told me today he thought it was really good!)
This was a cool puzzle.
>! I decided not to approach it mathematically and just come up with a pattern. Once I put the top row and the right column in I found the “cage” in the top right and the 2 space “cage” below both worked. I just continued the pattern and solved it. Took me just over 2 mins. I think it would have taken longer with no mention of a pattern.!<
Lots of fun thanks!
|>!1!<|>!2!<|>!5!<|>!4!<|>!3!<|
|:-|:-|:-|:-|:-|
|>!3!<|>!5!<|>!2!<|>!1!<|>!4!<|
|>!5!<|>!4!<|>!3!<|>!2!<|>!1!<|
|>!4!<|>!3!<|>!1!<|>!5!<|>!2!<|
|>!2!<|>!1!<|>!4!<|>!3!<|>!5!<|
Did bindings on the possible values for each based on size, then had ways to assign such that each section had a distinct prime. Removed number sizes that were impossible to reach based on arrangement, ended with 16. The >!3 and 5!!17!!7!<- which in turn let me determine the size of the size >!13!< cage. Top right and the 5-square one took longer, even though I knew the 5-square one was size >!19!
Please remember to spoiler-tag all guesses, like so: New Reddit: https://i.imgur.com/SWHRR9M.jpg Using markdown editor or old Reddit, draw a bunny and fill its head with secrets: \>!!< which ends up becoming \>!spoiler text between these symbols!< Try to avoid leading or trailing spaces. These will break the spoiler for some users (such as those using old.reddit.com) If your comment does not contain a guess, include the word **"discussion"** or **"question"** in your comment instead of using a spoiler tag. If your comment uses an image as the answer (such as solving a maze, etc) you can include the word "image" instead of using a spoiler tag. Please report any answers that are not properly spoiler-tagged. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*
Very Nice! I did indeed find a fully logical path. No guesses needed. >!I solved it using a combination of methods. With the total of all 5 rows set at 75 and primes known you can come up with a list of primes, then strike 23 because the c-shape doesn’t allow 23 even though a 5-square could get to 23 in a different shape. Then sum the primes and strike 2. The two-cell box is then 3, with a 1 and 2. Now figure out where the 5 box can be. Must be the 7-shaped one in the middle. Only one way to place the 2,1,2 to get 5. And you’re on your way…There is a neat bit of reasoning on a sum to 8 placing a 4 somewhere else.!<
Thank you! >!Yep, you can deduce that the middle 3-cage is 5! Maybe you were being brief with your comment and already dealt with this, but 5 could also be 1+3+1 instead of 2+1+2. In my "intended solution path" one first needs to figure out that the bottom left 4-cage sums to 7, and use that to figure out the placements of 1s and 2s in the 2-cage and central 3-cage. Did you "disambiguate" the positions of the 1 and 2 in the 2-cage some other way?!<
That's how I did it and I must admit I had too much fun tonight. Took me a good 20 minutes but I was watching hockey. Thank you for designing such an intriguing puzzle.
Thank you! Hope hockey was good haha
Not as great at the puzzle ;)
Superb! The way I came to that same conclusion >!was by way of figuring out that the 3 cell box with the 113/212 cage made it impossible to have 1s anywhere along the lines that cross it, making the bottom right box disambiguate from its possibilities of being both 7 or 17 as it would definitely need a repeating 1 to sum to 7 (both my two bottom boxes were able to sum to both 7 and 17 in the way I solved it) so that made the bottom left a 7 sum box!< Right after that, everything unfolded beautifully.
Nope. I mucked it up. Though I did do the bottom left cage as you describe.
Small flaw in your logic there >!5 can also be made with 1 3 1!<
Yeah - I realise that now. >!But you can use the sum of a column to tie down the bottom-left box. Then the sum to 7 in the z-shaped box to prove where the 1s and 2s go. Which is what I did next, so the solution still worked out. But if luck on my part, but it can be reached by logic.!<
This is such a brilliant style of logic that probably would have taken me far too long to realise. Bravo.
38 minutes, 15+ of which were me trying to think of any possible way to get into the puzzle after figuring out what the left side had to be. But I eventually figured out >!all the possible primes were on the board, 2 is impossible, so there were exactly 7 in total and 7 boxes. Just needed to find which box was which prime, and everything fell into place from there.!< Nice puzzle! I was surprised it was indeed solvable logically!
First off, let's name the pieces to make it easier to refer to them: the domino (2 squares), the inner and outer hooks (the 3-square pieces), and the C, S, T and I (like in Tetris, except the C). >!The total value of all the numbers is (1+2+3+4+5)×5 = 75. If each cage sums to a distinct prime number, then it can't include 2 (since an even plus six odds is even, and 75 is odd), and the seven smallest other primes are 3+5+7+11+13+17+19 = 75 exactly (since they're the smallest, anything else would be larger), so those must be the sums of the cages.!< >!Looking at what sums can go in what cages, remember that squares in the same row or column must have distinct numbers. This means the domino can have anything from 1+2=3 to 4+5=9, the hooks 1+2+1 = 4 to 5+4+5 = 14, the I 1+2+3+4 = 10 to 2+3+4+5 = 14, the T 1+2+3+1 = 7 to 5+4+3+5 = 17, the S 1+2+1+2 = 6 to 5+4+5+4 = 18, and the C 1+2+3+1+2 = 9 to 5+4+3+5+4 = 21.!< >!Some of the primes from before immediately fit into certain ranges; 3 can only be the domino and 19 the C, 17 must be either the T or S, and 5 can only be a hook from what's left.!< >!Then we can start filling in the puzzle; the domino with 3 can only be filled with 1 and 2. This stops the outer hook from being the one with 5 (since two of its squares can't have a 1 or 2), so it must be the inner hook (filled with either 212 or 131). The other hook has at minimum a 3 and 4 in the right column and a 2 in the remaining square (the inner hook has taken the 1), meaning it has a minimum of 9. That means it can't have the 7, which can only otherwise go in T or S; since one has 7 and the other 17, the I and outer hook must have the remaining two of 13 and 11.!< >!Since the T must have at least a 3 in the right, and can't have a 1 in the middle due to the inner hook having the 1, it can't total 7; that means it must be 17, and the S is 7. The only way the S can have 7 is 1-2-1-3, which means it has the 1 in the bottom two columns; that lets us collapse several choices in the domino and inner hook. The domino has 2 on bottom, 1 top, and the inner hook is 212. We also must have 13 on the upper half of the S (since the 2 in that row is in the domino) and 12 in the bottom.!< >!Since 12 is taken in the bottom row, the bottom of the T must have 345, meaning the top bit must have 5 to total 17, and the remaining square in that column on the left must be 4. In order for the I to total 11 or 13, we must leave the 2 or 4 out of 15 in the remaining cell; 4 is not available in that cell, so we can have 13 in the I and 11 in the outer hook, and collapse the rest of the S.!< >!Since we have 1 and 3 in the S, the left side of the C must have 245 in some order, but the top row is the only one that has a 2 free, leaving 45 in the other two. This totals 11, so the other two cells in the C must total 8 to make 19, which can only be 35; this eliminates some possibilities in the T, letting you narrow down its bottom row to 435 left to right.!< >!With 125 established in the right column, the other two cells in the outer hook must have 34; these sum up to 7, so making 11 requires a 4 in the last cell, forcing the other 4 into the lower column. This causes a chain reaction that collapses the rest of the C, and we can fill in the rest of the I as only one is left in each row.!< So the final solution is left to right, top to bottom >!12543 35214 54321 43152 21435!<.
Very cool! I was a little thrown at first by allowing duplicate digits in cages but obviously it doesn't work if you don't
Really cool puzzle, first thing I've seen on here that I've felt compelled to sit down and solve, thanks for sharing!
This is the coolest puzzle I've seen on this subreddit very fun to solve and an interesting twist on the KenKen format.
Had to take 1 clue from here but I finally managed it. Turns out that 2+2+1 is also 5... I was trying 3+1+1...
Nice, fun one. Was weird using duplicate digits in a cage. I kind of half and halfed figuring which cages go where while putting in digits. The final slotting in of everything was very satisfying. >! First is identifying the primes that can be made in these cages. 3, 5,7,11,13,17 and 19. 19 has to go in the five cage as there are no four cages that could support three fives. 3 has to go in the two cage as none of the three cages can support three ones.!< >!Top left four cage has to be 11 or 13 due to its shape, bottom left corner cell thus has to be a two or a four. !< >!The only way for the bottom left four cell to be the 17 cage would be to have two fives, which can't be done now as the bottom left cell needs to be two or four, and the 19 cage needs at least one five to sum to 19. This means the bottom right cage must be the 17 cage. This can again only be made with two fives, one of which must be placed in the top cell of the cage- first placed digit!!< >!This then places two and one in the first two cells of row five, disambiguating the top left four cage as the 13 cage. !< >!The central three cage must now be the 5 cage as the top right cage needs to have two cells be three four or five. Now the bottom left four cage must be the 7 cage as it cannot contain a five or two fours to make it the 11. Top right three cage becomes the 11 cage, and then it's just a little sudoku to fill out the digits!!< >!!< Very cute, thanks for the fun after work puzzle OP
This was really fun! Stayed up past my bedtime to finish it, thank you for a great puzzle :)
best 36 minutes of my whole week! congrats on making this killer sudoku, it was really such an interesting logic path it took me into :)
Great puzzle!
Excellent puzzle, thank you very much for sharing.
Oh I had fun with this, and even though I’m late to the party I need to share my logic because it was very unnecessarily complicated but also very fun. >!I made a crucial mistake right away that made solving the puzzle much harder: I counted six cages instead of seven, so I assumed one of the primes between 3-19 was missing. Obviously I could have noticed that six odd numbers sum to an even number, and the grid’s total of 75 is odd, but I didn’t think to check that either for some reason.!< >!Consider which primes any 4-cell cage can sum to: it must be 7, 11, 13, or 17 (notice that 19 is impossible because you cannot fit three 5s in 4 connected cells). But also notice that the Z-cage and the T-cage together sum to 15 (the entire bottom row) + an odd number, either 1/3/5 or 2/4/(1,3,5). The possible sums here are 22 (15+1/2/4), 24 (15+1/3/5 or 2/3/4), or 26 (15+2/4/5). Comparing this with the possible sums of two of (7,11,13,17) makes 24 the only possible sum for these two cages: 7+17 or 11+13.!< >!But now notice that the I-cage on the left must sum to 11 or 13! You can’t have three different cages summing to only two primes, so the T-cage and Z-cage must sum to some combination of 7 and 17.!< >!Next, pay attention to the two 3-cell cages: now that they can no longer sum to 7, they must sum to 5, 11, or 13. That forms a triple with the I-cage we just discussed; those three prime sums are taken by those three cages, in some combination. At long last, I have shown that 5 and 7 must be used elsewhere, and so I have in the most roundabout way possible proven that the 2-cell cage must sum to 3.!< >!Finally, to start putting numbers in the grid, pay attention to the parity of various cells in the I-, T-, Z-, and 2-cages. A cursory glance at the I-cage shows that the bottom left corner must be even, and therefore the remaining 3 cells in the Z-cage must be odd (since three evens and one odd cannot sum to 7 or 17 without three 2s or three 4s, which are both impossible). We also know by the same logic that the T-cage contains three odds and one even, and we furthermore know that the even number in that cage must be in the bottom row, so we have now successfully placed all three odds numbers in row 4. Put the 2 in column 5 since we know that cell sums to 3, the 4 in column 1 by process of elimination, and the 2 in the bottom-left corner again by process of elimination, and now you know the Z-cage sums to 7; the remaining three digits are 1/1/3. Fill those in, put the 5 in the T-cage in row 4 and the 1 in the 2-cage, and you’re off to the races.!<
this was my favorite puzzle i've ever done and i encouraged multiple friends to try it are there more puzzles like this? Thank you so much!! <3
Thank you! I think there are a lot of good puzzles on the website i linked. The YouTube channel “cracking the cryptic” also has good puzzles.
Discussion: OP, I LOVED this concept! You should [submit this to CtC!](https://crackingthecryptic.com/#submissions) [Here's a video of it being solved, which I hope you enjoy.](https://www.twitch.tv/videos/2132681582?t=0h30m13s) (I suggested it to him, and he told me today he thought it was really good!)
Omg haha thank you!
Just saw this, haven’t read the comments. Can the prime numbers repeat?
No, the digits inside the cages can repeat, but the resulting primes of each cage can’t!
Genius
"... *distinct* prime numbers"
That was the key for me. Once you assign each cage a distinct prime number based on math, it's pretty straightforward.
It says distinct prime numbers, and highlights it, so no.
This was a cool puzzle. >! I decided not to approach it mathematically and just come up with a pattern. Once I put the top row and the right column in I found the “cage” in the top right and the 2 space “cage” below both worked. I just continued the pattern and solved it. Took me just over 2 mins. I think it would have taken longer with no mention of a pattern.!< Lots of fun thanks!
I needed to look up what a prime number was first before I could proceed with the puzzle.
Very fun puzzle!
I tried to solve it for a solid hour before I came back and read that each cage must have a different number. Thought it was impossible at first
Neat.
That was a really enjoyable puzzle! And a great lesson in R-ing TFQ for m’y other half 😂
|>!1!<|>!2!<|>!5!<|>!4!<|>!3!<| |:-|:-|:-|:-|:-| |>!3!<|>!5!<|>!2!<|>!1!<|>!4!<| |>!5!<|>!4!<|>!3!<|>!2!<|>!1!<| |>!4!<|>!3!<|>!1!<|>!5!<|>!2!<| |>!2!<|>!1!<|>!4!<|>!3!<|>!5!<| Did bindings on the possible values for each based on size, then had ways to assign such that each section had a distinct prime. Removed number sizes that were impossible to reach based on arrangement, ended with 16. The >!3 and 5!!17!!7!<- which in turn let me determine the size of the size >!13!< cage. Top right and the 5-square one took longer, even though I knew the 5-square one was size >!19!
:) I knew you’d get more engagement over here than in r/sudoku — this sub has a quarter million followers!