Rule : fill in the grid with all different positive integers

Enjoy

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Rule : fill in the grid with all different positive integers

Enjoy

Enjoy

Last edited by Jean-Christophe on Thu Oct 05, 2006 3:02 am, edited 2 times in total.

- Jean-Christophe
**Posts:**149**Joined:**22 January 2006

I though it was clear enough.

Anyhow, it's nine numbers out of the "natural numbers" : 1,2,3,4.... infinity

No zero, fractions, decimals..., just plain positive integers

And as you can see, these are not the traditional 1..9

"All different" means : numbers cannot be repeated, must be distinct.

It's a killer, so the cells in the cage must add up to the sum displayed. And no number may be repeated within a cage, even if it would be otherwise permitted by rules for regular sudoku.

It can be solved using logic. No need of guess or T&E.

BTW notice the void/empty cell in R5C5

Hint : Think in terms of "45". How many "45" make the whole puzzle ?

Anyhow, it's nine numbers out of the "natural numbers" : 1,2,3,4.... infinity

No zero, fractions, decimals..., just plain positive integers

And as you can see, these are not the traditional 1..9

"All different" means : numbers cannot be repeated, must be distinct.

It's a killer, so the cells in the cage must add up to the sum displayed. And no number may be repeated within a cage, even if it would be otherwise permitted by rules for regular sudoku.

It can be solved using logic. No need of guess or T&E.

BTW notice the void/empty cell in R5C5

Hint : Think in terms of "45". How many "45" make the whole puzzle ?

- Jean-Christophe
**Posts:**149**Joined:**22 January 2006

Thanks.

after udosuks explanation I was able to solve it.

My key steps were:

N14: 127+R1C3=2 * sum of values => R1C3 odd, with respect to cage R1C34 => R1C3={1,3,5} => sum of values is 64, 65 oder 66

N69 129 +R9C7 = 2 * sum of values, 2 * 64 = 128 < 129 => sum of values <> 64, so 65 oder 66

Innie N 7 => R9C3 = 7 (if sum 65), 8 (if sum 66)

Outie R9 => R8C1 = 9 (if sum 65), 8 (if sum 66)

R9C3<>R8C1 => sum 65, R9C3=7, R8C1=9

R1C3 = 3 (Innie N14), R1C7= 9 (Innie N3), R1C6 = 7 (cage sum), R2C9 = 3 (Outie R1), Innie N8, R9C7= 1 (Innie N69), R9C4 = 6 (cage sum), R7C6 = 9 (Innie N8, R9C56=22, R9C4=6), R3C4 = 12 (Innie N2, R1C45=5,R1C6=7), R5C5= 7 (innie N123456789)

possible combinations cage 8/3 in N12: R1C45 = 5, R1C3=3 =>R1C4, R1C5={1,4}

known values: 1, 3, 4, 6, 7, 9, 12, sum of other both values 23

cage R78C4=20 => one of this values in {8,11,13,14,16,17,19},

the other would be in {15,12,10,9,7,6,4}

all 9 values different => the first in {8,13}, the second in {15,10}

cage R1C12 = 19 with naked pair 1,4 in R1 and naked singles in R1 =>

R1C1,R1C2={6,13}, so 13 and 10 are the other values,

values: 1,3,4,6,7,9,10,12,13

the rest uses basic techniques.

Pyrrhon

after udosuks explanation I was able to solve it.

My key steps were:

N14: 127+R1C3=2 * sum of values => R1C3 odd, with respect to cage R1C34 => R1C3={1,3,5} => sum of values is 64, 65 oder 66

N69 129 +R9C7 = 2 * sum of values, 2 * 64 = 128 < 129 => sum of values <> 64, so 65 oder 66

Innie N 7 => R9C3 = 7 (if sum 65), 8 (if sum 66)

Outie R9 => R8C1 = 9 (if sum 65), 8 (if sum 66)

R9C3<>R8C1 => sum 65, R9C3=7, R8C1=9

R1C3 = 3 (Innie N14), R1C7= 9 (Innie N3), R1C6 = 7 (cage sum), R2C9 = 3 (Outie R1), Innie N8, R9C7= 1 (Innie N69), R9C4 = 6 (cage sum), R7C6 = 9 (Innie N8, R9C56=22, R9C4=6), R3C4 = 12 (Innie N2, R1C45=5,R1C6=7), R5C5= 7 (innie N123456789)

possible combinations cage 8/3 in N12: R1C45 = 5, R1C3=3 =>R1C4, R1C5={1,4}

known values: 1, 3, 4, 6, 7, 9, 12, sum of other both values 23

cage R78C4=20 => one of this values in {8,11,13,14,16,17,19},

the other would be in {15,12,10,9,7,6,4}

all 9 values different => the first in {8,13}, the second in {15,10}

cage R1C12 = 19 with naked pair 1,4 in R1 and naked singles in R1 =>

R1C1,R1C2={6,13}, so 13 and 10 are the other values,

values: 1,3,4,6,7,9,10,12,13

the rest uses basic techniques.

Pyrrhon

- Pyrrhon
**Posts:**240**Joined:**26 April 2006

Well done Pyrrhon

You didn't followed the way I designed it, but could find the key info.

BTW forget about "mistery", I mispelled it (myspelled, mispeled ?) I must be dyslecsic, sic

You didn't followed the way I designed it, but could find the key info.

BTW forget about "mistery", I mispelled it (myspelled, mispeled ?) I must be dyslecsic, sic

- Jean-Christophe
**Posts:**149**Joined:**22 January 2006

Here is my walkthrough

Step 1

We have to find out the 9 numbers and their sum (equivalent to the "45")

All cages cover all cells but one in R5C5. This covers 9 rows (or columns or nonets)

The sum of all cages (578) + R5C5 must be a multiple of 9. Nine times S, the "45" equivalent for this grid.

578/9 = 64.22... -> round up to 65 -> S = 65, 66, 67...

-> R5C5 (X) = (S*9) - 578 = 7, 16, 25...

Step 2

X <= 16 because 25 or any higher number can't go in any cage of R1

-> X = 7|16, S = 65|66

If X were 16, S = 66

In N9, 16 can't go in Cage 15/2 nor 13/2

Outies of N6 -> R7C89 = 28+7+18+29 - 66 = 16, can't include a 16

Innies of N9 -> R9C7 = 66 - (15+16+19+13) = 3

-> Cage 19/2 can't be 16+3

There is no place in N9 where we can put a 16

-> X = R5C5 = 7, S = 65

From here on, the rest is much easier

Step 3

65 on R1 -> R2C9 = 3, R1C89 = 22

65 on N3 -> R1C67 = [7,9]

65 on N4 -> R3C12 = 8

65 on N1 -> R1C3 = 3, R1C45 = 5 = {1,4}

65 on N2 -> R3C4 = 12

65 on N5 -> R7C6 = 9

65 on R9 -> R8C1 = 9, R9C12 = 16

65 on N7 -> R9C34 = [7,6]

65 on N6 -> R7C89 = 17

65 on N9 -> R9C7 = 1, R9C56 = 22

Step 4

12 of N8 must go in R9C56 = {10,12}

We know all numbers but one, so we can deduce the missing number : 65 - (1+3+4+6+7+9+10+12) = 13

Step 5

Cage 9/2 in N2 -> R23C6 = [6,3], Cage 32/3 = {9,10,13}

Cage 25/3 in N3 = {3,10,12}

Cage 19/2 in R1 = {6,13}

R3C12 = 8 = {1,7}

Cage 16/2 in N1 = {4,12}

Cage 19/2 in N1 = {9,10}

Step 6

R2C8 = 1, R3C89 = 17 = {4,13}

Cage 13/2 in N3 = [7,6]

Cage 15/2 in N9 = {3,12}

Cage 19/2 in N9 = {6,13}

Cage 13/2 in N9 = {4,9}

R7C89 = {7,10}

...

Step 1

We have to find out the 9 numbers and their sum (equivalent to the "45")

All cages cover all cells but one in R5C5. This covers 9 rows (or columns or nonets)

The sum of all cages (578) + R5C5 must be a multiple of 9. Nine times S, the "45" equivalent for this grid.

578/9 = 64.22... -> round up to 65 -> S = 65, 66, 67...

-> R5C5 (X) = (S*9) - 578 = 7, 16, 25...

Step 2

X <= 16 because 25 or any higher number can't go in any cage of R1

-> X = 7|16, S = 65|66

If X were 16, S = 66

In N9, 16 can't go in Cage 15/2 nor 13/2

Outies of N6 -> R7C89 = 28+7+18+29 - 66 = 16, can't include a 16

Innies of N9 -> R9C7 = 66 - (15+16+19+13) = 3

-> Cage 19/2 can't be 16+3

There is no place in N9 where we can put a 16

-> X = R5C5 = 7, S = 65

From here on, the rest is much easier

Step 3

65 on R1 -> R2C9 = 3, R1C89 = 22

65 on N3 -> R1C67 = [7,9]

65 on N4 -> R3C12 = 8

65 on N1 -> R1C3 = 3, R1C45 = 5 = {1,4}

65 on N2 -> R3C4 = 12

65 on N5 -> R7C6 = 9

65 on R9 -> R8C1 = 9, R9C12 = 16

65 on N7 -> R9C34 = [7,6]

65 on N6 -> R7C89 = 17

65 on N9 -> R9C7 = 1, R9C56 = 22

Step 4

12 of N8 must go in R9C56 = {10,12}

We know all numbers but one, so we can deduce the missing number : 65 - (1+3+4+6+7+9+10+12) = 13

Step 5

Cage 9/2 in N2 -> R23C6 = [6,3], Cage 32/3 = {9,10,13}

Cage 25/3 in N3 = {3,10,12}

Cage 19/2 in R1 = {6,13}

R3C12 = 8 = {1,7}

Cage 16/2 in N1 = {4,12}

Cage 19/2 in N1 = {9,10}

Step 6

R2C8 = 1, R3C89 = 17 = {4,13}

Cage 13/2 in N3 = [7,6]

Cage 15/2 in N9 = {3,12}

Cage 19/2 in N9 = {6,13}

Cage 13/2 in N9 = {4,9}

R7C89 = {7,10}

...

- Jean-Christophe
**Posts:**149**Joined:**22 January 2006

PaulIQ164 wrote:My brother came up with one of these a while back. Can't remember if I ever posted it here.

Yes you have...

- udosuk
**Posts:**2698**Joined:**17 July 2005

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