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Hexagonal Grid Filling
A Trip to Ripple IslandsA spartan skeleton SudokuSkeleton sudoku, the secondA Knight's KuromasuThe Ludicrous Loop: over a thousand cells of circular logic!RPS Trinity - Teddy and Her RibbonHow does one go about creating a new grid-deduction puzzle?The Terrible Twos of Slitherlink Part DeuxFixing a Binairo MistakeIntersection Puzzle
$begingroup$
Fill in each blank white hexagon with a digit from 1 to 9. Each of the grey hexagons is the sum of the white hexagons around it, and all of the white hexagons that are around the same grey hexagon must be distinct.
The answer's already available, but I would like to see how people would logically solve this.
Source: USAMTS
grid-deduction
$endgroup$
add a comment |
$begingroup$
Fill in each blank white hexagon with a digit from 1 to 9. Each of the grey hexagons is the sum of the white hexagons around it, and all of the white hexagons that are around the same grey hexagon must be distinct.
The answer's already available, but I would like to see how people would logically solve this.
Source: USAMTS
grid-deduction
$endgroup$
add a comment |
$begingroup$
Fill in each blank white hexagon with a digit from 1 to 9. Each of the grey hexagons is the sum of the white hexagons around it, and all of the white hexagons that are around the same grey hexagon must be distinct.
The answer's already available, but I would like to see how people would logically solve this.
Source: USAMTS
grid-deduction
$endgroup$
Fill in each blank white hexagon with a digit from 1 to 9. Each of the grey hexagons is the sum of the white hexagons around it, and all of the white hexagons that are around the same grey hexagon must be distinct.
The answer's already available, but I would like to see how people would logically solve this.
Source: USAMTS
grid-deduction
grid-deduction
asked 2 hours ago
Jason KimJason Kim
3149
3149
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To get it started:
$21$ has to be $1+2+3+4+5+6$
$39$ has to be $9+8+7+6+5+4$
$21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
To get it started:
$21$ has to be $1+2+3+4+5+6$
$39$ has to be $9+8+7+6+5+4$
$21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)
$endgroup$
add a comment |
$begingroup$
To get it started:
$21$ has to be $1+2+3+4+5+6$
$39$ has to be $9+8+7+6+5+4$
$21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)
$endgroup$
add a comment |
$begingroup$
To get it started:
$21$ has to be $1+2+3+4+5+6$
$39$ has to be $9+8+7+6+5+4$
$21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)
$endgroup$
To get it started:
$21$ has to be $1+2+3+4+5+6$
$39$ has to be $9+8+7+6+5+4$
$21$ already has a $4$, so the overlap between $21$ and $39$ has to be $5$ and $6$, but $29$ already has a $5$, so it must get the $6$, while $30$ shares the $5$. (The hex shared by $21$, $30$, and $39$ is $5$, and the hex shared by $21$, $29$, and $39$ is $6$.)
answered 1 hour ago
shoovershoover
2,090614
2,090614
add a comment |
add a comment |
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