12
$\begingroup$

Help me to decorate my Christmas tree with garlands of 7 bulbs.

I want to fill in all the blank bulbs with numbers 1 to 7, so that each garland contains 7 different numbers. To make the tree more beautiful, any straight line of bulbs parallel to the side of the tree (horizontal and diagonals) should contain only different numbers as well.

enter image description here

$\endgroup$
  • $\begingroup$ so the bottom left is a 7? $\endgroup$ – JonMark Perry Dec 6 '16 at 6:33
  • 1
    $\begingroup$ I've filled in six circles and it doesn't seem like there's anywhere I can go from there without guessing and checking... $\endgroup$ – Deusovi Dec 6 '16 at 7:10
  • $\begingroup$ @JonMarkPerry : Yes indeed $\endgroup$ – Jamal Senjaya Dec 6 '16 at 7:22
6
$\begingroup$

Made a couple lucky guesses, but I'm pretty sure I got it:

rough ASCII sketch:

______1
_____2_7
____3_4_6
___4_6_7_5
__5_2_1_3_4
_6_1_3_7_5_2
7_5_4_1_2_6_3

Broken down by color:

Red from left to right: 4321765
Green from left to right: 1264735
Blue from left to right: 5431726
Yellow starting from 5 and following along its line: 5671324

$\endgroup$
4
$\begingroup$

My solution:

xmasuko

After the $7$, add the $1$ on yellow - the remaining reds are $567$, so the yellows on the diagonal are $234$.

The only place for $1$ in blue is now row $5$, because of the yellow $1$.

The blue $2$ tells us that the row $5$ and row $7$ yellows can't be $2$, so $2$ is row $6$.

The red, blue and yellow $2$s now force the green $2$.

This is where I guessed the $3$ on row $6$, which gives you the green $3$, the yellow $3$ and therefore the yellow $4$. The blue $7$ is also forced.

The blue $7$ forces the red $7$ by removing all but two of the green garland slots, leaving only the two top right. But the top green slot blanks both remaining red slots, giving us the red and green $7$s.

The bottom row needs $456$. The $4$ is the green garland is on the left hand side, and by associations, the leftmost blue cannot be $123467$ and so is $5$.

So the green $5$ is bottom right. The green row $4$ left is now forced to be $6$, which in turn forces the red $56$, the blue leftmost remaining slot, the blue $6$, and we are done.

$\endgroup$

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.