This December I participated in Advent of Code.

Days 7-10 saw me really trying (most days) to get a good ranking. (A good ranking turned out to be “a 4 digit number”.) I would be ready when the new puzzle dropped, sitting at my PC. And I would get the programming done! This later flagged, when the puzzles got harder, and I had to spend time simply trying to figure out what coding to do. But in this period, it was very much possible to get a good ranking. (From day 16 and forward, I was trying to catch up. It was a struggle to get the program done within 24 hours, and I wanted to solve the puzzles in the right order.)

      --------Part 1--------   --------Part 2--------
Day       Time   Rank  Score       Time   Rank  Score
 25       >24h  22113      0       >24h  14148      0
 24       >24h  23368      0       >24h  15862      0
 23       >24h  23812      0       >24h  21472      0
 22       >24h  25204      0       >24h  21799      0
 21       >24h  19594      0       >24h  16355      0
 20       >24h  24815      0       >24h  21577      0
 19       >24h  28005      0       >24h  25333      0
 18       >24h  27991      0       >24h  27338      0
 17       >24h  28700      0       >24h  20591      0
 16       >24h  24090      0       >24h  25303      0
 15   01:19:07   5843      0   07:18:20   8861      0
 14   01:28:12   7258      0   02:13:42   6000      0
 13   03:04:50  11545      0   03:08:52   7714      0
 12   03:55:52  14298      0   04:04:26   7259      0
 11   00:24:58   6479      0   01:09:10   5330      0
 10   00:44:34   6515      0   00:48:19   6013      0
  9   01:26:25   9059      0   02:59:32   8039      0
  8   01:53:25  10087      0   02:04:40   9294      0
  7   17:55:39  47349      0   18:02:03  44377      0
  6   00:45:55   8375      0   02:18:28   8078      0
  5   13:07:46  55110      0   13:13:45  44839      0
  4   01:51:36  15622      0   02:06:52  13570      0
  3   16:51:27  91593      0   17:04:46  80436      0
  2   12:11:26  78108      0   13:30:49  60978      0
  1   17:12:43  92846      0   17:24:33  87414      0

Day 9 I had a question. The answer turned out to be: I had a check in part 1, that I lost when I restructured the code for part 2. Oops.

Day 7: wonderful recursion! Day 10: more wonderful recursion!

A new December and a new bunch of puzzles from mscroggs.co.uk.

The candidate numbers all have 2 digits, ab. A palindrome number of this kind has a = b. So the palindrome numbers are 11, 22, …, 99.

11 + 22 + … + 99 =

(1+2+…+9)*11 =

9*5*11 =

495

A new December and a new bunch of puzzles from mscroggs.co.uk.

Oh goody, time to use my fairly new cross sum skills.

Going left to right, top to bottom, name the boxes a-i.

• R2, C3 (*) = f is an integer dividing 9, therefore one of 1, 3, 9.
• R3, C3 = i, same argument, same result.
• Still R3, C3 = i: the result of a division, c/f, divided by i, is 3. If i = 1, c/f = 3, it’s one of 3/1, 6/2, 9/3. First would require duplication, throw it away. If i = 3, c/f = 9, that would be 9/1. If i = 9, c/f = 27, impossible. New result: one of 1, 3.
• R3, C1 = g is an integer dividing 12, therefore one of 2, 3, 4, 6. (Can’t be 1, that would require the subtraction above, a/d to be 12; can’t be 12, that’s not a digit.)
• R3, C2 = h has to combine with 2, 3, 4, 6 to the left and 1, 3 to the right. g/h*i = 9. g/h must be 3 or 9. 9 is impossible, so it must be 3. Therefore i = 3. Therefore g/h must be 6/2. (It can’t be 3/1 because of duplication.) So g = 6, h = 2.
• Going back to C3, c = 9, f = 1.
• R2, d + e = 9. Of the digits left, it must be 4 and 5.
• R1, a – b = -1. The last digits are 7 and 8. a = 7, b = 8.
• C1, (7-d)*6 = 12. d = 5, e = 4.

(*) Row 2, column 3.

The digits in the boxes are 7, 9 and 4. 7*9*4 = 252.

A new December and a new bunch of puzzles from mscroggs.co.uk.

Official answer: “By adding a point inside a triangle, you can turn one triangle into three triangles. By adding a point outside all the current triangles, you can add either one or two more triangles. As we are after the maximum number of triangles, we will choose to add two triangles for each point we add. Four points make 3 triangles. Adding 286 more points will add 572 more triangles, giving a total of 575.”

I think, but I can’t quite prove, the construction of the most triangles possible goes like this:

1. Begin with 2 points and the line between them.
2. Add 2 points close to this and construct the 3 triangles possible.
3. Choose one of the outside lines of the figure.
4. Go back to #2 and repeat.

This means: 2 points, 0 triangles; 4 points, 3 triangles; 6 points, 6 triangles; … ; 2(n+1) points, 3n triangles.

290 = 2(n+1)

145 = n+1

144 = n

432 = 3n

The answer is 432.

This week the question is: Can You Squeeze the Sheets?

Two large planar sheets have parallel semicircular cylindrical ridges with radius 1. Neighboring ridges are separated by a distance L ≥ 2. The sheets are placed so that the ridges extrude toward each other, and so that the sheets cannot shift relative to each other in the horizontal direction, as shown in the cross-section below:

Which value of L (again, that’s the spacing between ridges) maximizes the empty space between the sheets?

To be clear, you are maximizing the volume of empty space per unit area of one flat sheet. In the cross-section shown above, that’s equivalent to maximizing the area of empty space per unit length of one flat sheet.

And for extra credit:

Instead of cylindrical ridges, now suppose the sheets have any number (greater than zero) of hemispherical deformations with radius 1 that extrude toward each other. This time, the sheets need not be the same as each other.

As before, the distance between the centers of any two deformations on the same sheet must be at least 2. What is the minimum empty space, again expressed as volume per unit area of one flat sheet?

Highlight to reveal (possibly incorrect) solution:

Video. Figure 1, 2, 3 and 4.

Construct (in the cross section of the 2 sheets) a triangle. 1 side is horizontal, 1 side vertical, and 1 side connects the centres of 2 neighboring ridges-semicircles.

The horizontal side is L/2. The hypotenuse is exactly 2. Therefore the vertical side is √(4 – L²/4). Let’s call that h for height.

Looking at a rectangle containing 2 of these triangles, its area (again, a cross section of the real 3d situation) is hL/2. Some of this area is taken up by circle parts; exactly π/2. The rest is empty: hL/2 – π/2. We want to normalize this, to correspond to a unit step in the horizontal direction. This gives (hL/2 – π/2)/(L/2). This function has a maximum for about L = 2.6581.

And for extra credit:

Figure 1, 2, 3 and 4.

Option a: This is the same solution as the optimal stacking of balls. In my figures there’s a triangle between the centres of the 3 blue hemispheres. The base of the triangle is 2, and the height is √3. Adding the centre of the red hemisphere, we get a tetrahedron. This has height 2√2/√3.

Looking at the volume with the 2 green triangles as the base, we have a complete red hemisphere (the bits chopped off have friends included elsewhere) and a complete blue hemisphere (2 x 1/3 and 2 x 1/6). The space taken up is π4/3, appr. 4.19. The complete space is 2*√3*2*√2/√3 = 4*√2, appr. 5.66. The empty space is appr. 1.45. The base of the volume is 2*√3, appr. 3.46. The empty volume pr. unit area is 1.45/3.46, appr. 0.42.

Calculations and animation.

Option b: Let L be the distance between 2 centres of hemispheres on a sheet. Vary L and find a minimum. Oh! That minimum occurs when L = 2, so we get the same solution, appr. 0.42.

Option c: Defies the my imagination.

A new December and a new bunch of puzzles from mscroggs.co.uk.

The circle all around is 360°.

The angle to the minute hand, corresponding to 22 minutes is 360 * 22/60 = 132.

The angle to the hour hand, going clockwise, corresponding to 8 hours and 22 minutes is 360 * (8 + 22/60) / 12 = 251.

The angle between the hands is 251 – 132 = 119.

This December I participated in Advent of Code.

Days 1-6 happened without a lot of drama (only 1 question on the forum). This phase also mostly saw me coding in a sandbox (exception: day 3, part 2). Day 7 I permanently changed to my PC. The tasks were rather easy for me. I didn’t really try to come up with a fast answer, in part because the sandbox + my tablet was a slow way to code. My examples below manipulate the test data, but ran on the actual data as well.

A new December and a new bunch of puzzles from mscroggs.co.uk.

n = 99999…999

= 10⁵⁵ – 1

n³ = (10⁵⁵ – 1)³

= (10⁵⁵)³ + 3*(10⁵⁵)²*(-1) + 3*(10⁵⁵)*(-1)² + (-1)³

= 10¹⁶⁵ – 3*10¹¹⁰ + 3*10⁵⁵ – 1

 10000...0000...0000...000
-         300...0000...000
+                300...000
-                        1
==========================
  9999...9700...0299...999

So now we just have to count the 9s and 7s and the 2.

The 1 lives in position 1. The 3 above it has a 3 in position 56. The next 3 is in position 111. Finally the 1 above it is in position 166.

The result of adding and subtracting is 9s from position 1 to 55, a 2 at position 56, then some 0s, a 7 at position 111 and finally 9s from position 112 to 165. 55*9+2+7+54*9 = 990. Which happens to be 55*18.

Just to test my logic here: if n had been 9999 (4 digits), n³ would be 999,700,029,999. 4*9+2+7+3*9 = 72 = 4*18. Checks out.

A new December and a new bunch of puzzles from mscroggs.co.uk.

n + n+1 + n+2 + … n+10 = 2024

Let m = n+5

m-5 + m-4 + … m + … m+5 = 2024

11m = 2024

m = 2024/11 = 184

n = m-5 = 179

A new December and a new bunch of puzzles from mscroggs.co.uk.

I think the example is also a hint.

A number with 1 factor would be 1. And that’s not 13.

A number with 2 factors would be a prime. Number p, factors 1 and p. The geometric mean would be (1*p)^1/2. We can’t get 13 to fit there.

A number with 3 factors has a special property. There’s a square here somewhere. The number, p², has factors 1, p and p². This accounts for p² = 1 * p² = p * p, the 2 different ways to write p². Like, 9 = 1*9 = 3*3. The geometric mean works out to be p, isn’t that neat? So my guess is that we’re looking for 13² = 169.