This week the #puzzle is: Will You Be Right on Time? #angles #clock #RightAngles #geometry

A standard analog clock includes an hour hand, a minute hand, and 60 minute markers, 12 of which are also hour markers.

At a certain time, both the hour hand and minute hand are both pointing directly at minute markers (either of which could also be an hour marker). The hour hand is 13 markers ahead (i.e., clockwise) of the minute hand.

At what time does this occur?

And for extra credit:

At various times of day, the minute and hour hands form a right angle. But is there a time of day when the hour hand, minute hand, and second hand together form two right angles, as illustrated below? If you can find such a time or times, what are they?

If you cannot find any such times, suppose the measures of the two nearly right angles formed by the three hands measure A and B degrees. What time or times of day minimize the square error function f(A, B) = (A − 90)2 + (B − 90)2?

Either way, your answer should be precise to at least a thousandth of a second.

Will You Be Right on Time?

Solution, possibly incorrect:

We can have any time from 00:00 to 11:59.

The time:

So many minutes have passed since 00:00:

It takes 720 minutes for the hour hand to go all the way around the clock.

When the hour hand points directly at the nth minute marker:

We also have some extra knowledge about the distance between the hands:

There are 2 ways to compute where the hour hand is.

---

Method 1, example:

---

Method 2, example:

---

As these 2 methods don’t result in the same n for the example, this isn’t the time we’re looking for.

---

As we have, from method 1:

n is an integer, so m is a multiple of 12. Further, from method 1:

This leads to MM being a multiple of 12. So:

Then we use method 2 to find n:

MM	00	12	24	36	48
n	13	25	37	49	1

And method 1, backwards, to find m:

MM	00	12	24	36	48
n	13	25	37	49	1
m	156	300	444	588	12

And then find HH:

MM	00	12	24	36	48
m	156	300	444	588	12
HH	2.6	4.8	7	9.2	11.4

Only the time 07:24 gives an integer HH. So this is the solution.

And for extra credit:

Program

I write a program to go through a lot of different options. I fine tune the program to find better and better solutions, with better and better error functions. The results are:

Good time, in seconds..:    13744.07855
 -- Good time..........: 03:49:04.07855
 -- Error function.....: 0.0079550
Good time, in seconds..:    29455.92145
 -- Good time..........: 08:10:55.92145
 -- Error function.....: 0.0079550

So the best possible times are 03:49:04.079 and 08:10:55.921.

I have watched #ProjectHailMary! Boy, did that director send small hails in all directions, primarily to other #movies ? Fun! #references

Rocky has 3 fingers, Grace has 5. Zammis has 3 fingers, Davidge has 5. Enemy Mine.

Sitting on a beach, looking out on the water. Looking to the right. Somebody’s coming! Contact.

A big structure of rods, parallel to each other or perpendicular to each other. Interstellar.

The last one is a nod (I think) to “Story of Your Life”. The story behind Arrival.

“Then Raspberry brought the gourd down between its legs, a crunching sound resulted, and the gourd reemerged minus a bite…”

This week the #puzzle is: Have You Heard the Buzz? #counting #frequency #combinations

I recently introduced my children to a game called “Buzz” (also known as “Fizz buzz”), in which players take turns reciting whole numbers in order. However, in one particular variant of the game, anytime a number is a multiple of 7 or at least one of its digits is a 7, the player must say “buzz” instead of that number.

For example, here is how the first 20 turns of the game should proceed: 1, 2, 3, 4, 5, 6, buzz, 8, 9, 10, 11, 12, 13, buzz, 15, 16, buzz, 18, 19, 20.

How many times should “buzz” be said in the first 100 turns of the game (including those mentioned above in the first 20 turns)?

And for extra credit:

As we just saw, in the first 20 turns of the game, 15 percent of the numbers were “buzzed.” But as the game proceeds, an increasing frequency of numbers get buzzed.

There is a certain minimum number N such that, for the Nth turn in the game and for every turn thereafter, at least half the numbers up to that point have been buzzed. What is this value of N?

Have You Heard the Buzz?

Solution, possibly incorrect:

Program

Method 1: I write a program to produce all the buzzes and count them. There are 30.

Method 2:

• We are looking at the numbers 1-100.
• If the last digit is 7, it’s a buzz: 7, 17, 27, … 97. 10 of these.
• If the next to last digit is 7, it’s a buzz: 70, 71, 72, … 79. 10 of these.
• If the number is a multiple of 7, it’s a buzz: 7, 14, 21, … 98. 14 of these.
• But if we just say 10 + 10 + 14, we’ve counted something twice. 77 will be counted 3 times!
• The first 10, we keep.
• The next 10, we throw 77 away, because we already had it. 10 + 9. (Among the 10 numbers beginning with the digit 7, exactly 1 also ends with a 7, and we already covered that.)
• The next 14, we throw 7, 70 and 77 away, because we already had them. 10 + 9 + 11. (Among the 14 numbers that are multiples of 7, exactly 3 are multiples that “keep” the 7. 1 * 7, 10 * 7, 11 * 7.)
• 10 + 9 + 11 = 30.

And for extra credit:

Desmos

Method 1: I expand the program from above. I look for a frequency of about 0.5 and then zoom in to learn more about this general area. The result is illustrated in the Desmos. Look for the clump at 0.5 and zoom!

Method 2: The graph has a distinctive shape. Every time we run into a whole bunch of numbers, all beginning with 7, the frequency will grow. Afterwards it will go up and down a bit, until the next bunch, but on the whole it will keep growing. For our purposes, once we hit the 700000-799999 bunch, it grows past 0.5 and never dips below again.

I confirm this result via the program, looking at different thresholds. E.g. the 0.45 threshold is 71574. Look for the red stars in the Desmos file.

Anyway. N = 708588.

Method 3: A similar counting argument as above. Left as an exercise to the reader 😉

This week the #puzzle is still unknown to me, as I write this. Because I am busy. But here’s some other content for you.

Some other very fine fiddlers:

• Benjamin Phillabaum (bphillab.github.io)
• Ed Nigma’s Answer Emporium
• Jamshrivelson
• Rohan Lewis (humanrickshaw)
• Sameer Gauria

If you want to compare 2 large numbers, and remember to carry the one(s), this is how you do it.

First row in the table below: Our units. In the middle, there’s a familiar system: 1, 10, 100. This is how our own numbers are constructed. There’s a position for 1s, a position for 10s, etc. To the left and to the right, the units are not the familiar ones, but hopefully we can understand this anyway.

Second row: A sum of a lot of other numbers. The sum hasn’t been normalized to its shortest from yet. We want to normalize and then compare with the last row, another number from the same source.

Normalization example: In the second row, 15 and 15 are bolded. The jump from the first to the second column is to multiply by 2. (1/120 * 2 = 1/60.) So to normalize, we look at the first 15 and divide by 2.

• a = 1/120, b = 1/60
• 2a = b
• 15/2 = 7.5
• 15 = 1 + 7 * 2
  • 15a + 15b
  • = (1 + 7 * 2)a + 15b
  • = a + 7 * 2a + 15b
  • = a + 7b + 15b
  • = a + 22b

So in the next row, instead of 15 and 15, we have 1 and 22.

Move to the next pair of columns and repeat.

1/120

1/60

1/30

1/10

1/5

1

10

100

300

3000

18000

15	15	24	14	19	26	39	11	7	8	5
1	22	24	14	19	26	39	11	7	8	5
1	0	35	14	19	26	39	11	7	8	5
1	0	2	25	19	26	39	11	7	8	5
1	0	2	1	31	26	39	11	7	8	5
1	0	2	1	1	32	39	11	7	8	5
1	0	2	1	1	2	42	11	7	8	5
1	0	2	1	1	2	2	15	7	8	5
1	0	2	1	1	2	2	0	12	8	5
1	0	2	1	1	2	2	0	2	9	5
1	0	2	1	1	2	2	0	2	3	6
1	0	2	1	1	2	2	1	1	3	6

And yes, the end result is, that the 2 numbers aren’t quite the same.

Background: Recently I read this, in From One to Zero, Georges Ifrah :

This week the #puzzle is: Can You Rile and Grace the Polyhedron? #geometry

An alien—more specifically, an Eridian—named “Rocky” needs to pass a solid xenonite crystal to his human friend in a neighboring spaceship. The crystal is shaped like a regular tetrahedron, as shown below, and all its edges have length 1.

To safely transport the crystal, Rocky needs to create a long cylindrical tunnel between the spaceships, and then orient the tetrahedron so that it fits through the tunnel. It’s okay if the crystal fits snugly inside the tunnel—in this case, it can slide along without any friction.

What is the minimum possible radius for my tunnel so that the crystal will fit through it?

And for extra credit:

Next, Rocky wants to transport a solid crystal shaped like a regular dodecahedron to his human friend. As before, each edge has length 1.

This time, the long tunnel between the space ships can be any right prismatic shape, not necessarily a cylinder. Once again, Rocky needs the crystal to fit through the tunnel, and it’s okay if that fit is snug.

What is the minimum possible cross-sectional area for the tunnel so that the crystal will fit through it?

Can You Rile and Grace the Polyhedron?

Intermission

The results from Q1 are in, and the finest of Fiddlers, who each solved all 24 puzzles, are: 👑 Adnan Haque 👑 from New York 👑 David Kong 👑 from Toronto, Canada 👑 Josh Silverman 👑 from Glen Head, New York 👑 Lise Andreasen 👑 from Valby København Danmark 👑 Peter Ji 👑 from Madison, Wisconsin 👑 Seth Cohen 👑 from Concord, New Hampshire

Woohoo!

Solution, possibly incorrect:

Desmos

My only option this week is to guess. I guess the circle shown on Desmos. Radius 0.5.

Given more time, I would have coded a tetrahedron, that could rotate, so that I could test different options.

And for extra credit:

I don’t even have a guess.

This week the #puzzle is: Can You Pile the Primes? #primes #coding #BruteForce #oeis #subsets

Suppose you want to make two groups with equal sums using the first N2 prime numbers. What is the smallest value of N2 for which you can do this?

The answer is three! (Clearly, that wasn’t actually the puzzle.)

The first three primes are 2, 3, and 5, and you can split them up into two sets: {2, 3} and {5}. Sure enough, 2 + 3 = 5.

Your puzzle involves making three groups with equal sums using the first N3 prime numbers. What is the smallest value of N3 for which you can do this?

And for extra credit:

Now you want to make six groups with equal sums using the first N6 prime numbers. What is the smallest value of N6 for which you can do this?

Can You Pile the Primes?

Intermission

I am very excited, because this is the last fiddler of Q1, and I still have maximum points. Wish me luck.

Solution, possibly incorrect:

Program

Method 1: Write a brute force program. This actually finds a solution rather quickly. N₂ is confirmed, among other things.

Testing N1 =   1. Testing these primes: 2
................. Group 1: 2
................. Sum of groups: 2
................. N1 = 1

Testing N2 =   2. Sum of primes    5 isn't a multiple of 2, rejected.
Testing N2 =   3. Testing these primes: 5,3,2
................. Group 1: 5
................. Group 2: 3,2
................. Sum of groups: 5
................. N2 = 3

Testing N3 =   3. Sum of primes   10 isn't a multiple of 3, rejected.
Testing N3 =   4. Sum of primes   17 isn't a multiple of 3, rejected.
Testing N3 =   5. Sum of primes   28 isn't a multiple of 3, rejected.
Testing N3 =   6. Sum of primes   41 isn't a multiple of 3, rejected.
Testing N3 =   7. Sum of primes   58 isn't a multiple of 3, rejected.
Testing N3 =   8. Sum of primes   77 isn't a multiple of 3, rejected.
Testing N3 =   9. Sum of primes  100 isn't a multiple of 3, rejected.
Testing N3 =  10. Testing these primes: 29,23,19,17,13,11,7,5,3,2
................. Group 1: 29,7,5,2
................. Group 2: 23,17,3
................. Group 3: 19,13,11
................. Sum of groups: 43
................. N3 = 10

Result: N₃ = 10.

And for extra credit:

OEIS

Method 1: The same program.

N₆ has to be at least 57. But then my program gives up. Heat death of the universe kind of thing.

Testing N6 =  57.
Testing these primes:
269,263,257,251,241,239,233,229,227,223,
211,199,197,193,191,181,179,173,167,163,
157,151,149,139,137,131,127,113,109,107,
103,101,97,89,83,79,73,71,67,61,59,53,
47,43,41,37,31,29,23,19,17,13,11,7,5,3,2
Or rather, test left as an exercise for the reader.

Method 2: Try to find a solution with the first 57 primes. And I do!

• 269,199,197,139,137,89,59,37,19
• 263,211,181,163,109,103,53,47,13,2
• 257,223,173,167,107,101,71,41,5
• 251,227,179,157,127,83,61,31,29
• 241,229,193,151,113,97,67,43,11
• 239,233,191,149,131,79,73,23,17,7,3

N₆ = 57.

Method 3: Look for an OEIS sequence. Found. Results confirmed.