Puzzles

[ByteInBits]

It has been 4 weeks now so time for the E-Bots answer.

Answer : 20 MPH
For this forum’s audience I was seeking this simplified answer.

Now before you throw your arms in the air screaming but, but, but what about. .
I will accept 20.05 as the more complex math answer.

Consider:
Speed-E travels a total distance equivalent to twice the column’s length (2 × 10 = 20 miles) in 1 hour, and the effect of the column’s motion (1 mph) cancels out due to the distance “lost” going to Head-E being balanced by the distance “gained” returning to End-E. This results in a constant speed of exactly 20 mph in the ground frame, as verified by the total distance travelled (10.5263 miles forward + 9.4737 miles back = 20 miles) in 1 hour.

The calculation yielding ( 10 + sqrt(101) approx 20.05 mph ) arises from a more complex approach that strictly accounted for Speed-E’s exact position matching End-E’s at ( t = 1 ), but my approach shows the problem intends a simplification where

the total distance

governs the speed, making 20 mph the correct and intended answer.

[ByteInBits]

@@John_Musgrave Informational Post

Seems like both your question and answer were put up!
Maybe you did not indicate to the keymaster which is which?

Tip for every one, at the top of you answer post put the following in capital letters

########## ANSWER – DO NOT POST THIS ##########

You will be posting your answer again in response later.

[_madness_]

I have a solution to the math puzzle in today’s news (2025-10-19).
Harry and or an elf can edit this out so as not to spoil it for anyone else.
He or elf can replace it with “[correct solution]” or “[utter garbage]”, as the case may be.
(I’m really only posting this to find out which.)

The solution rests on the assumption that the puzzle-creator is not an idiot.
This is a big assumption, when comparing to some other puzzles that have passed our way.
The line segment seems to meet the top of the rectangle at a random point, but since
the puzzle-creator is not an idiot, that point must not be relevant to finding the answer.
The segment also passes through a corner on the left-most square, so that must be the relevant thing.
There are no dimensions, absolute or relative, given in the figure. Since the puzzle-creator
is not an idiot, this means that they are not needed to find the answer. So consider the limiting
case in which the rectangle is 3 squares wide and 2 squares tall. In that case, three of the squares
lie flat against the bottom of the rectange, and the line segment is at a 45° incline.
Therefore, the answer is 45°.

(There is another limiting case, in which the rectangle is twice as wide as it is tall.)

[ByteInBits]

Ho RATS, just what is the next number in the sequence?

1,2,4,8,16,?

If you say 32 then you could be right.

However, if I said the next number is 77 becuase the sequence I have in mind
continues 145 668. . . so we have:

1,2,4,8,16,77,145,668,?

Then what is the next number?

I will be very interested to see if anyone gets this! Harry

[Robb27]

@ByteInBits
1345

But without the RATS clue (which nudged a memory that I’d seen this before and then I had to think a bit to remember) that would be a difficult sequence to figure out!

[ByteInBits]

@Robb27 you seem to be well versed in these things, it is of course the correct answer.

Now that Robb27 has given the hint anyone else needs to give the next in the sequence as their answer.

1,2,4,8,16,77,145,668,1345,?

[DrMontysCiphers]

EDITED BY HARRY

Thanks Dr Monty, would you mind reposting with the solution and the method of encryption? Harry

[F6EXB_the_frenchy]

ByteInBits
#111827

1,2,4,8,16,77,145,668,1345,? ==> 1345 + 5431 = 6776 ==> 6677

1,2,4,8,16,77,145,668,1345,6677

[ByteInBits]

@F6EXB_the_frenchy Your answer is of course correct.

That is amazingly well done assuming you had no online help!
As you have given the workings that brings it to an end.

Thanks for your participation 🙂

RATS INFORMATIONAL
==================
The RATS sequence, named after mathematician John Horton Conway’s acronym
for “Reverse, Add, Then Sort,” is a divergent sequence of positive integers
in base 10, generated iteratively from an initial number (typically starting
with 1). It is conjectured that, regardless of the starting positive integer,
the process either enters this specific divergent sequence or converges to a
finite cycle.

### Iterative Generation Process
Given the current term a_n (a positive integer with no zero digits, sorted
in non-decreasing order):

1. Compute the reverse of a_n, denoted rev(a_n).
2. Add them: s = a_n + rev(a_n).
3. Remove all zero digits from s.
4. Sort the remaining digits in non-decreasing order to obtain a_{n+1}.

This yields the sequence OEIS A004000:
1, 2, 4, 8, 16, 77, 145, 668, 1345, 6677, 13444, 55778, 133345,
666677, 1333444, 5567777, 12333445, 66666677, 133333444, 556667777, …

### Example Iteration
Starting from 668:
– Reverse: 866
– Sum: 668 + 866 = 1534
– Remove zeros: 1534 (none present)
– Sort digits: 1345

Thus, the next term is 1345.

The sequence grows indefinitely, with each term adding roughly one more
digit, alternating patterns of increasing 1s/2s/3s and 5s/6s/7s/8s/9s.
No closed-form formula exists for the n-th term, as it is defined
purely by this recursive digit-manipulation rule.

============================================= Prove RATS conjecture

The RATS conjecture posits that for any positive integer with no zero digits,
the RATS sequence—generated by reversing the digits, adding the number to its
reverse, removing zeros, and sorting the digits in non-decreasing order—either
enters the divergent sequence (1, 2, 4, 8, 16, 77, 145, 668, 1345, 6677, …)
or converges to a finite cycle. As of October 23, 2025, the conjecture remains
unproven, with no complete proof or counterexample identified in mathematical
literature. Below, I outline the challenges, known results, and approaches to
tackling the conjecture, rather than a definitive proof, since none exists.

### Understanding the RATS Sequence
The RATS sequence (OEIS A004000) is defined iteratively:
1. Start with a positive integer a_n (no zero digits, digits in
non-decreasing order).
2. Compute its reverse rev}(a_n).
3. Add: s = a_n + rev}(a_n).
4. Remove all zero digits from s.
5. Sort the resulting digits in non-decreasing order to get a_{n+1}.

For example, starting with 1:
– 1 \to rev}(1) = 1 \to 1 + 1 = 2 \to 2 (no zeros, sorted).
– 2 \to rev}(2) = 2 \to 2 + 2 = 4 \to 4.
– 16 \to rev}(16) = 61 \to 16 + 61 = 77 \to 77.
– 77 \to rev}(77) = 77 \to 77 + 77 = 154 \to 145.

The sequence 1, 2, 4, 8, 16, 77, 145, 668, 1345, … appears to diverge, with
terms roughly doubling in digit length periodically.

### Known Results
Computational evidence suggests that most starting numbers either:
– Join the divergent sequence (e.g., starting at 1, 3, 5, 9, …).
– Enter a finite cycle, e.g., 89 ? 88 (cycle of length 1),
or 1189 ? 112 ? 224 ? 448 ? 899 ? 889 ? 888 ? 1189 (cycle of length 7).

Key observations:
– **Divergent Sequence**: The sequence starting from 1 grows indefinitely, with
patterns like increasing counts of 1s/2s/3s followed by 5s/6s/7s/8s/9s.
– **Cycles**: Some numbers lead to cycles (e.g., 88, or the 7-term cycle above).
These are rare but documented.
– **No Counterexamples**: Extensive computations (up to millions of starting values)
have not found a sequence that neither diverges nor cycles, supporting the
conjecture.

### Challenges in Proving the Conjecture
Proving the RATS conjecture is difficult due to:
1. **Nonlinear Dynamics**: The process involves arithmetic (addition), digit
manipulation (reversal, zero removal, sorting), and base-10 dependence, making it
resistant to standard number-theoretic tools.

2. **Divergence vs. Cycles**: The conjecture requires showing that *every* starting
number either joins the known divergent sequence or enters a cycle, which demands
analyzing all possible digit combinations.

3. **Growth Patterns**: The divergent sequence’s growth (approximately doubling
digits) complicates analytical bounds, as carry-overs in addition create
unpredictable digit changes.

4. **Cycle Detection**: Identifying all possible cycles is challenging, as their
lengths and structures vary unpredictably.

### Approaches to a Proof
While no proof exists, here are strategies mathematicians might pursue:

1. **Invariant Analysis**:
– Seek invariants or semi-invariants in the sequence. For example, the number of
digits or digit sums might provide clues, though sorting disrupts simple patterns.
– Consider the effect of carries in a_n + rev}(a_n). Carries propagate
leftward, potentially stabilizing into patterns (e.g., repeated 5s/6s in divergent
terms).

2. **Cycle Characterization**:
– Prove that all cycles have a specific form. For instance, known cycles
(like 88 or the 7-term cycle) have symmetric or near-symmetric digit structures.
– Hypothesize that cycles arise when a_n + rev}(a_n) produces a
number with digits that, after sorting, loop back to a previous term.

3. **Divergent Sequence Dominance**:
– Show that the divergent sequence (1, 2, 4, 8, …) is an “attractor” for most
starting numbers. This could involve analyzing the probability that random starting
numbers merge into this sequence via repeated iterations.
– Study the transition from numbers like 16 to 77, where carries and sorting
shift digit patterns significantly.

4. **Digit Pattern Analysis**:
– Model the sequence as a dynamical system on digit strings. For example, treat
numbers as sorted strings and analyze transformations under reversal and addition.
– Use combinatorial arguments to bound the number of iterations before a number
enters a known sequence or cycle.

5. **Computational Exhaustion**:
– Extend computational searches to identify all cycles or a counterexample
(a sequence that neither diverges nor cycles). This is impractical for large numbers
but useful for small cases.
– For instance, all single-digit numbers (1–9) have been tested: 1, 3, 5, 9 join
the divergent sequence; 2, 4, 6, 8 merge at 2; 7 hits 88; 8 hits 88; 9 joins at 145.

### Example: Testing a Starting Number
Consider 89:
– 89 -> rev(89) = 98 -> 89 + 98 = 187 -> 178.
– 178 -> rev(178) = 871 -> 178 + 871 = 1049 -> 149.
– 149 -> rev(149) = 941 -> 149 + 941 = 1090 -> 19.
– 19 -> rev(19) = 91 -> 19 + 91 = 110 -> 11.
– 11 -> rev(11) = 11 -> 11 + 11 = 22 -> 22.
– 22 -> rev(22) = 22 -> 22 + 22 = 44 -> 44.
– 44 -> rev(44) = 44 -> 44 + 44 = 88 -> 88.
– 88 -> rev(88) = 88 -> 88 + 88 = 176 -> 167 -> 88 (cycle).

This shows 89 enters a cycle at 88, consistent with the conjecture.

### Why No Proof Yet?
The conjecture’s reliance on base-10 digit operations makes it resistant to
algebraic or analytic methods. Unlike similar problems (e.g., the Collatz conjecture),
the sorting step introduces combinatorial complexity, and zero removal disrupts
continuity. The divergent sequence’s growth suggests most numbers are “attracted”
to it, but proving this for all cases requires ruling out undiscovered cycles or
divergent paths.

### Current Status
As of 2025, the conjecture is supported by extensive computational evidence but
lacks a formal proof. Research continues in number theory and dynamical systems,
with parallels to problems like the Lychrel process (reverse-and-add without sorting).
A proof would likely require a breakthrough in modeling digit transformations or a
novel invariant that captures the sequence’s behavior.

[bridges]

While trying (unsuccessfully) to solve your puzzle I discovered multiple sequences beginning 1,2,4,8,16,77.
I thought the next term was supposed to be 145,668 with the comma separating thousands (rather than two terms 145 and 668) and even found such a sequence 1,2,4,8,16,77,145668 but the next term wasn’t 1345. Maybe I should post those sequences as puzzles here.

[Kingswinford-Warriors-Alumni]

===== EXCURSIONS IN GEOMETRY =====

(This is my first post to the Cipher Challenge forum this year, so hello everyone! I’m looking forward to the Challenge getting started in earnest in two weeks’ time!)

I really liked last week’s Sunday puzzle about the geometry diagram. I actually had a slightly different solution to that given on the Mathforge website (https://notes.mathforge.org/notes/published/four+squares+inside+a+rectangle+solution), so I thought you might like to see it!

Using their notation, here is my way of proving that ∠FCE is 45°, using circle theorems. [Hopefully the HTML code is correct!]

• Let M be the midpoint of BE.
• Since MB = ME = MF, the circle Γ through B, E and F has centre M, which also means that Γ has diameter BE.
• However, ∠BCE is 90°. Thus, the circle through B, C and E also has diameter BE, which means that this circle is also Γ.
• Hence, B, C, E and F all lie on a circle centred at M. This means that ∠FCE is half of ∠FME.
• Now we just use the fact that ∠FME is 90° to get the final answer of 45°.

(And then we use parallel lines to conclude that this is also the answer to the question given.)

Anyway, in the same spirit, here is a geometry puzzle that I thought up of. See what you make of it:

Points A, B and C lie in the plane, with M being the midpoint of BC. Given that the lengths of AB, AM and AC are 3, 4 and 5 respectively [in whatever units you choose], find the area of the triangle ABC [in said units squared].

(As a bonus question, calculate the area of ABC if AM=2 instead.)

Good luck!

[F6EXB_the_frenchy]

@Kingswinford-Warriors-Alumni
#112062

Knowing two sides of a triangle and the median relative to the third side, we can calculate that side. We note that when the median measures 4, the triangle is special.
Then, Heron’s formula allows us to calculate the area of this triangle.

[ByteInBits]

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Post #107332 And so the contestants each walk away with an empty bucket as a constellation prize?!
😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉 😉

[F6EXB_the_frenchy]

Were is #107332?
In which topic ?

[upsidedown]

@F6EXB_the_frenchy: it’s on the first page

@ByteInBits: answer for #107332

It depends whether the contestants round the 1/5 that’s put in the hole up or down.

1. If they round up, 3128
2. If they round down, 3122

We’re after the smallest number of coins, so I’ll say 3122.

Supposing the coins were a continuous quantity, I have a closed form for the initial number of coins c₀ in terms of: n, the final number of coins per player; p+1, the total number of players; and 1-r, the ratio which is put in the hole.

The minimum number of coins before n of the players went into the room together was np + 1

I’m now only going to consider the iterative process of each player going into the room alone. If the number of coins before going in the room was cᵢ, then cᵢ₊₁ is defined by:

cᵢ₊₁ = r(cᵢ + 1)

The closed form for cᵢ is:

cᵢ = rⁱc₀ – (1 – rⁱ)r / (1 – r)

So if cₚ = np + 1,

rᵖc₀ – (1 – rᵖ)r / (1 – r) = np + 1

c₀ = (np + 1 + (1 – rᵖ)r / (1 – r)) / rᵖ

…although this is useless because the coins are not, in fact, a continuous quantity 🙂

[Author]

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