Maths
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 This topic has 13 replies, 12 voices, and was last updated 6 days, 5 hours ago by KingswinfordWarriorsAlumni.

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25th October 2024 at 10:04 am #97971frazza238Participant
USBC_is_supreme
@Astralica has exhausted all solutions as:
As 2^n is a logarithm, it will double for every increase of n.
1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120,
If a(or b)! > 4, the final digit will always be 0. 2^n cycles final digits in the order 2, 4, 8, 6. Therefore, both a and b cannot be greater than 4.
When a(or b) is greater than 4, the other must be 2, 4, 6. No factorial exists that is either 2, 6 or 24 less than any 2^n when n > 3.
This means that the resulting possibilities are:
1! + 1! = 2^1: 1 + 1 = 2 {1:1:1}
2! + 2! = 2^2: 2 + 2 = 4 {2:2:2}
3! + 2! = 2^3: 6 + 2 = 8 (and viceversa) {3:2:3} {2:3:3}So:
a, b <= 3 for valid solutions. If a, b > 4: will always end in 0 which does not occur in the 2^n series. For values a,b where a <= 3 and b>= 4:
a = 1: 2^n – 1 does not exist as a factorial as always odd. ( also followed for a=3).
a = 2: 2^n – 2 = 2(2^(n1)1) which is always singly even (only 1 factor of 2) therefore cannot be factorial >= 4(This final bit i was assisted with. This is not entirely my own work.)
25th October 2024 at 10:08 am #97975USBC_is_supremeParticipantThis seems like more of a brute force method @ByteInBits @Astralica What I thought is that for the current possibilites, 2 is the highest common factor for a! and b!, so you can factorise out 4: 4(…) If 3 is a common factor, then you can factor out 9, making it impossible for any other triples.
26th October 2024 at 2:08 pm #98001OliversParticipantHow would you methodically go about finding every possible permutation given the bits you want to rearrange?
26th October 2024 at 2:11 pm #98009RickOSheaParticipantExcellent question USBC_is_supreme. Here are my thoughts along the same lines as the previous solutions provided…
Given a! + b! = 2^n
1. If a = b, then:
2a! = 2^n
a! = 2^(n1)No solution for a >= 3, as 3 is not a factor of 2^(n1).
2. If a < b, then:
a!(1 + b!/a!) = 2^n, as a! is a factor of b!.No solution for a >= 3, as 3 is not a factor of 2^n.
3. If a > b, then:
The same applies as 2 above, but with a and b swapped.
28th October 2024 at 9:59 am #97722ByteInBitsParticipant@CaldewBAAlfie You are indeed correct. A=2,B=3,C=6,D=8 You gave a nice logical workout!
For me it was the C – B = B that gave a quick solve as C must = 2 * B and A * B must = C,
with 2 ways of calculating D from A and C for checking values it just fell together ðŸ˜‰28th October 2024 at 10:00 am #97788RickOSheaParticipantFollowup post to #97505 – coin placing puzzle. It is relatively easy to find a solution to the puzzle by trial and error, although it does take me a while. I haven’t found a strategy that always works without some iteration.
Here is one solution:
00X00000
0000X000
0X000000
0000000X
X0000000
000000X0
000X0000
00000X00In this case, there will be just one rotation of 90 degrees providing a another solution. Rotation of 180 degrees returns to the original pattern.
Having attempted to simulate all possible coin positions and checked for compliance with the rules, I believe there are 92 different solutions to the puzzle, but they are grouped with up to three rotations and up to four reflections of the same basic solution and then additional reflections of each rotation. It’s looking more like nine unique solutions excluding symmetries.
28th October 2024 at 10:00 am #97792OtocinclusParticipantI took a similar approach, but started by rearranging CB=B to give C=2B, then because A*B=C, A*B=2B and A must equal 2.
If 4*A=D then D=8.
If A+C=D then C=82 which means C=6.
And lastly if A*B=C, then B=6/2 and so B=3.
So multiple ways to approach it – I really like the puzzle!
Perhaps there is a way of setting it slightly differently so you can only find one piece of information first?28th October 2024 at 10:00 am #98066KingswinfordWarriorsAlumniParticipantNice question @USBC_is_supreme! Here is a question I thought of that is similar to yours:
Find all triples of positive integers a,b,p, with p being a prime, such that a^p + b^p = p!. (Please tell me that my formatting is correct!)
[Note: You will (probably) need to look up Fermat’s Little Theorem, if you don’t know it already. It will prove quite useful. No other weird background knowledge is necessary.]
I hadn’t seen this problem before and it looks really interesting. Am very much enjoying the discussion on this thread! Keep them coming. Harry

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