Maths

[frazza238]

USB-C_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 vice-versa) {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^(n-1)-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.)

[USB-C_is_supreme]

This 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.

[Olivers]

How would you methodically go about finding every possible permutation given the bits you want to rearrange?

[RickOShea]

Excellent question USB-C_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^(n-1)

No solution for a >= 3, as 3 is not a factor of 2^(n-1).

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.

 

[ByteInBits]

@CaldewBA-Alfie 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 😉

[RickOShea]

Follow-up 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
00000X00

In 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.

[Otocinclus]

I took a similar approach, but started by rearranging C-B=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=8-2 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?

[Kingswinford-Warriors-Alumni]

Nice question @USB-C_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

[Kingswinford-Warriors-Alumni]

@ByteInBits, #97270
I’m surprised no-one has said the other (trivial) solution A=B=C=D=0 yet…!

[ByteInBits]

@Kingswinford-Warriors-Alumni 🙂 🙂 🙂 Sure there was nothing to it!!!

[Kingswinford-Warriors-Alumni]

#98066 (Fermat’s Little Theorem conundrum)

So… has no-one found a solution yet? Is my question too devious? (Cue evil laugh here)

[frazza238]

Ive had it for a while but didnt upload because i might be missing a solution. Currently I have:

a^p + b^p = p!
a^p + b^p = p(p-1)!

Therefore,
(a^p – a + b^p – b) / ((p-1)!) = p
a^p + b^p = a^p – a + b^p – b
0 = -a -b
so, -a=b and -b=a therefore a=0,b=0 BUT as a,b ^p, even powers ignore the negative effect so p=2
a^2 + b^2 = 2 which gives:
a=1, b=1, p=2

This is my only solution so far, nice problem though

[Author]

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