Maths
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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.)
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.
How would you methodically go about finding every possible permutation given the bits you want to rearrange?
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.
@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 😉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
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.
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?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
@ByteInBits, #97270
I’m surprised no-one has said the other (trivial) solution A=B=C=D=0 yet…!@Kingswinford-Warriors-Alumni 🙂 🙂 🙂 Sure there was nothing to it!!!
#98066 (Fermat’s Little Theorem conundrum)
So… has no-one found a solution yet? Is my question too devious? (Cue evil laugh here)
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=2This is my only solution so far, nice problem though
@frazza238, #98502
Well done for getting this far! Some comments:
- You are correct that the only solution in (a,b,p) is (1,1,2). Congratulations!
- You are also correct that p divides a+b – or, if you like, that a+b=0 modulo p. (I assume this is what you mean by 0=-a-b.) However…
- …the point above does not necessarily imply a=b=0. (e.g. (a,b,p)=(2,4,3) satisfies said conditions.) For this bit you can definitely assume p>2 (you did the p=2 case!), and see if you can put some bounds on a and b. [Hint: Why must a<p? And can you strengthen this?]
Good luck!
#98066
So, is no-one up to the challenge?Since this challenge has been out for a while now, here are some hints. All have been encrypted with the ROT47 cipher (I am using ROT47 rather than ROT13 because of the abundance of punctuation involved). Good luck!
[*I will use the acronym FLT throughout to mean Fermat’s Little Theorem, not Fermat’s Last Theorem! Even I can agree that the latter would be a bit much…]
HINT 0 (solutions)
%96 @?=J G2=F6 @7 W2[3[AX E92E H@C<D :D W
[[aX]HINT 1 (FLT)
u6C>2EVD {:EE=6 %96@C6> DE2E6D E92E[ 7@C 2?J AC:>6 A 2?5 :?E686C < 4@AC:>6 E@ A[ </WA\X 92D C6>2:?56C@? 5:G:5:?8 3J A] W~C </WA\Xl>@5 A]X |@C6 FD67F==J 7@C E9:D AC@3=6>[ :E 2=D@ >62?D E92E 7@C 2?J AC:>6 A 2?5 p}* :?E686C <[ 6 C6>2:?56C @? 5:G:5:?8 3J A] %9:D >62?D E92E H6 >2J C6A=246[ D2J[ 2/A 3J 2 :? E96 AC@3=6> 2?5 DE:== AC6D6CG6 E96 C6>2:?56C @? 5:G:5:?8 3J A]]]HINT 2 (first steps)
pD 9:?E65 2E :? w:?E `[ FD6 u{% E@ >@5:7J E96 =67E\92?5 D:56 @7 E96 6IAC6DD:@?[ H9:=DE AC6D6CG:?8 :ED C6>2:?56C @? 5:G:5:?8 3J A] W%92E :D[ 4@?D:56C 3@E9 D:56D >@5 A]X s65F46 E92E 2Z3l_ W>@5 AX[ @C E92E A 5:G:56D W2Z3X] </pre> <strong>HINT 3</strong> (bounds on a,b,a+b) <pre> w:?E a DE2E65 E92E A 5:G:56D 2Z3] x7 Ala E96? H6 92G6 2/aZ3/ala[ H9:49 :D 62DJ E@ 4964<[ D@ 2DDF>6 Ama] }@E6 E92E A :D @55] {6E ?lWAZ`X^a] (C:E6 ?/A 2?5 AP 2D 7@==@HD WG:6H :? 2 >@?@DA2465 7@?EXi ?/A l ,?Y?. Y ,?Y ? . Y ]]] Y , ? Y ? . Y ? AP l ,YA. Y ,aYWA\X. Y ]]] Y ,W?\XYW?ZX. Y ? !C@G6 E92E 6249 @7 E96 3C24<6ED 7@C ?/A 2C6 =2C86C E92? E9@D6 7@C AP] w6?46[ D:?46 2/A k AP k ?/A[ 565F46 E92E 2klW?\`X] $:>:=2C=J D9@H 3klW?\`X[ 2?5 96?46 D9@H E92E 2Z3kA]HINT 4 (final steps)
WpDDF>:?8 Ama 96C6]]]X w:?E a D2:5 E92E A 5:G:56D 2Z3] w:?E b D2:5 E92E 2Z3kA] 2 2?5 3 2C6 A@D:E:G6[ D@ H6 2=D@ 92G6 2Z3m_] s6C:G6 2 4@?EC25:4E:@?]
Who said that geometry was boring? Here are three interesting questions related to a triangle ABC. Enjoy!
First we assume that ABC is an equilateral triangle of side length 1.
QUESTION 1
A farmer, Kim, wishes to choose a point P inside ABC, and connect P to the three sides AB,BC,CA using three pieces of fencing. Where should Kim place P, and how much fencing will he need?I am guessing Kim wants to do this as cheaply as possible? Harry
QUESTION 2
Kim the farmer (see Question 1) is now chosen to connect A,B,C to three points X,Y,Z (in that order), which are on the interiors of sides BC,CA,AB (in that order). He decides to do this using three paths. However, the paths cannot cross any of the sides of ABC, due to walls (or some other contraption that blocks travel) being present, and they cannot cross each other. Can he lay down the three paths?[Just to make it clear: the three paths must be in the plane of triangle ABC, and take the form of continuous curves which do not touch any part of triangle ABC apart from their two endpoints. All three paths must be pairwise disjoint. They can arrive at A/B/C/X/Y/Z from the inside or the outside of ABC. OK? Good.]
Now we let ABC be any triangle. The only theorem you need (which is basically stolen from GCSE circle theorems) is:
Let W,X,Y,Z be a convex cyclic quadrilateral. (That is, let W,X,Y,Z lie on a circle in that order.) Then:
- Angles WXZ and WYZ are equal.
- Angles WXY and WZY add to 180°.
What isn’t taught at GCSE is that the converse of this is true also. So, for example:
If WXYZ is a convex (informally: nice) quadrilateral, and angles WXZ and WYZ are equal, then W,X,Y,Z lie on a circle in that order.
And here is the question!
QUESTION 3 (Harder!)
Let Γ be the circumcircle of ABC. (That is, let Γ be the (unique) circle which passes through all of A,B,C.) Consider the three circles Γ_A, Γ_B, Γ_C created by reflecting Γ over BC,CA,AB respectively.Prove that Γ_A, Γ_B, Γ_C all concur at a point. (Extra marks if you can prove what that point is!)
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