Math Is Fun Forum

Here's what I do:

6 and 10 both have more than two factors so there are several alteratives to try.

6  10              6   5             3   10         3   5

1   1               1   2             2    1          2   2

By trial I found that 3 times 5 - 2 times 2 gives 11

(3x     )(2x     )

The 3x must multiply the 5 so

(3x   2)(2x   5)

I want + 11 so that fixes the signs

(3x -2)(2x + 5)

If you want to check first if an integer solution is lurking there you can check B^2 - 4AC to see if it has an integer root:

B^2 - 4AC = 121 + 240 = 361. This has root 19 so if I used the quadratic formula a simple 'solution' is available.  That check saves wasting time looking for a factorisation if there isn't one.

Bob

Hello Tony.

Welcome to the forum.

I'm pretty certain it's a German Shepherd.

Bob

Bob

You're welcome.  I enjoyed working it out so you did me a favour by setting the problem.

My 'complete proof' is meant to cover all cases, so if I'm right it does carry on doing this as infinitum.  The problem with proves is, sometimes, they have flaws in them.  Andrew Wiles surprised the mathematical world by coming up with a proof of Fermat's last theorem.  His proof was taken away by other mathematicians for close scrutiny. They did find some points where he had jumped ahead too fast and he went back to the drawing board to clean up the criticisms (which he did!).

The common multiple bit is fairly robust.  Then I had to show the lowest common multiple.  I think it's ok too but maybe others will spot a weakness.  When I was at university we spent a lot of time showing even the most obvious results are generally true so I've had practice.

You have to be careful assuming results just because they work for a few easy numbers. Have a look here:

http://www.mathisfunforum.com/viewtopic.php?id=30690

Another example is the statement N^2 + N + 41 is a prime number generator.  That is if you substitute N = 1, 2, 3 etc
each time you get a prime number.

If you try some N values it look promising. But the statement is FALSE. Can you find a value of N for which it fails?

Has someone already come up with your theorem? Not as far as I know. I tried an internet search and came  up with nothing.  So I suggest you claim it as yours and give it a fancy name.  The history of maths is full of disputed claims (eg. Leibnitz and Newton over calculus) so you'll be in with a select crowd.

But now, can you pull this out of the sphere of 'pure maths' and find a useful application?

Best wishes,

Bob

Odd cases.

I used MS WORD and lots of copy and paste to construct some evidence:

N = 3
123 bar 2 sem12 bar 32 sem1 bar 232 bar sem1232

repeats after 4 bars

N = 5
12345 bar 432 sem12 bar 34543 bar 2 sem1234 bar 5432 sem1 bar 23454 bar 32 sem 123 bar 45432 bar sem12345432

repeats after 8 bars

N = 7
1234567  bar 65432 sem12 bar 3456765 bar 432 sem1234 bar 5676543 bar 2 sem123456 bar 765432 sem1 bar 2345676 bar 5432 sem123 bar 4567654 bar 32 sem12345 bar 6765432 bar sem123456765432

repeats after 12 bars

N = 9
123456789 bar 8765432 sem12 bar 345678987 bar 65432 sem1234 bar 567898765 bar 432 sem123456 bar 789876543 bar 2 sem12345678 bar 98765432 sem1 bar 234567898 bar 765432 sem123 bar 456789876 bar 5432 sem12345 bar 678987654 bar 32 sem1234567 bar 898765432 bar sem1234567898765432

repeats after 16 bars

So the repeat starts after the point where bar and sem occur next to each other .

N = 3  at bar   5
N = 5  at bar   9
N = 7  at bar 13
N = 9  at bar 17

formula 2N - 1 which is what you have stated.

Complete proof:

If N is the bar length then new bars start every N notes.

The 1 to N and back down sequence is always 2N - 2 that's every note twice less once for each end point.

So the bars will start again at every common multiple of N and 2(N-1).  But is that the first? ie Is it the lowest common multiple?

When N is even, no it isn't because N and N-1 have no common factor (except 1) and N is even so it does have a common factor with 2(N-1) namely 2. So the LCM is N(N-1) leading to a repeat bar at N-1 plus 1 ie at N.

When N is odd, N and 2(N-1) have no common factors (except 1) so the reset is at N.2(N-1) and so is after 2(N-1) bars. So the repear bar occurs at 2(N -1) plus 1 ie at 2N -1.

Bob

hi Frank b1

Welcome to the forum.

I'd better start by saying I cannot play a piano.  But we have one in the house and others can play it.

123, 212, 321, 232, 123

I'm not following how each pattern progresses to the next. Why not 312 or even 333 ?

Please would you explain how you are going from one to the next and why 212 follows from 123 for example.

Bob

I find 'man in lift' problems take some thinking about.  I'd like to chat about the general principles before taking on your problem.

If the lift isn't moving then 10 m/s/s is all there is and the reaction upwards opposes -10 x 90 so 900N up.

But reactions are not always up as the following will hopefully show.

Suppose the lift cable breaks and the lift tumbles down with only gravity acting. The man is now in free fall, effectively weightless and he could float off the floor so the reaction of the floor on him is zero.  If we compare with a parachute drop, the wind rush tells us we're falling. In our imaginary lift drop the air moves with the man so he'd have no way of telling that he wasn't able to float.

Now if we imagine a lift that is so powerful it can accelerate downwards at 20 m/s/s. The lift would drop faster than the man because he's already floating up, until he bangs his head on the ceiling. The reaction of the ceiling on him is definitely downwards.

But how much?  He is trying to fall at 10 m/s/s but the lift is pushing him at 20.  Effectively the force on him is 90 x (20-10) downwards.  So you need to subtract the lift's acceleration from gravity to determine the effective acceleration in the frame of reference.  It's like the moving train frames of reference.  If you 'bring the lift to rest' by applying an upwards acceleration of 2 m/s/s the gravity is similarly given that upwards acceleration so -10 + 2 = -8 m/s/s.

So I think the reaction (upwards) is ( 10 - 2 ) x 90 N

In really fast lifts you will notice the effect as your internal organs rise up inside your body.  Many years ago I stayed in the Luxor hotel in Vegas. The hotel is in the shape of a square based pyramid and the lift shafts follow the sloping line so not ony do you notice the stomach effect but also you tend to sway inwards because the motion is no longer just up/down but includes a sideways component. Really strange but then it was Vegas.

Bob

Wrong! It doesn't. The bot is confusing actual velocity with how much we want to alter the velocity.  It takes more effort if we want to alter velocity a lot, but Newton's law just says velocity stays constant until we apply an external force.  The amount of force determines how much the velocity changes but the start velocity is irrelevant.  That's why we have frames of reference; otherwise we've got to take account of all the possible velocities (orbits, movement of galaxies etc.)

It takes the same force to change velocity from say 10 m/s to 5 m/s whether the start velocity is 10 m/s or 10 million m/s.

Conclusion: Don't trust bots to tell you anything.

Bob

hi polarisgleam

Welcome to the forum.

If you look back through this thread you'll see that it's not just a case of adding 13. That's too easy for this forum.  Members try to find a calculation (the more interesting, the better) to make the next number.

Such as 2 x 2 x 7 x 13 x 83 = 30212.

Bob

It has to be like that because of all the other motions that we're all involved in.  eg. the Earth is hurtling through space (could estimate the speed but hopefully my point stands without that).  If you had to stop that speed too you'd need a lot of force just to take the dog for a walk!

Bob

There is a formula for the sum of n terms.  The sum to infinity will be infinite so the formula for that is

Do you know the finite sum formula? Would you like me to derive it?

Bob

I did! That looks very good.  I cannot install code; only MIF can do that.  Try emailing him at funmath@gmail.com

Bob