Go up one from k1, and test the block from there to F, using no more than k-1 drops.

k2 = k1 + 1

Shouldn't that be k2 = k1 -1 ?

This generates the series k1, k1+1, k1+2, ...

which has to sum to F.

Sum: k*(k+1)/2 + k*k*(k+1)/2

= (k+1)*(k+1)/2

so F = (k+1)^2/2

so k = sqrt(2F)

I think that is Sum = k(k+1)^2 over 2

Bob

]]>We're testing in blocks, in effect.

:

o 0-k1

o k1+1 - k1+k2

, etc.

Once we've finished the first block, we know that floors 1-k are no-break - and so that the break floor occurs above there. We're into the second block. k2 is analogous to k1 in its block.

But where, exactly, is k2?

We have now dropped one ball, so we can't take as many drops for this block as we did for the first - so let's try just one less.

Go up one from k1, and test the block from there to F, using no more than k-1 drops.

k2 = k1 + 1

This generates the series k1, k1+1, k1+2, ...

which has to sum to F.

Sum: k*(k+1)/2 + k*k*(k+1)/2

= (k+1)*(k+1)/2

so F = (k+1)^2/2

so k = sqrt(2F)

In my solution I got to the point where I wanted

b + (b-1) + (b-2) + ... + (a+3) + (a+2) + (a+1) = 99

If sum(n) means the sum of the natural numbers up to n then my formula can be re-written as

sum(b) - sum(a)

There's a formula for the sum of n natural numbers which can be worked out as follows:

sum(n) = n + (n-1) + (n-2) + ... + 3 + 2 + 1

sum(n) = 1 + 2 + 3 + ... + (n-2) + (n-1) + n

adding

2 x sum(n) = (n+1) + (n+1) + (n+1) + .... + (n+1) + (n+1) + (n+1) = n x (n+1)

so

Bob

]]>Let F be the number of floors, and k be the number of the floor that it's optimal to drop the first ball from.

Drop a ball from the kth floor

(We don't know what the value of k is yet; it's an unknown with only the property that we've set for it (it's the optimal first-drop floor).)

If the ball doesn't break, we still have two balls - and only F-k floors to cover.

(We've dropped a ball once so far.)

If it does break, we have k-1 floors left to test, with only one ball left to test them with.

And the only way to do that is to start at the bottom floor and test each floor above that successively until either it breaks or we get to the (k-1)th floor (we've tested the kth floor)

(We could just continue on up testing with the second ball - but testing only one floor at a time is not efficient enough for the optimal solution.)

State: 2 balls, F-k floors to test.

We'll climb up a few floors to do our next test drop -- but how many?

If we go up too much farther, we won't be able to test all the remaining floors below that with a single surviving ball - which violates a condition of our effort.

Yes, I agree that we should end up with a nice clean solution.

And it'd be nice to connect it up to the old problem statement, somehow.

]]>There was a command square brackets code that allowed a poster to show their line of LaTex code completely without activating the interpreter. I don't know why that doesn't now work. But I've found a way round it. I've deliberately left out a close bracket. Then the code command shows the rest so you just have to re-insert the missing bracket.

Here's the first example using that trick.

`[math V_{sphere}=\frac{4}{3}\pi r^3[/math]`

But there's an easy way to see what the code is.

In the first example I clicked on the LaTex and got this: V_{sphere}=\frac{4}{3}\pi r^3

You can always do this to see what the code was.

I've decided to re-write post 4 properly in case anyone else is misled by it.

Bob

]]>Is there a switch for that that should be used on that page?

]]>Don't worry about taking finite time to reply!

]]>Apologies for thinking 9 is a prime. Still it doesn't alter the proof. I'm wondering whether to make it a bit neater now I've finished developing it.

Also sorry that I've taken so long to reply. I had forgotten I was still logged in from earlier so I've only just spotted your reply.

The maths characters are done using a code called LaTex. The FluxBB engine that drives the forum will jump to the interpreter when it meets the command open square bracket math close square bracket. The interpreter switches off again in a similar way with the command /math

There's a tutorial here: http://www.mathisfunforum.com/viewtopic.php?id=4397

The first 20 or so posts will be enough to get you started. Then you can dip in when you need a specific symbol.

Also useful: http://latex.codecogs.com/legacy/eqneditor/editor.php

You can create code there to copy and paste into your post.

Furthermore, if you see some LaTex that you want to investigate, just click on it to see how it was done.

Other square bracket type commands you might find useful:

quote url hide img b (for bold) u (for underline)

Hope that helps,

Bob

]]>How do you make the nice math characters that are in your first reply?

I used to be able to do them in Word, but never in a context like this one.

]]>Whoops! I feel like such a fool. I was so pleased I'd found a method, I overlooked this detail. Which is easier: rework the whole of maths defining 9 to be a prime, or change my earlier post?

Think I'll make the edit.

Thanks,

Bob

]]>I was just quickly scanning recent posts, and spotted this:

bob bundy wrote:

198 has 4 prime factors, 1, 2, 9, and 11.

According to my understanding of primes, 198 has 3 prime factors: 2, 3 and 11.

I didn't look to see if this makes any difference to your work there.

]]>You will see from the next post by Phrontister that I made a careless mistake and said that 9 was a prime. I've realised that the solution doesn't need primes anyway so here's a complete revision.

We need:

Clearly b - a < b + a + 1 and this pair must be a whole number factor pair multiplyig to give 198, so the possibilities are:

1 x 198, 2 x 99, 3 x 66, 6 x 33, 9 x 22, 11 x 18.

So I'll investigate b - a being each of the first of these pairs:

If b - a = 1, then a = 98 and b = 99

If b - a = 2, then a = 48 and b = 50

If b - a = 3, then a = 31 and b = 34

If b - a = 6, then a = 13 and b = 19

If b - a = 9, then a = 6 and b = 15

If b - a = 11, then a = 3 and b = 14.

You will notice that b gets smaller as b - a gets larger.

This follows because to make b smallest we want b + a + 1 to be smallest, and that will happen when b - a is largest.

So I could have left out all those trials and jumped straight to b - a = 11 immediately.

I'm feeling old and useless.

That's why joining MIF was a good move. We'll have you back to your prime in no time.

Bob

]]>The seemingly obvious "try n+1" was new to me - and so, appreciated, even though it's not algebraic.

Yes, I'm wanting an algebraic solution.

Sadly, although I've found two in the past, trying for several days hasn't reproduced them - so I'm feeling old and useless.

]]>