[DJMayes]

For the skyscraper question with the glass balls, it has been observed that by going in increments of 2 reduces the maximum number of trials to 51. Going in 4, you could reduce this to 28 trials. I think this could be reduced to an equation:

Let f (x) = number of trials, and x = the floor intervals (e.g. when x = 2 you are doing every 2 floors)

Our objective is to minimise f (x) = 100/x + x - 1

Differentiate to obtain f'(x) = -100/x^2 + 1

At a stationary point, 0 = -100/x^2 + 1

x^2 = 100

x = 10 (Not -10 as you don't test floors as negatives)

You can find a second derivative and thereby confirm that this is a minimum of the above equation. As such you would drop the balll every 10 floors and, when it breaks, drop the ball from the lowermost untested floor and so on until it breaks. This results in a maximum of 19 drops, a significant advance on the current 51.

[Carolus]

(Original post by DJMayes)
For the skyscraper question with the glass balls, it has been observed that by going in increments of 2 reduces the maximum number of trials to 51. Going in 4, you could reduce this to 28 trials. I think this could be reduced to an equation:

Let f (x) = number of trials, and x = the floor intervals (e.g. when x = 2 you are doing every 2 floors)

Our objective is to minimise f (x) = 100/x + x - 1

Differentiate to obtain f'(x) = -100/x^2 + 1

At a stationary point, 0 = -100/x^2 + 1

x^2 = 100

x = 10 (Not -10 as you don't test floors as negatives)

You can find a second derivative and thereby confirm that this is a minimum of the above equation. As such you would drop the balll every 10 floors and, when it breaks, drop the ball from the lowermost untested floor and so on until it breaks. This results in a maximum of 19 drops, a significant advance on the current 51.

Yes, Brit_Miller did this on the last page.

[TheMagicMan]

(Original post by Quantaˌ)
Care to explain? How would 2 cover 3?



No, it's written by him.

So when the reader reads 3, the reader thinks that 2 is covered in the credits as it is the previous one referred to. However, 2 now takes on the meaning that the previous footnote i.e. 3 is covered.

Not very well explained I know

[TheMagicMan]

By the way you need 5 different weight to get every possible weight between 1 and 121...can you work out what they are?

[mattmattmatt]

Pretty sure i can do it in 14 tries.

If you drop from the following floors, derived from n+(n-1)+(n-2)...

14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100

If it breaks on 14, you then have to test the 13 floors below, 14 attempts MAX.

This works for the rest too. If for example it broke on the 95th floor, you have already had 10 attempts and will use 4 more from 91-94 to work out the actual floor with your remaining egg.

Not sure if that made sense but 14 is the best i can do.

[Quantaˌ]

(Original post by TheMagicMan)
So when the reader reads 3, the reader thinks that 2 is covered in the credits as it is the previous one referred to. However, 2 now takes on the meaning that the previous footnote i.e. 3 is covered.

Not very well explained I know

I see what you mean. However, I would assume that 3 would supersede 2. Also, it was not the answer I was told. The answers given to me was that the fourth footnote was the same as the third but written in English since Littlewood was a British mathematician.

This one is more mathematical:

Give 1000 consecutive integers, none of which are prime numbers.

[Hopple]

(Original post by Quantaˌ)
One of the papers by the famous mathematician Littlewood published in a French journal concludes as follows (The sentences in the journal were, of course, in French, but below is the English translation).

"1. I am greatly indebted to Prof. Risez for translating the present paper.

2. I am greatly indebted to Prof. Risez for translating the previous foot note.

3. I am greatly indebted to Prof. Risez for translating the previous foot note."

Littlewood is completely ignorant of French language; so how did he avoid infinite regression?

"2. I am greatly indebted to Prof. Risez for translating all footnotes." ?

[KyraBloke]

(Original post by Quantaˌ)

This one is more mathematical:

Are negative numbers prime?

[Hopple]

(Original post by DJMayes)
For the skyscraper question with the glass balls, it has been observed that by going in increments of 2 reduces the maximum number of trials to 51. Going in 4, you could reduce this to 28 trials. I think this could be reduced to an equation:

Let f (x) = number of trials, and x = the floor intervals (e.g. when x = 2 you are doing every 2 floors)

Our objective is to minimise f (x) = 100/x + x - 1

Differentiate to obtain f'(x) = -100/x^2 + 1

At a stationary point, 0 = -100/x^2 + 1

x^2 = 100

x = 10 (Not -10 as you don't test floors as negatives)

You can find a second derivative and thereby confirm that this is a minimum of the above equation. As such you would drop the balll every 10 floors and, when it breaks, drop the ball from the lowermost untested floor and so on until it breaks. This results in a maximum of 19 drops, a significant advance on the current 51.

Why must you drop it at regular intervals?

If I drop first at floor 18, where either it smashes and I have a maximum of 18 drops, or I then go on to floor 35, where either it smashes and I again have a maximum of 18 drops, or I go on to floors 51, 66, 80 and 93. Clearly there's room for improvement here but I've already got a new upper limit of 18 drops.

(Original post by mattmattmatt)
Pretty sure i can do it in 14 tries.

If you drop from the following floors, derived from n+(n-1)+(n-2)...

14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100

If it breaks on 14, you then have to test the 13 floors below, 14 attempts MAX.

This works for the rest too. If for example it broke on the 95th floor, you have already had 10 attempts and will use 4 more from 91-94 to work out the actual floor with your remaining egg.

Not sure if that made sense but 14 is the best i can do.

Yeah, that's what I was getting at but didn't follow it through completely. It's sort of finding the smallest triangle number that adds to its number (i.e. 6 is the third triangle number, so 6+3=9) to give more than 100.

[DJMayes]

(Original post by Carolus)
Yes, Brit_Miller did this on the last page.

He cited the answer but did not show why it was the answer. Formulating the question as f(x) = 100/x + x - 1 and then using calculus to get a minimum shows why you cannot complete the task in fewer than 19 tries.

[TheMagicMan]

(Original post by Quantaˌ)
I see what you mean. However, I would assume that 3 would supersede 2. Also, it was not the answer I was told. The answers given to me was that the fourth footnote was the same as the third but written in English since Littlewood was a British mathematician.

This one is more mathematical:

Easy: 999! +k, where k runs between 0 and 999

[TheMagicMan]

(Original post by DJMayes)
He cited the answer but did not show why it was the answer. Formulating the question as f(x) = 100/x + x - 1 and then using calculus to get a minimum shows why you cannot complete the task in fewer than 19 tries.

You can do it in 14 I think

[L'Evil Fish]

Can you not just drop it off 50 and go higher or lower if it breaks. And then use the midpoint between 0-50 or 50-100 and continue until it gets to the maximum?

[Joshalos]

(Original post by Quantaˌ)
Thought I would join the thread. Here is a good teaser:

As soon as I saw it, I thought 'easy!'

Upside-down 4 and then a normal 4 = 64

[DJMayes]

(Original post by TheMagicMan)
You can do it in 14 I think

I see what you mean if you don't drop it at constant intervals the way mattmattmatt did.

[Hopple]

(Original post by shadab786ahmed)
Can you not just drop it off 50 and go higher or lower if it breaks. And then use the midpoint between 0-50 or 50-100 and continue until it gets to the maximum?

What if it breaks at 50? Then you're looking at doing all the floors from the ground to 49.

[TheMagicMan]

(Original post by DJMayes)
I see what you mean if you don't drop it at constant intervals the way mattmattmatt did.

The real answer is actually 0...there's no way a glass ball can survive even a one storey drop, so you might as well save yourself the trouble (and the broken glass)

[Hopple]

(Original post by Quantaˌ)
Give 1000 consecutive integers, none of which are prime numbers.

1000! onwards. 1000!+n will be divisible by n.

[Carolus]

(Original post by Quantaˌ)
This one is more mathematical:

I suppose you could always go for -1000 ... -1

[Quantaˌ]

(Original post by TheMagicMan)
Easy: 999! +k, where k runs between 0 and 999

I got 1001!+2, 1001!+3, ... 1001!+1000, 1001!+1001