jim

Correction. Let h(k,m) be the height that can be handled with k eggs and up to m drops. Above I used a recursion of h(k,n) = h(k,n-1) + h(k-1,m) when I should have used a slightly different formula. Let the base at any time be the highest known safe floor. With k eggs and m drops left we can afford to skip h(k-1,m-1) floors and do our next drop at base + h(k-1,m-1) + 1. When an egg breaks the base remains the same and both k and m decrease by 1. If the egg doesn't break, increase the base by h(k-1,m-1) + 1 and then decrease just m by 1. h(1,m) = m where start dropping at the lowest floor and work upwards one at a time. h(2,m) = (m-1)+1 + (m-2)+1 + ... + (m-m)+1 = SUM of m, m-1, m-2, ...1 = m(m+1) / 2 h(3,m) = (m-1)m/2 +1 + (m-2)(m-1)/2+1 + ... = SUM (1/2)j^2 - (1/2)j +1 for j from 1 to m and we get (1/12)m(m+1)(2m+1) - (1/4)m(m+1) + m = (1/6)(m^3 -m) + m = (1/6)(m-1)m(m+1) + m. so 8 is not enough bu 9 drops would let us go as high as 120 + 9 which is enough.

Reverse the problem and let m be the number of drops, and then n is the tallest building that can be done using this number of drops. If you only have one egg n=m. With two eggs n = 1 + 2 +...+m = m(m+1)/2---call this b(m). With three eggs we get n = b(1) + ...+b(n) which we can call c(n) and so on. Now c(m) is half the sum of 1+...+m and 1+...+m^2 or half m(m+1)/2 and m(m+1)(2m+1)/6 so we get c(m) =(1/6)m^3 + (1/2)m^2 + (1/3)m or c(m) = (m^3 + 3m^2 +2m) / 6. For three eggs we get c(8) = 120 so the answer to the basic problem is 8.

If x=0.999... and y=1.000... then x, if it exists, is a number that differs from a finite sequence of 0.99...99 by an amount that is smaller than any positive number if we carry the expansion out far enough. For example, it will be less than one millionth if we go out 7 decimal places or more. The difference y-x, if it exists, is an approximation for 0.00...001 that has an error that can be made smaller than any positive nuimber if we go to a large enough number of decimal places This is the definition in standard math. If you wanted to define two real numbers as equal only if the rational approximations used to define the numbers were eventually identical you could do so and get two distinct numbers infinitesimally close together--but that is not the definition that is used in standard math.

There are infiniesinally small real numbers in the non-standard real number system of Abraham Robinson which he used in his non-standard analysis. That system allows a form of calculus where the derivative is the ratio of two infinitesimals.

In standard mathematics 1.000... and 0.999... are considered to be two different ways to express exactly the same number.

As of June 30, 2012 is a child born in 2010 at least x yearts old?

442. Since we can solve this with either Fibonacii numbers or combinations, we can conclude that F10 = C(10,0) + C(9.1)+ ...+c(5.5). More generally Fn = C(n,0) + C(n-1.1) + C(n-2,2) + ... as long as the first number is no les than the second. NOTE If you interpet the question to mean 10 steps plus the upper landing the answer would be F11, a point one poster raised. In addition if we distinquish between first strride right foot and first stride left we would double our answer.