64. For each positive integer n, let f_n be the function defined on the interval [0,1] by f_n(x) = x^n /(1+x^n) which of the following statements is true?
I. The sequence converges pointwise.
II. The sequence converges uniformly
III. lim_n^Inf Integral_0^1 f_n(x) dx = Integral_0^1 lim_n^Inf f_n(x) dx
65. Which of the following statements are true about the open interval (0,1) and the closed interval [0,1]?
I. There is a continuous function from (0,1) onto [0,1]
II. There is a continuous function from [0,1] onto (0,1)
III. There is a continuous bijective function from (0,1) to [0,1]
Q. 55: Simple: Number of integers with one zero in their factorial
1 ! = 1 (No zeros)
2 ! = 2
3 ! = 6
4 ! = 24
5 ! = 120 (One zero)
6 ! = 720
7 ! = 5040
8 ! = 40320
9 ! = 362880
10 ! = 3628800 (Two zeros)
Obviously as soon as your factorial gets multiplied by a 5 or a 10 (or their multiple) one more zero is added to the end. So for example 5! - 9! end in 1 zero, 10! - 14! end in two zeros...
The answer is 5.
Hi Nameless,
In my solution, I meant that, suppose k! ends in 99 zeros, where k is divisible by 5 (k mod 5 == 0). Then (k + 1)!, (k + 2)!, (k + 3)! and (k + 4)! will also end in 99 zeros since no 5 is being added to the multiple. As soon as you reach (k + 5)!, more zeros will be added to your multiple.
Regarding your solution, the equation is more precise than my inductive solution but the solution to it is 400 <= x <= 404.
Of course one flaw in my solution is that it doesn't take into account jumps in the number of zeros that occur at every power of 5. So for example number of integers "k" such that k! has exactly five zeros = 0.
But if there is at least one integer such that its factorial ends in k zeros, then the number of integers whose factorials end in "k" zeros is exactly 5.
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