Friday Quiz, Take 7
This quiz was brought to you by the Fairfax County Mathematics League. This is an extracurricular math activitiy for high school students. And these questions go to the varsity teams. (That usually starts around preCalculus.)
1. Assuming that the Earth is a sphere with a radius of 4000 miles, how far is it, in miles, from an astronaut seated in a horizontal spacecraft 100 miles from the Earth's surface, to the Earth's visible horizon?
2. In how many zeroes does (5^10)! terminate?
If you have a favorite math problem you would like to submit to the Friday Quiz, please send it my way: chandlerm@washpost.com
By
Michael Alison Chandler

January 23, 2009; 12:30 PM ET
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Posted by: tomsing  January 23, 2009 1:39 PM  Report abuse
i think i might be falling down on the definition of 'the horizon' ... because as i conceived the problem, it is not a right triangle, so Pythagoras might not be as operative as one might think ...
Posted by: BillthePMP  January 26, 2009 10:55 AM  Report abuse
Pythagorean theorem says a^2 + b^2 = c^2.
900 mi doesn't make sense because that would be located in the middle of the earth.
4000^2 + 4100^2 = c^2, therefore c = 5728 mi
As for the second question, it doesn't really make sense because factorials don't terminate zeros. If anything it would produce a number with a lot of zeros.
Posted by: PorterPotty  January 27, 2009 4:41 PM  Report abuse
R = radius of Earth = 4000 mi
h = altitude of astronaut (above Earth's surface) = 100 mi
R+h = distance of astronaut from center of Earth = 4100 mi
The astronaut's horizon is defined by the farthest point on the Earth she can see. Imagine the astronaut looking straight down. Her "line of sight" extends 100 mi down to the surface of the Earth. If she sweeps her gaze outward toward the visible limb (edge) of the Earth. Her "line of sight" to the Earth's surface will get longer & longer until (when she's looking exactly at the limb) the line will just barely touch the Earth's surface and will be as long as possible. That's her horizon. Geometry tells us that a line that touches the surface of a sphere at only one point (a tangent line) is perpendicular to the line from the center of the sphere to that point (imagine the sphere rolling on a horizontal line).
So we have a right triangle:
side 1 from Earth center to horizon point (length R)
side 2 from astronaut to horizon point (length x)
hypotenuse from Earth center to astronaut (length R+h)
The Pythagorean theorem says: R^2 + x^2 = (R+h)^2
so x^2 = (R+h)^2  R^2 = 2*R*h + h^2 = 810000 = 900^2
and x = 900 mi
Posted by: rbtorrance  January 27, 2009 6:47 PM  Report abuse
If a number, M, terminates in (some number of = n) zeros, then you can divide it by 10 (some number of times = n). That means that 2 and 5 each divide M (at least n times).
Since
2! = 2*1
3! = 3*2*1 = 6
4! = 4*3*2*1 = 24
5! = 5*4*3*2*1 = 120
10! = 10*9*8*7*6*5*4*3*2*1 = 3628800
etc., you can see that any factorial, M > 4! will have many more factors of 2 than factors of 5. So, for each factor of 5, a factor of 2 can be "paired" with it to give a factor of 10, which corresponds to a terminating zero. To find the number of terminating zeros in a factorial, all we have to do is count the number of factors of 5 it has. This would be really easy except for the fact that you can get "extra" factors of 5 if, e.g., 25 = 5^2 or 125 = 5^3, etc., are included in M.
In our case M = (5^10)!.
Now M is the product of every number in the sequence of integers from 1 through 5^10. Every 5th number is a multiple of 5; every 25th number is a multiple of 25; every 125th number is a multiple of 125; etc. So the number of factors of 5 in (5^10)! is given by dividing 5^10 successively by 5, 25, 125, ..., 5^10:
5^9 + 5^8 + 5^7 + 5^6 + 5^5 + 5^4 + 5^3 + 5^2 + 5 + 1
This geometric series has the sum ( 5^10  1 ) / ( 5  1 ), so the number of factors of 5 in (5^10)!, and hence the number of terminating zeros, is 2441406.
Posted by: rbtorrance  January 27, 2009 6:54 PM  Report abuse
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The first problem is pretty straightforward  a triangle is formed by the line from the spacecraft to the center of the Earth (4000 + 100 = 4100 mi), the line from the center of the Earth to the horizon (at the surface of Earth, so 4000 mi), and the line from the spacecraft to the horizon (d). Pythagoras says that 4100^2  4000^2 = d^2, which gives the nice round solution of d = 900 mi.
For the second problem...I had to cheat. But once I did, the answer made sense. I'm curious to see how it gets explained here.