Math Is Fun Forum

Macy

Guest

Geometry and factorial problem

1. A rectangular prism 6cm by 3cm by 3cm is made up by stacking 1cm by 1cm by 1cm cubes. How many rectangular prisms, including cubes, are there whose vertices are vertices of the cubes, and whose edges are parallel to the edges of the original rectangular prism? (Rectangular prisms with the same dimensions but in different positions are different)

2. The number 2008! means the product of all the integers 1,2,3,4,...., 2007,2007. With how many zeroes does 2008! end?

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Geometry and factorial problem

For number 2 consider what is required for a number to end with a 0.  All numbers ending in a 0 are divisble by 10, which is 2 X 5.  100 is 2^2 X 5^2, etc.  That means that in order to determine the number of 0's at the end of 2008! you need to calculate the number of pairings of 2 and 5 you can make with all of the numbers from 1 to 2008.

Jane had a nice algorithm for just this sort of problem a couple weeks ago, if you search the forums you can find it.  If you'd like to figure it out on your own first you need to realize that there are more 2's in the prime factorization of 2008! than 5's, so we can focus just on finding the number of 5's.  In order to do that first determine the amount of numbers from 1 to 2008 that have 5 as at least one of their prime factors.  Then you figure out how many numbers in that list have 5 as at least 2 of their prime factors, and you add them together.  You then repeat this step with numbers that have 5 as at least 3 factors, then 4, and so on until you reach the point that there are no numbers with more factors of 5.

Edit: In case the explanation was unclear, which I think it was, here is an example.  How many 3's are in the prime factorization of 30!

First we figure out how many numbers have at least one 3 in their factorization.  In this case we have 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, which is ten.  We then find how many numbers from 1 to 30 have 3 as two of their prime factors: 9, 18, 27, so there are three of them.  Add this to the original ten to get 13.  Now we check for numbers with three 3's: 27.  That brings us to fourteen 3's in the prime factorization of 30!.  If we try checking for numbers with four 3's we find that there are none since 81 is the smallest such number, so we know that there are exactly fourteen 3's in 30!.  Now just do the same thing for 5's in 2008!.

Last edited by TheDude (2008-10-20 00:03:08)

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Geometry and factorial problem

There's a way to figure it out without listing numbers though, which is lucky since listing the multiples of 5 up to 2008 would be some work.

Basically, you divide 2008 by 5 and round down, divide that answer by 5 and round down, divide that by 5 and round down, and keep going until you get zeroes. Then add up all the intermediate numbers and you have your answer.

Using this on TheDude's example, you get:
30/3 = 10
10/3 = 3
3/3 = 1
1/3 = 0

10+3+1=14, which agrees with what he got.

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Macy

Guest

Re: Geometry and factorial problem

Thank you all. You guys are such wonderfull people !!!

ZHero

Real Member

Registered: 2008-06-08

Posts: 1,889

Re: Geometry and factorial problem

Another way to state that is..
No. of 5's in 2008! is
[2008/5]+[2008/5^2]+[2008/5^3].... where [..] is a Greatest Integer function.

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Macy

Guest

Re: Geometry and factorial problem

Based on The Dude and Mythsyperson idea, I figured out a simple way to work out this problem:

Take 2008 divide by 5 until cannot do any further then add up all the divisors you have. This method is actuall based on the way to work out the base number system, if you know about how to convert the number in base 10 to any other base, say, base 2, base 4 or base 8, then you will get this idea.
So the answer is 500 zeros in 2008!

Thank you for all your help and i have another question cannot figure out how to draw the diagram and solution. Appreciate any help from you
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Lloyd starts a bushwalk from the base of a wind generator, which is 55m tall and located at point A, which is at sea level. He walks for 4km on the bearing 0600T up a slope that is inclined at 100 to the horizontal to position B, where a fire tower of height 80m is located.

a. Find to the nearest degree and minute, the angle of depression from the top of the fire tower to the top of the wind generator.

Lloyd then treks 10km on the bearing 1500T down a slope to reach point C at sea level

b. Find the angle of depression from the base of the fire tower to position C.
c. Find the angle of elevation from C to the top of the wind generator
d. Find, to the nearest degree, the bearing of position C from A.

Macy

integer

Member

Registered: 2008-02-21

Posts: 79

Re: Geometry and factorial problem

...
Lloyd starts a bushwalk from the base of a wind generator, which is 55m tall and located at point A, which is at sea level. He walks for 4km on the bearing 0600T up a slope that is inclined at 100 to the horizontal to position B, where a fire tower of height 80m is located.
Macy

Please explain what these two phrases mean in your question:

bearing 0600T

What is 0600 T ?
Is it to be interpreted something similar to  0 600 hours military time?

I cannot make any sense of this:
a slope that is inclined at 100 to the horizontal

Does that mean the slope is 100 degrees from the horizontal?