1. Do there exist irrational numbers r and s such that r^s is rational? Do there exist irrational numbers r and s such that r^s is irrational?

Solution: Yes and yes. Assume e and ln 2 are irrational. r^s = e^{ln 2} = 2, which is rational. r^s = e^{ln 2e} = 2e, which is irrational.

In this problem, I started with my intuition. I know that exponential functions are continuous and so yield both rational and irrational values. e is a natural irrational to use as a base. My intuition tells me that e raised to any rational number will yield an irrational number. So if e^s is to be rational, s must be irrational, and the irrationalities of e and s will complement each other to produce a rational. I asked myself: e raised to what number is, say, 2? No sooner had I asked the question than I realized that I was describing a natural logarithm! So of course, r^s = e^{ln 2} = 2. e and ln 2 are both known to be irrational, and 2 is rational.

The second half of this problem seems easier: One could hardly expect e^π to be rational, and so that could be my answer. But I decided to go further and see if I could prove that r^s is irrational. I considered e^{ln π} = π. e and π are known to be irrational, but how do I prove that ln π is? I got no further along that path. I played around with the properties of logarithms, until I stumbled on the simple solution: r^s = e^{ln 2e} = 2e. Consider the exponent: ln 2e = ln 2 + ln e = ln 2 + 1. ln 2 is known to be irrational. An irrational number plus a rational number is always irrational, so therefore ln 2 + 1 is irrational, so therefore the exponent ln 2e is irrational. r^s = 2e is also irrational, since a rational times an irrational is always irrational.

2. Does there exist an infinite collection of sets such that the intersection of every two distinct sets in the collection is nonempty, but the intersection of every three distinct sets in the collection is empty?

Solution: Yes: {{(x, y) | x = A}Ç{(x, y) | y = A}}, for all real A. In other words, each set is the union of (1) all ordered pairs with A as the first number, and (2) all ordered pairs with A as the second number. The infinite collection is the collection of all such sets for every real number A.

This problem did not at first seem to be a mathematical one as much as a logical one: How to organize a collection of elements into an infinite number of sets. It is soon clear that there must be an infinite number of elements. For otherwise, there would be a finite number of distinct sets. Furthermore, each set must have an exclusive intersection with every other. Therefore, the sets themselves must be infinite (in order to have so many intersections). I soon made a geometric analogy: Each set is a line, each point an element. Each line intersects an infinite number of other lines:

But how will each vertical line intersect the others? If only the vertical lines were also horizontal! But they can be, if I allow a set to be represented by two lines, one vertical and one horizontal:

By the way, this problem is the same as asking for a telephone network such that every telephone line has a direct, exclusive connection to every other one. The solution is similar. Every telephone has two lines that run throughout the town: One for calling and one for incoming calls. The calling line is connected to every other incoming line, and the incoming line crosses every other calling line. The result is a grid.

This grid can be described in terms of the coordinate plane, each set A containing all points on the lines x=A and y=B. To simplify the system and avoid two sets with one common line, each set A contains the points on x=A and y=A. This translates into set notation as {{(x, y) | x = A}Ç{(x, y) | y = A}}, which means "the collection of all sets A such that each set is the union of (1) all ordered pairs starting with A and (2) all ordered pairs ending with A."

3. Consider the number 2^9876543. Determine (1) the number of digits it has; (2) its last three (rightmost) decimal digits; and (3) its first three (leftmost) decimal digits.

Solution: (1) It has 2973136 digits.

(2) It ends in 208.

(3) It starts with 497.

For question (1) I soon realized that the question is really asking: What is the greatest integer in log₁₀ 2^9876543 +1? One is added because log₁₀ 2^9876543 would give the number of digits after the first. Calculation of this is easy with a scientific calculator: log₁₀ 2^9876543 +1=9876543 log₁₀ 2 + 1=2973136.

For questions (2) and (3) I used the computer program Mathematica to compute 2^9876543. After a minute of computation I was able to see all 2000 pages of digits. Later I decided to try to find a pattern in the last three digits of successive powers of 2. I know that the last three digits of 10ⁿ are 000 for n>2. I know that the last three digits of 5ⁿ alternate between 125 and 625 for n>2. I already knew that the last digit of powers of 2 cycle through the sequence 2-4-8-6-2. I proceeded to record the last two digits in 2¹, 2⁵, and so on. I knew that the last digit would be 2 in each case. I found that the last two digits returned to 04 every 20 exponents. Thus 2², 2²², 2⁴², and so on all end in 04. I looked for a three-digit cycle by testing 2², 2²², 2⁴², etc. I found that the last three digits of 2²² are 304 and they are also 304 every 100 exponents. That means that 2^9876543 has the same last three digits as 2⁴³, which is 208. Extending this investigation, I found that for each successively higher digit, the length of its cycle is generally five times the length of the cycle of the previous digit. In other words, the last four digits of an exponent of 2 will remain unchanged if the number is multiplied by 2⁵⁰⁰, the last five digits will remain unchanged if multiplied by 2²⁵⁰⁰, the last six will remain unchanged if multiplied by 2¹²⁵⁰⁰, and so on, seemingly forever. I also found that 2²² = 4194304 is a special number in that it is the lowest number that will always be repeated at the end of exponents of 2, complete with zeroes. For instance, 2⁶²⁵²² ends in 4194304, and 2^1562522 ends in 004194304. I could not prove these last two hypotheses, and I was not able to use the results of these studies to analytically solve part (3) of the problem.

4. Find the sum of the 4536 numbers from 1,000 to 10,000 which have all their digits distinct. This is really two problems: Find a solution by using brute force on the computer; find a solution by doing it analytically by hand.

Solution: The sum is 24,917,760. The sum can be shown to equal (1+2+3+4+5+6+7+8+9)((9*8*7-8*7)111 + (9*8*7)1000) = 24,917,760.

To find the sum, I wrote the following program for the TI-86 graphing calculator:

1000ÞI	I is the index number; it runs from 1000 to 10000
0ÞN	N counts the numbers with distinct digits
0ÞX	X is the running sum
Lbl A	Part A: This is the start of the main loop
If IŠ10000:Goto Z	When it come to 10000, the loop terminates
int(I/1000)ÞA	This stores the first digit of the number in A
int(I/100)-10AÞB	This stores the second digit of the number in B
int(I/10)-100A-10BÞC	This stores the third digit of the number in C
int(I)-1000A-100B-10CÞD	This stores the fourth digit of the number in D
If AŠB:Goto B	If any of the digits are the same, skip to Label B
If AŠC:Goto B
If AŠD:Goto B
If BŠC:Goto B
If BŠD:Goto B
If CŠD:Goto B
N+1ÞN	Now we can assume I has distinct digits; increment N...
X+IÞX	... and add the number I to the running total X
Lbl B	Part B: The following is executed no matter what I is
If DŠ0:Disp I	This will indicate the program's progress every 10 numbers
I+1ÞI	Now, increment I...
Goto A	... and repeat the loop
Lbl Z	Part Z: after the loop terminates
Disp N	Displays the number of numbers added
Disp X	Displays the final total

The program takes about 45 minutes to run on a TI-86. After running, N=4536, so I knew all the necessary numbers were added. X=24,917,760.

A simple way to describe all those numbers is: All the permutations of 4 digits from the set {0,1,2,3,4,5,6,7,8,9}, excepting those beginning with 0 (because they are less than 1000). The number of such permutations is: 10!/6! - 9!/6! = 4536, which agrees with the directions.

After that realization, I was thinking of the sum not as a sum of large numbers, but a sum of many individual digits. Their order does not matter in calculating the sum. All I need to do is figure, out of the 4536 numbers, how many 1s there are in the 1s place, how many 2s, etc., and calculate the same thing for the 10s, 100s, and 1000s places. The number of four-digit number ending with 1 is the number of permutations of 4 digits from the set {0,1,2,3,4,5,6,7,8,9}, ending in 1 and excepting those beginning with 0. The number is 9!/6! - 8!/6!. That number is the same for all four-digit numbers ending in 2, 3, etc., and it is the same for all four-digit numbers containing a certain digit in the 10s or 100s place. The 1000s place is different, though: The number of four-digit numbers starting with a certain digit (not 0) is 9!/6!. The term 8!/6! is not necessary because there are no four-digit numbers that begin with, say, 1 and are less than 1000. The preceding paragraph is summarized in this table:

	1000s place	100s place	10s place	1s place
Sum of 0s	0	0(9*8*7)	0(9*8*7)	0(9*8*7)
Sum of 1s	1(9*8*7)	1(9*8*7-8*7)	1(9*8*7-8*7)	1(9*8*7-8*7)
Sum of 2s	2(9*8*7)	2(9*8*7-8*7)	2(9*8*7-8*7)	2(9*8*7-8*7)
Sum of 3s	3(9*8*7)	3(9*8*7-8*7)	3(9*8*7-8*7)	3(9*8*7-8*7)
Sum of 4s	4(9*8*7)	4(9*8*7-8*7)	4(9*8*7-8*7)	4(9*8*7-8*7)
Sum of 5s	5(9*8*7)	5(9*8*7-8*7)	5(9*8*7-8*7)	5(9*8*7-8*7)
Sum of 6s	6(9*8*7)	6(9*8*7-8*7)	6(9*8*7-8*7)	6(9*8*7-8*7)
Sum of 7s	7(9*8*7)	7(9*8*7-8*7)	7(9*8*7-8*7)	7(9*8*7-8*7)
Sum of 8s	8(9*8*7)	8(9*8*7-8*7)	8(9*8*7-8*7)	8(9*8*7-8*7)
Sum of 9s	9(9*8*7)	9(9*8*7-8*7)	9(9*8*7-8*7)	9(9*8*7-8*7)
Subtotals	22680	20160	20160	20160
Total	22680*1000 + 20160*100 + 20160*10 + 20160 = 24,917,760

5. Three positive integers a, b and c form a Pythagorean Triple if a²+b²=c². It is primitive if they have no common factors. For example, 3,4 and 5 form a primitive Pythagorean Triple, 5,12 and 13 form another primitive Pythagorean Triple. Find two more primitive Pythagorean Triples.

Solution: I found the triples 8-15-17 and 9-40-41 in a publication. I discovered the following primitive Pythagorean triples independently:

7-24-25

11-60-61

13-84-85

17-144-145

19-180-181

23-264-265

29-420-421

31-480-481

37-684-685

43-924-925

Before I started thinking about this problem, I noticed a solution in a brochure of the University of Michigan. It casually mentioned the primitive Pythagorean triples 8-15-17 and 9-40-41. It also claimed that there is a method for easily finding more. Emboldened by this, I came up with a practical method for finding them with a TI-86 calculator. I stored the value 1 in the variable N

1®N

and iterated the following command:

N+1®N: Ö(p²+N²)

where p is a prime number. This command increments N while displaying the length of the hypotenuse of a right triangle with bases p and N. Eventually the command would display an integer, in which case I knew that p, N, and the resulting integer constituted a Pythagorean triple. Checking it for primitivity was easy, as the numbers were small enough to check for prime factors easily. I used prime numbers for the first of each triple because a composite number would often result in an undesired non-primitive triple.

While finding these triples I discovered some interesting behaviors of the numbers. Each prime number p representing a base of a right triangle seemed to form a Pythagorean triple with only one pair of numbers. For example, I could not find a Pythagorean triple with base (not hypotenuse) 13 other than the triple 13-84-85.

Even more astonishing is that for each triple I found, the two numbers besides the prime number I chose were consecutive. A quick glance at the list above will affirm this. These consecutive numbers have some sort of pattern to them. They are generally multiples of 20. However, a scanning of the list will reveal that the second number of each triple exhibits strange behavior, skipping one or more multiples of 20 or sometimes being 4 greater than a multiple of 20. There is probably some sort of pattern in these Pythagorean numbers, but I cannot discern it.

6. Suppose your worst enemy gives you n red points and n blue points in the plane, no three of them collinear. (Your enemy picked them, so they are not nicely arranged.) Show that it is possible to draw n line segments, each segment joining a red point to a blue point (so that each red point is paired with a unique blue point, and vice versa) in such a way that none of the line segments intersect.

Solution: Label the lowest point A. If a coordinate plane is imposed, A is the point with the lowest y-coordinate. If two points share the lowest y-coordinate, rotate your frame of reference or your coordinate just enough so that one point (point A) is lower, but not enough so that another point is just as low. Define any point of point A's color to be worth +1. Any point of opposite color is worth -1. Now set the origin of the coordinate plane at point A's position.

Now imagine that a ray radiates out from (but does not include) the origin and sweeps through a semicircle above the x-axis. The angle the ray makes with the x-axis ranges from 0 to π. Since the origin is at the lowest point in the array, the ray will pass over all other points. The ray will never cross more than one point at a time because that would require a line to pass through three points (including point A) at once, which is a mischief my enemy cannot manage (Fig. 1).

The ray sweeps over each point besides point A one at a time in a certain order. Now count up the values of the points (worth +1 or -1 each) as the ray passes over them. Since A is worth +1, and there are n points worth +1 and n points worth -1, the total value that the ray passes over is (+1)(n-1) + (-1)n = -1. During the ray's passing, the count may reach different integers and fluctuate, but it will end up as -1.

Now stop the ray at the first instant the count drops from 0 to -1. Since the count starts at 0 and ends at -1, this must occur at one time or another. The drop in the count can only be caused by the ray crossing a negative point, which we call point B. The ray now divides the upper plane into two sectors. Since the ray passed through a total value of 0 before touching point B, the right sector of the plane (not including points A and B) must contain a total value of 0. If the ray would continue its sweep of the upper plane, it would have to end its count with -1. Since the count is already -1 at point B, the total value of the left sector must be 0.

Now draw a line segment from A to B. The only way for each sector to have a total value of 0 is for each sector to contain exactly as many red points as blue points. Now reorient the coordinate plane and repeat this process for the points in each sector (one or both sectors may be empty, in which case you may move on). For each sector, ignore the points A and B and all the points in the other sector. The line segment finally drawn as a result of this method will connect C and D. It will not cross the first ray because both points will be in one sector and therefore on the same side of the first ray. Since CD will not cross the first ray, CD will not cross AB, which lies on the first ray.

Continue this process, drawing line segments to partition the plane into isolated domains, thus partitioning the point array into subarrays, until at last a line segment will join the two last disconnected points, and no more line segments remain to be drawn.

The above is a method for drawing n line segments so that each red point is paired with a unique blue point, and vice versa, in such a way that none of the line segments intersect. The above method makes no reference to the specific nature of the configuration of points beyond the requirements that there are as many red points as blue points and that no three are collinear. Therefore, the method works for all such arrays of points and thus it is always possible to draw n such line segments.

After reading this problem, I drew several random arrays of points and found that it was always quite easy to connect them. I tried to construct an array of points that could not be connected as follows: I placed a red and blue point A and A. To prevent them from being joined by a line segment, I placed two more points B and B and connected them with a line segment that separated A from A. To prevent A from joining with B and A with B, I had to place four more points (Fig 2). Thus I needed an infinite array of points. I did not know how to make an airtight proof of this, though.

I instead decided instead to construct a method for drawing line segments in any array. Starting with point A defined as the bottom or rightmost point was natural, as it is more isolated. Then I had to decide on a way of deciding which point to connect to point A. It couldn't be just the closest blue point, because that might have the effect of closing a red point off from all other blue points. So I had to draw a line segment such that there are as many red points as blue points on each side. I soon made the above method, which is so precise that a computer could follow it.

7. Show that the expressions 2x+3y and 9x+5y are divisible by 17 for the same set of integral values of x and y.

Solution:

· Expression A is 2x+3y. It is divisible by 17, which means that 2x+3y=17n, where x, y, and n are integers. x and y represent all integers that make expression A divisible by 17.

· Rearranging the equation gives 2x=17n-3y.

· Since x is an integer, and 2 times an integer is even, 17n-3y is even.

· Since n & y and therefore 17n & -3y are both integers, and since an even number is the sum of either (1) two odd integers or (2) two even integers, then 17n and -3y are both odd or else they are both even.

· Since n and y are both integers, and since an odd number multiplied by an odd number is odd and an odd number multiplied by an even number is even, then in order for 17n and -3y to be odd n and y must be odd, and in order for them both to be even, n and y must both be even. Therefore, n and y are both odd or else they are both even.

· Solving the equation for x yields x=(17n-3y)/2

· For the same x and y, consider expression B: 9x+5y. Substituting for x yields 9(17n-3y)/2+5y.

· Simplifying yields 17(9n-y)/2.

· Now consider 9n-y. Recall than n and y are both odd or both even. Since an odd number times an odd number is odd, and an even number times an odd number is even, then 9n and -y are either both odd or else they are both even.

· Since an odd number plus an odd number is an even number, and an even number plus an even number is even, 9n-y is even.

· Now consider (9n-y)/2. Since the numerator 9n-y is even, the expression (9n-7)/2 is an integer.

· Now consider the modified expression B: 17(9n-y)/2. It is the product of 17 and (9n-y)/2, which is an integer. Therefore expression B is divisible by 17.

· Therefore, all x and y that make expression A divisible by 17 also make expression B divisible by 17.

· Now let x and y represent all integers such that 9x+5y (expression B) is divisible by 17.

· Rearrange the expression: 17(9(2x+3y)/17-y)/2.

· Since the expression is divisible by 17, the other factor (9(2x+3y)/17-y)/2 must be an integer.

· Since that expression and its denominator are integers, the numerator 9(2x+3y)/17-y must be integer.

· Since 9(2x+3y)/17-y and y are both integers, 9(2x+3y)/17 must also be an integer.

· Because of that, the numerator 9(2x+3y) must be an integer divisible by 17.

· Since x and y are integers, 2x+3y is an integer.

· Therefore, since 9(2x+3y) is divisible by 17, and since 9 is not divisible by 17, 2x+3y (expression A) must be divisible by 17.

· Therefore, all x and y that make expression B divisible by 17 also make expression A divisible by 17.

· In conclusion, 2x+3y and 9x+5y are both divisible by 17 for the same set of integer values for x and y.

I had several false starts in working on this one. Algebraically, I discovered that if x and y make both expressions divisible by 17, then if I increase x and y so that the first expression is still divisible by 17, then the second expression will still be divisible by 17.

I also subtracted the two expressions and found a third expression, 7x+2y, that is also supposedly divisible by 17. Finally, when I set the first expression equal to 17n and solved for x, I realized that y and n had to be both even or odd. That provided the essential key to the above proof.

I proved that if x and y make the first expression divisible by 17, then the second expression is also divisible by 17. Next I had to prove that there were no values of x and y that made the second expression divisible by 17 but not the first. In other words, I had to prove that if x and y make the second expression divisible by 17, then the first expression is also divisible by 17.Working backwards was tricky. I could not just reverse the proof because the logic I used concerning the evenness or oddness of the integers is not simply reversible. By only looking at the second expression, I could not draw any conclusions except that if 9x+5y=17m, (m being an integer) then x, y, and m cannot all be odd. That did not seem helpful. Finally, by doing my first proof backwards, I could use logic concerning the integrality (integer-ness) of expressions. I constructed the steps of both proofs in about the same chronological order as they appear above.

8. A round-robin tournament was held among 13 players. Each player played one game against every other player. Each player won six games. How many triples (K,L,M) of players are there such that K beat L, L beat M, and M beat K?

Solution: 273 triples.

I initially thought that there could be many different complicated results of this tournament. However, the question implies that if there are, any result I consider will yield the single, correct answer. I first drew this diagram:

Here player K is at the bottom. Above him are the 12 other players. Each player beats the 6 players to his right and loses to the 6 players to his left. Next I considered choosing a triple (K,L,M). I choose K to be the one at the bottom of the diagram. L loses to him, and so must be to K's right. M must be further to the right and have K on his right. So (K,L,M) forms a circle, going counterclockwise. L can be any of the 6 players above K and to the right of the line. L cannot be to the left of the line, or else he would beat K. M must be no more than 6 seats to the right of L, but must be on the left side of the line (or else he would not beat K). If L is the player immediately to K's right, then M can only be the 7th player to K's right. If L is the 2nd player to L's right, M may be the 7th or the 8th player to K's right.

Continuing in this way, there are 1+2+3+4+5+6=21 possible locations of L and M for a given K. Since there are 13 possible locations of K, there are 21*13=273 possible triples (K,L,M).

After reaching this number I wondered if I had counted a triple more than once, but I did not. For each unordered set {K,L,M} I had counted three triples: (K,L,M), (L,M,K), and (M,K,L). That is what the problem asks for.

9. Suppose n children holding loaded water pistols are standing in an open field, no three of them in line, such that all the distances between pairs of them are distinct. At a given signal, each child shoots the child closest to him or her with water. Show that if n is an even number, then it is possible (but not necessarily the case) that every child gets wet. Show that if n is odd, then necessarily at least one child stays dry.

Solution: Suppose there are two children. They will shoot each other, and both will end up wet. Suppose there are four children arranged according to Fig. (1). Children A, B, and C will shoot D, and D will shoot C, but no one will shoot either A or B. Therefore, in n is even, it is possible, but not always the case, that every child gets wet.

Now consider a group of n children, containing at least one subgroup of m children. They shoot no one outside of their subgroup and no one outside of their subgroup shoots them. Assume that every member of the subgroup gets wet. In that case, every child must have at least one shot directed at him. Since there are m shots, each child may be shot by only one other. The subgroup must contain at least one member: Child B. He will remain dry unless there is another member of the subgroup, child A. A is defined as the one who shoots B. B may shoot A back. In this case, no children may shoot either of them because A and B are already being shot. Therefore, A and B may form an all-wet subgroup of 2. Now suppose there are more than 2 children in this subgroup. Since a child must shoot and cannot be shot twice, the only configuration possible in which all children are shot is a loop: A chain of water shots in which the last child, defined to be Z, shoots A, thus ensuring that A is wet (Fig. (2)). Now if B is to shoot C (who may be Z), the distance between them BC must be less than AB, or else B would shoot A. The distances cannot be the same, as required by the directions. By similar arguments, BC>CD (if there is a child D), etc., YZ>ZA. Therefore, AB>ZA. But that would require A to shoot Z, which by definition he does not. Therefore, there can be no subgroup of more than 2 children that all end up wet. Therefore, a full group of children will only be all wet if the group is composed of 2-child subgroups. Therefore, for an all-wet group, n must be a multiple of 2 and therefore even. Therefore, if n is odd, at least one child must remain dry.

Working with a geometric analogy was a natural first step. Playing with the formations of nodes and arrows in my head, I soon discovered the pair and Fig. (1), which I used to prove the first part of the problem. I couldn't see a connection between the nature of n and the wetness of the children, so I started with a general, hypothetical group of children. After some thinking, I classified all "wet" groups into pairs or loops. Then I realized that a loop cannot function if every child shoots the closest one because you can't have a loop of successively shorter segments. Since a "wet" group must be a pair (or a collection of pairs), only even groups are all wet. I then simplified the proof.

10. How many zeros are there at the end of the number 1000!=(1)(2)(3)...(999)(1000)? How many for 1001!, 1002!, 1003!, 1004!, and 1005!?

Solution: 1000! ends in 249 zeroes.

1001! ends in 249 zeroes.

1002! ends in 249 zeroes.

1003! ends in 249 zeroes.

1004! ends in 249 zeroes.

1005! ends in 250 zeroes.

The first question is really asking, "How many times is 1000! divisible by 10?", or "What is the divisibility of 1000! by 10?". Instead of imagining 1000! as the product of 1000 integers, I thought of it as a product of a huge number of prime factors. Then I realized that to find how many times 1000! is divisible by 10 is to find the lesser of: (1) The number of times it is divisible by 2, or (2) The number of times it is divisible by 5. I found part (1) first. I classified all factors of 1000! divisible by 2 as multiples of powers of 2. I constructed the following table to calculate the number of 2s that are factors of 1000!.

Power of 2 = A	Number of factors of 1000! divisible by power of 2 = B	Number of divisibilities by 2 in these numbers = C
2^9=512	1 (being 512 itself)	9*1=9
2^8=256	3 (being 256, 512, and 768)	8*(3-1)=16
2^7=128	7 (being 128, etc.)	7*(7-3)= 28
2^6=64	15	6*(15-7)= 48
2^5=32	31	5*(31-15)= 80
2^4=16	62	4*(62-31)= 124
2^3=8	125	3*(125-62)= 189
2^2=4	250	2*(250-125)= 250
2^1=2	500	1*(500-250)= 250
Total divisibility of 1000! by 2:		994

Column A is a list of all the powers of 2 less than 1000. I then divided 1000 by each power of 2 and placed the greatest integer in column B. Column C multiplies (1) The divisibility of A by 2 and (2) B minus the previous value of B, which represents multiples of other values of A. Finally, the last row indicates that 1000! is divisible 994 times by 2. Here is a similar table for 5 as a prime factor:

Power of 5 = A	Number of factors of 1000! divisible by power of 5 = B	Number of divisibilities by 5 in these numbers = C
5^4=625	1	4*1=4
5^3=125	8	3*(8-1)= 21
5^2=25	40	2*(40-8)= 64
5^1=5	200	1*(200-40)= 160
Total divisibility of 1000! by 5:		249

Since 1000! is divisible more by 2 than by 5, the divisibility of 1000! by 10 is equal to the divisibility of 1000! by 5, which is 249. Therefore, there are 249 zeroes at the end of the number 1000!

1001! is simply 1000!*1001. 1001 is not divisible by 5, and so 1001! is divisible by 5 just as many times as is 1000!, and so it too ends in 249 zeroes. 1002! through 1004! follow the same logic. They all end in 249 zeroes.

1005!, however, is 1004!*1005. 1005 is once divisible by 5. Therefore, 1005! is divisible by 5 250 times. Therefore 1005! ends in 250 zeroes.

By Nathan Stiennon, 15 May 2002