Problem 1 How many zeros at the end of 30!
In class, we discussed how the number of zeros at the end of an integer is exactly the highest power of 10 that divides the integer evenly. To count the total number of powers of 10, we can count separately the number of powers of 2 and powers of 5. In factorials, we quickly realized that there are plenty of factors of 2 so we counted factors of 5 in 30!= 1 x 2 x 3 x 4 x ... x 28 x 29 x 30 The only factors that had powers of 5 were 5,10,15,20,25, and 30. Each of the those factors had one power of 5, except 25, which had two powers of 5 since 25= 5 x 5. That gave us a total of 7 fives in the factorization of 30!, so there would be 7 zeros at the end of 30!.

Problem 2 What is the next largest number after 5 which does not occur as the exact number of zeros at the end of some factorial?
This was assigned as a hw problem and we talked about it in class afterwards. Gaps in the number of zeros which occur have to do with getting more than one factor of 5 at a time. So there are gaps which come at 25!, since there are 2 powers of 5 in 25, and the next one is at 50!, which also has two powers of 5. We also noticed that there would be a double gap between 124! and 125!, since 125 = 5 x 5 x 5, the cube of 5, has 3 powers of 5 in its factorization.

Problem 3 How many zeros at the end of 200!
Assigned as homework, and discussed in class. For this, we counted the multiples of 5 (40 of them), the multiples of 25 (8 of them), and the multiples of 125 (just one of them.) That gives us a total of 49 powers of 5, so there are 49 zeros at the end of 200!. We did not need to consider the next power of 5 (which is 625= 5^4 = 5 x 5 x 5 x 5) since that was bigger than 200.

Problem 4 Describe an algorithm to find the number of zeros at the end of n!
Worked on in class, then assigned as homework, then discussed again in class. We have many different approaches. We realized that the number of 5's occuring in the complete factorization will determine the highest power of 10 that divides the number, and thus the number of zeros. To count the total number of 5's, we need to count multiples of 5, multiples of 25, multiples of 125, and so on until we are trying to count multiples of some power of 5 which is bigger than n. Here was a suggested algorithm:
1. Set an index r=1.
2. Compare 5^r with n. If it is bigger, stop increasing r and go to step 3. If it is less than or equal to n, increase r by one and do step 2 again.
3. Count the multiples of 5^1, 5^2, ... 5^(r-1) as follows:
n/5^1 + n/5^2 + ... + n/5^(r-1)
always rounding non-integer answers down to the integer part. That total will give the total number of 5's and thus the total number of zeros in n!.

Problem 5 How many zeros are there at the end of (5^15)* 10!
Assigned as homework and then discussed in class. For this one, we have plenty of 5's but not very many 2's. So to count the zeros at the end, we notice there are 17 fives and 8 twos. All the twos come from 1 x 2 x 3 x 4 x 5 x ... x 9 x 10, the 10! part and remember there are 2 twos in 4 and 3 twos in 8.

Problem 6 If the numbers between 1 and 100 are put into alphabetical order, what is the first prime on the list?
Assigned as homework, discussed in class. We realized that numbers that start with e like "eight", "eighteen" and "eighty three" are near the beginning of the alphabet and a little playing around revealed that the first prime was "eighty-nine"

Problem 7 What is the last digit of 3^2714?
Assigned as homework, discussed in class. We started by trying to find the last digits of 3^0, 3^1, 3^2, etc. until we noticed a pattern, which was that there is a cycle of length 4. So every four powers of 3 accomplishes nothing, and we are interested in what the remainder of our exponent is after dividing by 4.

Problem 8 44 students come to class and each of them shakes hands with every other student. How many handshakes take place? If there are twice as many people, are there twice as many handshakes?
Assigned as homework, discussed in class. Solved several ways. There are 43+42+41+...+1+0 total handshakes, since the first person shakes hands with every other person in the room, and then the second person shakes hands with everyone whose hand they haven't already shaken, and so on. Another way to think about it is there are 44 choices for the first person involved in a handshake, and then 43 choices for the other person. But that counts handshakes twice, so we divide by 2.

Problem 9 After detailing the rules of the TV game show ``Jeopardy'', the problem is: what is the theoretical maximum amount of money possible to win on one show?
Assigned as homework, discussed and solved in class. Important features: maximum possible winnings if we run into the daily doubles at the very end of each round and if we bet everything we can at each point. Since we don't actually get the money from the daily doubles, those should occur under the least valuable ($100 in the first round, $200 in the second round) rows. There were several ways to think about the total which was more than $283,000.

Problem 10 Compute the sum 1+2+3+4+ ... + 998 + 999 + 1000
Done in class after talking about sums of numbers. If we group 1000 and 1, 999 and 2, etc. we end up with 500 groups, each group totalling 1001, to give us a total of 500,500.

Problem 11 A golf store sells golf balls in sleeves of 3 and boxes of 12. The manager reports that 9052 balls have been sold. Should we ask for a recount? If so, why? What about if balls were sold only in boxes of 12 or 20?
We notice that no matter what combination of sleeves and boxes we sell, they will always be multiples of 3 balls sold. So since 9052 is not a multiple of 3, there is some mistake and we should ask for a recount. The general idea involves greatest common divisor, which we discussed. For the 12 and 20 case, the gcd is 4, which does go evenly into 9052. So there is a chance that it can work out. In fact, there are many ways it can work out, including some such as 9052= 20 (452) + 12 (1)

Problem 12 Write out the first 10 rows of Pascal's Triangle (described in lecture) What pattern occurs for the number of possible handshakes between n people?
Pascal's Triangle was discussed and drawn out. It begins as
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
...
where each row is obtained by adding sucessive entries on the row above, preceded by and followed by a 1. The number of handshakes for 2 people is 1, 3 people is 3, 4 people is 6, 5 people is 10, and we notice that these numbers occur in Pascal's triangle along a sloping diagonal line. In the version above, it appears as the third column, but that is because I have arranged the triangle to be flush to the left. Normally we draw it more symmetrically. We talked about "triangular numbers" which are the numbers that occur as possible numbers of handshakes as well as triangular arrangements of bowling pins.

Problem 13 What is the sum of all multiples of 17 between 1000 and 3000
We notice that 1003 and 2992 are multiples of 17, so we are looking for the sum 1003+1020+1037+...+2992. There are a total of 117 numbers to add up, which we can group evenly in to teams of 2 which add up to 3995. So the total sum is 3995*118/2.

Problem 14 How many different rectangle placements are there on an 8x8 checkerboard?
This one we discussed extensively in class. After trying some small examples to get a feel for the problem, we started thinking about this in terms of choices of corners of the rectangle. There are 9 rows and 9 columns of corners on an 8x8 rectangle, so there are 81 choices for the first corner. For the opposite corner, we want to pick something that makes a rectangle (not a stick or a dot) so we cannot choose in the same row or column as our first corner. There are 8 other rows and 8 other columns so there are 64 choices for the second corner. But we have counted each rectangle several times- exactly 4 times, so to get the total number of rectangles we divide by 4. That gives us 81^2 64^2 /4 rectangles.

These are the notes up to exam 1. Notes for the coursework following exam 1 will be posted only under the "Course Content" section in the WebCT page.