Solving Big Problems

Transcription

1 Solving Big Problems A 3-Week Book of Big Problems, Solved Solving Big Problems Students July 25 SPMPS/BEAM

2 Contents Challenge Problems 2. Palindromes Pick Your Color A Big Multiplication Solving Big Problems: Week 6 2. Sectors of a Circle Shifting Digits Coins in Frobenia to Solving Big Problems Week Stones Problem ! Checkerboard Dominoes Prime Triplets Solving Big Problems Week Candy Conundrum Listed Numbers Infinity Infinite Inn Problems Were Solved By

3 Challenge Problems. Palindromes A palindrome is a number that is the same when it is read forward or backwards (for example: 2332, 54645, or 858). Find a number that evenly divides every 4-digit palindrome, and prove that every 4-digit number is divisible by this number. Proof: So before everything we thought that the answer was, but we didn t know why. During our Challenge Problem activity we had to try and find the formula abba, and we found when a < b that abba = a(b a)a, where a and b are -digit numbers. The way we got this was by doing long division, but it only works when a < b. So we need another method to prove it in general. First, we used random examples of four-digit palindromes to test that they are divisible by. Then we needed to prove that these work on every four-digit palindromes. For example, 3223, 4554, and We tested if they were divisible by by breaking the palindromes up. For example, 3223 = = = The broken-down numbers were divisible by. For example, 33 = 273, and 22 = 2 44 = 364 and 55 = 5 77 = 637 and 77 = 7 So now we have to prove if that formula works for every palindrome, so we substituted the outer end of the palindromes with a and the inner numbers with b, where a and b are single digit numbers. We broke up abba as abba = aa + bb, and that will be the formula. So now we use a + b. Since = 9 and =, we use the sentence and the distributive property: aa + bb = a + b = (9a + b) This shows that aa + bb is a multiple of. Then when we distribute the answer will be a + b = abba. So abba is a multiple of. These proofs were written by Rebecca and Qige. The solution was joint work with Malachi, 3

4 Ghania, and Olivia..2 Pick Your Color 2 people stand in a line. Each person has a sign with a color on their back, either RED or BLUE. Everyone can see the color of everyone in front of them in line, but not their own color (so the person at the front of the line cannot see anyone s color). You can strategize before everyone has their own color, but once the people are lined up and everyone has a color, each person is allowed to say exactly one word, either BLUE or RED to guess their own color. Is it possible to make sure that people guess their own color correctly? Proof: Yes, it is possible. The person at the back of the line speaks first, and uses the following code. If he says RED, that means there are an even number of red people ahead of him in line (and an odd number of blue people). If he says BLUE, that means there are an odd number of red people ahead of him in line (and an even number of blue people). The person at the back of the line might not say his own color correctly, but everyone else can use this clue to figure out their own color. For example, if the person last in line says RED, the second-to-last person knows that there are an even number of red people in the line, including himself. So he looks at the people ahead of him: if an even number of them are red, he must be blue, so he says BLUE. If there are an odd number of red people in front of him, he himself must be red in order for there to be an even number of red people in line. So, he says RED. The next person in line now knows the color of the person behind him, as well as the color of all the people ahead of him, so he uses the same strategy to figure out his own color: if the number of red people ahead of him and behind him is even, he will say BLUE but if the number of red people ahead of him and behind him is odd, he must be RED in order for the total number of red people to be even. This works if the person last in line originally says BLUE too, you can use the same reasoning. This proof was based on a presentation given by Jay, summarized by the SBP instructor..3 A Big Multiplication The number 99 9 {{ 94 times 4

5 is multiplied by the number 44 {{ 4 94 times What is the sum of the digits of the product? Solution: The sum of the digits of the product is 94 9 = 846. When you look at smaller cases of the problem we have 9 4 = = = = As you can see from the pattern, this suggests that 99 {{ 9 94 times 44 4 {{ 94 times = 4 4 {{ 93 times 3 55 {{ times There are several ways to explain this. First, if we instead compute {{ 44 {{ 4 = 44 {{ 4 {{, 94 times 94 times 94 times 94 times then by the distributive property, we have 99 {{ 9 44 {{ 4 = {{ 44 {{ 4 44 {{ 4 94 times 94 times 94 times 94 times 94 times When we subtract , we get 99 {{ 9 44 {{ 4 94 times 94 times = 4 4 {{ 93 times 3 55 {{ times Now when you add up the digits, there are 93 pairs of and pair of 3 + 6, so the sum of the digits is 94 9 = 846. We can also see why this is the product by looking at what happens when we multiply In the first column, you will always bring down a 6. In the second column, you ll bring down a 5. For the next 92 columns, you will be adding more copies of 9 to 9 + 6, but we are also carrying over another larger digit, so you are always bringing down a 5. Then, when we get to the 95th digit, we are adding a 3 instead of a 9, so the sum is 3. After that, each column has one fewer 9, so the sum is 4 for the rest of the digits. These proofs are based on a presentation given by Rebecca, Ghania, and Chuka, summarized by the SBP instructor. 5

6 2 2. Solving Big Problems: Week Sectors of a Circle Given a circle with 6 segments, each segment filled with some whole number, we apply the following rule: at each step, we can choose a pair of adjacent segments, and add to each segment. We name the circles as follows: F A E B D C For example, we can get from the first circle below to the second circle below by adding s to A and B, then B and C, then D and E, then E and F. 2 2 We can not get from the first circle below to the second circle by ap- 2 2 Conjecture: plying this rule Proof: We can only add to a segment if we also add to an adjacent segment. The only 2 segments adjacent to B are A and C. But A and C need to stay the same, so there is no 6

7 way to make B into a 4 without making A or C bigger. Proof by Crystal, with input from Ghania and Khalil. Lemma: The numbers on the circle never get smaller. Malachi s Conjecture: You cannot start at the first circle below and get to the second circle: Proof: We can only add to pairs of adjacent sectors. So if we want B to become, we need to add to either A or C, which makes one of them 2. Proof was a collaborative effort of SBP Week Block 4-6 Claim: If you have 3 pairs of matching numbers that are adjacent to each other (see the diagram), then it is possible to get all segments equal. c a c a b b Proof: If you have three pairs of adjacent matching numbers, then you can get all segments to be equal because you can keep adding s to any matching pair as many times 7

8 as you want until all the pairs are the same number. Proof by Taco, with help from the 4-6 section. Keita s Conjecture: Starting at the first circle below, you cannot get the second circle below: Proof: We start with and 5 s. We need to add to the 5 s. After we add in 2 pairs of s, there is going to be a left over that is adjacent to s on both sides. If we add to that and an adjacent segment, the will be come a but the will become a 2. Proof written by the 4-6 section. Alternate proof to Keita s Conjecture: We are starting with a total sum of (which is odd). But the circle with all s has a total sum of 6, which is even. We cannot get from a total sum that is odd to a total sum that is even, so it is impossible to reach the second circle from the first. Proof written by the 4-6 section. Theorem: If we begin with a circle so that the sum of all the sectors is odd, we will never get all sectors equal to each other. Proof: Every time we apply the rule, we add 2 to the total sum, since we always add to sectors in pairs. So if we start with a circle that has a total sum that is odd, the total sum will always be odd. This is because whenever we add an even number (2) to an odd number, the result is odd. So the total sum can never be an even number. But if all of the sectors were equal to some number n, the total sum would be n + n + n + n + n + n = 6n, and 6n is even since 6 is even. So the sectors will never be equal to each other. Proof written by the 4-6 section. 8

9 Open Problem: An open problem is a problem that has not yet been solved. If we begin with: can we ever get all sectors equal to each other? We know we cannot make all the sectors equal to. We think it is impossible to get all sectors equal, but since the sum of the sectors is even, we have not yet been able to prove that it is impossible. If we begin with a circle so that the sum of all the sectors is even, will we always be able to make all sectors equal? 2.2 A MathPath Problem Students in the -3 section solved Problem 5 from the MathPath 25 qualifying Test. The solution has been omitted, as MathPath may reuse the problem in future years. 2.3 Coins in Frobenia In the land of Frobenia, their only currency is in the form of 5 and 7 cent coins. Using as many of these coins as you want, for what values of n can you make change? Conjecture (Ghania): We can make any multiple of 5 or 7. Proof: If we want 5n cents, I can use n 5 cent coins. If I want a number that is 7 something, it is a multiple of 7 and I can make it with 7 cent coins and forget about the 5 cent coins. Corollary: I can also make any multiple of 2, since 2 = cents, I can use n 5 cent coins and n 7 cent coins. So if I want 2n Conjecture: We can never make 6 cents. 9

10 Proof: If we only use one 5 cent coin, there is no single coin to get us to 6. use more than one 5 cent coin, or if we use a 7 cent coin, we are above 6 cents. Proof written by the -3 section. If we Conjecture: We cannot make 8 cents. Proof: Since 8 is not a multiple of 5 or 7, we won t be able to make it using only 5 cent coins or 7 cent coins. So, we need at least one of each: a 5 cent coin and a 7 cent coin. This brings us to 2 cents. Now we have to use coins to make the remaining 6 cents, and we already proved that this is impossible. So we cannot make 8 cents. Proof by Khalil and his table. Question from 4-6 section: Is there a largest number n that you cannot make with these coins? Taco s Method: Taco s method allows you to take a large number n, and make change for n using 5 and 7 cent coins. Here is an example: if n = , we first find a multiple of 7 that ends in 9: for example, 49. Now you subtract 49 from n, and so we can rewrite n = We can use 5 cent coins to make , and 7 cent coins to make 49. Putting these together gives us n. In general, if we start with a big number n, we find a multiple of 7 that has the same digit in the s place that n does. Now we subtract that multiple of 7 from n, so that you are left with a number that has in the s place. This is a multiple of, so we know we can make this amount using 5 cent coins. We can make the multiple of 7 using 7 cent coins. We can put all these together to make n. Room 238 Conjecture (Johnathan and Malachi): Taco s Method will allow us to make any number greater than or equal to 7 using 5 and 7 cent coins. Proof: Let e be a number greater than or equal to 7. Our goal is to use Taco s Method to use 5 and 7 cent coins to make e: e = 5n + 7m, where n is the number of 5 cent coins we use, and m is the number of 7 cent coins we use. Given e, we find a multiple of 7 that is less than or equal to 7 whose s digit matches e s

11 s digit. The first multiples of 7 give us each of the possible digits in the s place: 7 = = = = = = = 56 7 = = = 49 We will subtract this multiple of 7, 7m, from e, and end up with in the s place. The means e 7m is a multiple of 5. We divide this number e 7m by 5 to get the number of nickels we need: so n = e 7m 5. If we put together these n nickels and m 7 cent coins, we get e: 7m + 5( e 7m 5 ) = 7m + 5(e 7m) 5 = 7m + (e 7m) = e, so we can make e using 5 and 7 cent coins. Proof written by Malachi and Johnathan, with help from the 4-6 section. Theorem: We can get any number greater than 23. Proof: We can make a table to keep track of which ones we can and cannot make = = = = = = = = = = = = = = = = = = In any column, once we can get one of the numbers, we ll be able to get all the numbers below it in that column. For example, once we have 2 = 7 3, we can get 26 because we can just add a 5 cent coin to 2. Then we can get 3 by adding another 5 cent coin to 26. This works in any column. Any number bigger than 23 will be underneath in one of the columns, below one of the numbers we can make, and so we just add 5 cent coins to the one we can make until we get to the one we want..

12 Proof written by the -3 section, with guidance from Kaya. 2.4 to 5 Question: Is it possible to insert + and signs in the spaces below and have the total sum be? = Theorem: It is impossible to make this sum equal to. Proof: The sum of all the (positive) numbers from to 5 is = 275, which is an odd number. To get the total list of numbers to be, we need the sum of all the numbers being added, and the sum of all the numbers being subtracted to be equal. When the sum of the numbers being added and the sum of the numbers being subtracted are added together, we get 275 because every number on the list is either being added or subtracted. Since 275 is odd, we cannot break it in half without getting a decimal. However, since we are only using addition and subtraction, the sum of the numbers being added cannot be a decimal. So it is not possible. Proof written by the 4-6 section. 2

13 3 Solving Big Problems Week Stones Problem You play a game with a friend. There is a pile of stones in between you. Each turn, you can take either, 2, or 3 stones from the pile. You can t skip your turn. How can you make sure you always win the game? Claim: If the number of stones in the pile is a multiple of 4, you can always win by letting your friend go first. If the number of stones in the pile is not a multiple of 4, you can always win by going first. Proof: Suppose the pile has n stones. If n is a multiple of 4, say n = 4m, you should go second, and use Team Bun s Strategy, which tells you what to take based on what your opponent takes: If your opponent takes you should take n 4 n If you follow this strategy, each time you will remove 4 stones from the pile between you and your opponent. So if n = 4m, after one round there will be 4m 4 = 4(m ) stones left. After 2 rounds, there will be 4m 8 = 4(m 2) stones left. After m rounds, there will be 4m 4(m ) = 4m 4m + 4 = 4 stones left. Once your opponent is left with 4 stones, they will have to take, 2 or 3, and you will be able to take the rest and win. If n is not a multiple of 4, then it must be in between two multiples of 4, for example 4m and 4m + 4. Then n must be either 4m +, 4m + 2, or 4m + 3. You should go first, and take either, 2 or 3 stones to leave your opponent with 4m stones. Now the number of stones left is a multiple of 4 and it s your opponent s turn, so we are back to Case. This proof was a combination of the solution given in the -3 class, and the Losing Theorem/Victor Theorem/Karoline s Starburst Theorem in the 4-6 class. 3

14 3.2! Question The number! = must end in a, since when we multiply any whole number by, the number stays the same except with 2 extra digits at the end. How many zeroes appear at the end of!? Solution: There are 24 zero s at the end of!. We figured this out by looking at smaller values of n!. The first factorial that ends in a is 5! = = 2. We don t get another at the end until we reach! = The next new appears at 5!, which ends in 3 zero s, and the next new appears at 2! which ends in 4 zero s. We predicted that every multiple of 5 gives us a new at the end of the factorial. But when we get to 25!, it ends in 6 zero s instead of 5. We realized that 25 4 =, and when we multiply 25 by any multiple of 4, it will give us 2 extra zero s at the end of the number. Them we realized that since 25! = 25 24!, and since 24! is a multiple of 4, we will get two extra zero s when we get to 25!. Similarly, every multiple of 25 will contribute two new zero s. So each of the multiples of 5 between and will contribute one new zero, except for the multiples of 25, which will contribute 2 new zero s. There are 4 multiples of 25 between and, so in total we have = 24 zero s at the end of!. This solution was a joint effort of the -3 class, with a big conjecture from Daisy about multiples of Checkerboard Dominoes An 8 8 checkerboard has marbles covering two opposite corners. A domino can cover any two squares that share a side. Is it possible to cover the checkerboard with dominoes so that every square (other than the two corners) are covered with exactly side of a domino? Solution: No, it is impossible. In an 8 8 checkerboard, there are 64 squares total: 32 black squares and 32 white squares. If we cover two opposite corners, the marbles are covering 2 white squares, so we have to cover 3 white squares and 32 black squares with dominoes. But each domino can only cover two squares that share a side, so every domino must cover one black square and one white square. But we only have 5 pairs of black and white squares, and at the end we will always have 2 4

15 black squares leftover. Since two black squares never share a side, we cannot cover the leftover squares with one domino, so it is impossible to cover the whole board. Proof by Angela in the -3 class. Subdividing a Square Suppose we start with a square. It is possible to break up the square into smaller squares, not all necessarily the same size, but so that the original square is totally filled with squares. For what values of n can we subdivide a square into n smaller squares? Amadou s Theorem: We can make any multiple of 3 greater than or equal to 6. Proof: We start by subdividing a square into 6 smaller squares. We can do this by making 9 equal size squares, but turning 4 of the smaller squares into larger square. See Diagram A. Now we can take any of the squares we have, and subdivide it into 4 squares by cutting it in half vertically and horizontally. This gives us 3 new squares because we added 4 squares, but used one of the old ones to do it. So we have a total of 9. Now we can do this again to any of the squares, which gives us 3 new squares for a total of 2. We can keep doing this and get to any multiple of 3. Proof written by Amadou, with help from the -3 section. Crystal s Theorem: We can make any even number n bigger than 2. Proof: If I take the number (2N) that I want to get and divide it by 2 (I get N), I know how big my square has to be: N N. I fill it out but then erase everything except the left side of squares and the top row of squares. There is big square left in the middle. Then you count all the squares on the side and on top, making sure you don t recount any squares, and add that to the big square. That gives you N + (N ) + = 2N squares total. In diagram B below, you can see how to do this for N = 7, to get 2N = 4 squares. Proof written by Crystal. 5

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17 Johnathan s method: Using Crystal s Theorem, we can get all the odd numbers that are 3 greater than an even number bigger than 2 (so all odd numbers bigger than 5). Proof: If we want to make an odd number n of squares, and n is at least 7, we know how to make n 3 using Crystal s Theorem, because n 3 is an even number that is at least 4. We make the n 3 squares using Crystal s Theorem, and then pick one of the squares and cut it into 4 smaller squares by cutting it in half vertically and horizontally. This gives us 3 new squares (because we added 4 small squares but lost a bigger square) which gives us n = n squares in total. Proof written Johnathan with help from the 4-6 section. Corollary: We can subdivide a square for n =, n = 4, and any n greater than 5. Proof: This follows from Crystal s Theorem and Johnathan s method. 3.4 Prime Triplets A prime triplet is 3 consecutive odd numbers, all of which are prime. One example is 3, 5, 7. Are there any others? Solution: There are no other prime triplets. We claim that every three consecutive odd numbers must contain exactly one multiple of 3. This is because every third number starting at 3 is a multiple of 3, and every second number starting at 3 will be odd. Then every sixth number starting at 3 will be an odd multiple of 3. So 3, 5, 7 is a prime triplet, but any odd triplet that starts at 5 will contain an multiple of 3 that is bigger than 3, which is not prime. This proof was a summary by the SBP instructor of the solution given by the class, and inspired by a claim made by Magda. 7

18 4 Solving Big Problems Week 3 4. Candy Conundrum You have 3 orange starbursts and 3 yellow starburst, and you divide them between 3 bags as follows: 2 orange, 2 yellow, and orange, yellow. You label each bag with a post-it according to what is inside each bag. Then, your worst enemy comes along to ruin your day, and switches the post-its so that all of the labels are now incorrect. Your enemy now tells you that you can choose exactly piece of candy from exactly bag, look at it, and try to relabel all three bags correctly. If you succeed, you get to keep all the candy. If you fail to correctly label all the bags, your enemy keeps all the candy. Is there anything you can do to make sure that you keep your candy? Strategy: You can always win the game by picking candy from the bag labelled yellow-orange. Here is the strategy: If you pull out an orange, re-label bag 2 orange-orange. Now change the label on Bag to yellow-yellow and change the label on bag 3 to yellow-orange. If you pull out a yellow, re-label bag 2 yellow-yellow. Now change the label on Bag 3 to orange-orange, and change the label on bag to yellow-orange. Proof: Case : you take an orange from the bag labelled yellow-orange. We know this bag must contain two orange starburst, because we know the yellow-orange label is wrong, so it must contain two oranges or two yellows. Now we relabel bag from orange-orange to yellow-yellow. The first bag cannot be labeled orangeyellow, because if it were, then bag 3 keeps the same label yellow-yellow, and we know all the labels start wrong. Now we only have one label left, so we must label bag 3 as yellow-orange. Case 2 is analogous. If you choose orange-yellow from bag 2, we now know that bag 2 must be yellow-yellow, and you still need to switch the other labels because you can t leave any original labels the same. Proof written by the -3 section. 8

19 4.2 Listed Numbers Choose any odd number n. Take your number, double it (2n), and list all of the numbers from to 2n. Pick any 2 numbers on your list, cross them out, and replace them with the positive difference between your two numbers. For example, if you pick 2 and 5, you cross out 2 and 5 from your list, and replace them with 3, because 5 2 = 3. Repeat that step with your new list, and continue to repeat that step until there is only one non-zero number left on the list. What numbers can be left? Conjecture: We can only be left with odd numbers. Proof If we start with a list of numbers from to 2n, we will have n even numbers on the list and n odd numbers on the list. When we choose 2 numbers from the list, we must choose either 2 evens, 2 odds, or one of each. If we erase 2 even numbers, we replace them with another even number, so the amount of odd numbers on the list stays the same. If we erase 2 odd numbers, we replace them with an even number, because odd - odd = even. So the amount of odd numbers on the list goes down by 2. If we erase even and odd, we replace them with an odd number, because even - odd = odd. So the amount of odd numbers on the list stays the same. According to Joseph, this proves that if your list starts with 2 odds, you can never be left with odd. This is because every time we erase two numbers, the amount of odd numbers on the list stays the same or goes down by 2. Since we start with an odd amount of odd numbers, and remove only or 2 at a time, we will never get down to exactly odd numbers at the end, so we must be left with odd number at the end. This proof was based on a solution given by the -3 section, summarized by the SBP instructor. 4.3 Infinity Conjecture: If we add up an infinite amount of positive numbers, the sum goes to infinity. Proof: This is FALSE. As an example, we look at the sum:

20 we can show that the sum gets really close to, but never goes above. If we just add the first two terms, we get = 3 4. If we add the first 3 terms, we get = 7 8. If we add the first 4 terms, we get = 5 6. Every time we add a new term, we get closer to, but we don t ever reach. Proof written by the -3 section. 4.4 Infinite Inn At the Infinite Inn, there are infinitely many rooms, with each door labeled with a natural number. Every guest at the Inn must stay in their own room. The Infinite Inn is completely full of guests. If one new guest arrives, can they be accommodated? If 4 new guests arrive, can they be accommodated? If a bus full of infinitely many new guests arrive, can they be accommodated? If two buses arrive, each full of infinitely many new guests, can they be accommodated? Solution: The answer to all of these questions is yes. If one new guest arrives, the manager of the hotel will ask every guest in the hotel to come out of their room, and move to the next room in the hallway. So, the guest in room moves to room 2, the guest in room 2 moves to room 3, etc. In general, the guest in room n moves to room n +. Since there are infinitely many rooms, every guest will still have a room, and now room is empty, so that the new guest can stay there. If 4 new guests arrive, the manager asks each guest to move 4 rooms down the hall; so the guest in room n moves to room n + 4. Again, since there are infinitely many rooms, everyone still has a room, and the first 4 rooms are now empty so the new 4 guests can be accommodated. If a bus full of infinitely many new guests arrive, we can assign new rooms as follows: For every guest already in the hotel, the guest in room n will be asked to move to room 2n. So the people originally in the hotel will move to all of the evennumbered rooms. For every guest in the bus, the guest in seat n will move to room 2n. So the guests on the bus will stay in the odd-numbered rooms. 2

21 If two infinite buses arrive, we can assign new rooms as follows: For the guests in the hotel, the guest originally in room n will move to room 3n 2. For the guests on the first bus, the guest in seat n will move to room 3n. For the guests on the second bus, the guest in seat n will move to room 3n. These proofs are based on solutions given by the -3 section, summarized by the SBP instructor. Jesus (open) Question: Can we fit infinitely many infinite buses into the hotel? Ghania s Conjecture: The real numbers are a larger infinity than the natural numbers. Proof: We will show that a bus with a seat for every real number cannot be accommodated at the Infinite Inn. To prove this, we suppose that the real number bus can be accommodated at the Inn. Then we would be able to list all the real numbers. This list would look like: a. a a 2 a 3 a 4 a 5... b. b b 2 b 3 b 4 b 5... c. c c 2 c 3 c 4 c 5... d. d d 2 d 3 d 4 d 5... e. e e 2 e 3 e 4 e 5... We can find a real number that is NOT on the list. Here is the number:. M =.M M 2 M 3 M 4 M 5, where M is any digit other than a, M 2 is any digit other than b 2, M 3 is any digit other than c 3, etc. We will show that M is not on this list. If M were on the list, say it was the nth number on the list, then we would have chosen a different nth digit for M n, so M cannot be the nth number on the list. Then we know that the guest in seat M cannot be accommodated at the hotel, so the real numbers must be larger than the natural numbers. Proof written by the -3 section. 2

22 5 Problems Were Solved By... Week, -3 PM Amadou, Chris, Crystal, Ghania, Hilda, Joseph, Karoline, Kaya, Khalil, Nhyria, and Tiffani Week, 4-6 PM Angela, Briana, Daisy, Emmanuel (Taco), Jonathan, Keita, Magda, Malachi, Qige, and Sally Week 2, - 3 PM Amadou, Angela, Chris, Daisy, Hilda, Joseph, Malachi, and Sally Week 2, 4-6 PM Brianna, Crystal, Ghania, Johnathan, Karoline, Kaya, Keita, Khalil, Magda, Nhyira, Qige, Emmanuel (Taco), and Tiffani Week 3, - 3 PM Andrew, Ghania, Jesus, Jilanda, Joseph, Josue, Khalil, and Malachi Challenge Problems Chuka, Ghania, Jay, Malachi, Olivia, Qige, and Rebecca 22