Raymond EDCP 442

To solve this problem without algebra, I need to phrase this question. An unknown number of guests are handed a number of rice dish, a number of broth dish, a number of meat dish. The total number of dishes handed from rice, broth and meat are 65, and each dish is shared with 2, 3, 4 guests respectively. In this question, I can begin by trial and error. Starting with product of 2, 3 and 4 which is 24 guests, we would have 24/2 + 24/3 + 24/4 = 12 + 8 + 6 = 26 dishes. This is too low. Next I double the number of guests to 48, we would have 48/2 + 48/3 + 48/4 = 24 + 16 + 12 = 52 dishes. It is a little low. But I notice it increase with a linear trend with every 24 guests yielding 26 dishes which can be interpreted that every 1 additional guest contributes to 13/12 additional dishes. So I compute how many more guests from 48 (which yields 52 dishes) do I need to achieve 65 dishes. Note that, since you can't have fractions with number of guests, we would use a floor value. 48 + (65 - 52...

Our assignment group was responsible for presenting Ahmes' loaf sharing problem, which was a problem created for entertainment purposes in ancient Egypt. It had two conditions: 1) 100 loaves for 5 men, with the allocations being an arithmetic progression, and 2) 1/7 of the 3 largest portions combined equals the 2 smallest portions combined. For the presentation, we discussed the context leading up to the problem, organized a toy example activity, discussed our modern and ancient solutions, and proposed an alternative strategy as our extension. Click here to access the presentation slides Pictured: Ahmes' loaf sharing problem in the style of an ancient Egyptian mural For the background, Raymond was responsible for looking into the ancient Egyptian mathematical developments that were necessary for the solving of this problem and presenting the problem description to the class. As for Zain, he was responsible for researching and presenting the solution that would've been possi...

I first started on the one-pan scale problem first using 5 weights to sum up from 1 to 31. The trial and error process began with trying out sum of weights for the first few whole numbers. Since I know the highest number achievable should be 31, having two weights both using 1 is irrational. Hence, I started with the first two weights as 1 and 2 and was able to reach 3 through that. For next whole number of 4, I instinctively assigned the third weight as 4 and was able to reach whole number of 7. The problem begins to look like a combinations of weights using exponents of 2. Using this rationale, I assigned fourth and fifth weight as 8 and 16 respectively which the sum of five weights are 31. I knew then that the solution is a series of exponents of 2. I proceeded to validate this solution accordingly. Moreover, each weight has a binary status of 0 and 1, so the total combination is 2^5-1 which also comes out to be 31. Next, I worked on the two-pan scale problem with only 4 weights to ...