We know that every 2 guests shared a dish of rice, every 3 shared a dish of broth and every 4 shared a dish of meat. What is the smallest number of guests that can fit this scenario?
If we only take into account the first two conditions, that is, 1) every 2 guests shared a dish of rice and 2) every 3 guests shared a dish of broth, then we need at least 6 people for this to work (since 6 can be divided into both 2 and 3). Now, for the multiples of 2, 3, and 4, let's see where they share a common multiple!
Listing out the multiples:
2, 4, 6, 8, 10, 12, ...
3, 6, 9, 12, ...
4, 8, 12, ...
So 12 is our friend here. Let's consider the case where there are 12 guests. How many dishes of each type would there be?
12 ÷2 = 6 dishes of rice
12 ÷3 = 4 dishes of broth
12 ÷4 = 3 dishes of meat
6 + 4 + 3 = 13 dishes in total
The question has 65 dishes. (65 ÷ 13 = 5)
Since we multiply 13 dishes by 5 to obtain 65 dishes, we also multiply 12 guests by 5 to obtain our final answer: 60 guests in total.
I think cultural context certainly would have an effect on the problem or the attitude of students towards solving this problem. If put into a Chinese context, a student with a Chinese cultural background may be more likely to tackle the problem without much thought. Another student without this background may not be familiar to the idea of so many people sharing multiple dishes, therefore be a bit confused when solving this problem. I found this same question on the NCTM facebook page using Mexican food as the cultural context. This shows that math puzzles can be tailored to meet the specific needs of certain students/community.
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