Diary of Mathematical Thinking 1 I am doing a mathematical thinking problem this evening.
I supported my forehead with a pen and frowned. Here and there? Neither. Hey, how to write it? Suddenly a correct method flashed before my eyes, three times four equals twelve, sixty divided by five, ah! Before the answer of sixty divided by five was worked out, I immediately denied it again.
I didn't know I almost succeeded just now, so I called out to my mother and she asked me what I was doing. I said: I can't do this problem. Mom just asked me to try the draft paper. I multiply three times five to get fifteen, which is definitely not possible. The title stipulates that a number can only be used once, and the five in the formula is repeated. I will multiply two by five to get ten, but I can't, because no matter how the numbers six, four and three are combined, they can't be divided by ten. I multiply four by five to get twenty, which is no good, because I still can't work out twenty with the remaining numbers. Finally, I use three times four equals twelve equals sixty divided by five to work out the correct answer.
I told my mother that I almost got it out, and my mother said meaningfully, be careful! Don't be too impatient!
Diary of Mathematical Thinking 2 Today I'm going to tell you the last question of hat trick.
Look at the problem first: the hat shop owner brought two white hats and three red hats, and then put three of them on the heads of three people, at least one of whom was wearing a red hat. Ask Taro first: "What color is your hat?" Taro replied, "I don't know." Next, I asked Hanako, and Hanako replied, "I see it. It's red." So, what color is Mr. Shadow's hat?
If "Mr. Shadow" and "Mr. Shadow" are both white hats, then Taro can definitely guess that his hat is red, so that at least one of Hanako and "Mr. Shadow" is red. Then Hanako replied that it was a red hat. We know that if Mr. Shadow is a red hat, Hanako should not guess whether he is a red hat or a white hat. Therefore, Mr. Shadow is wearing a white hat.
This problem mainly adopts hypothesis method, reasoning step by step, looking at the problem from different angles, and finally finding the answer.
Diary of Mathematical Thinking 3 This is a book about permutation and combination. Generally, the knowledge of permutation and combination will be taught in the course of "Probability and Statistics" in high schools. I don't quite understand this book. Let me give a simple example first.
If there are three little pigs and you have to live in five houses, and the number of little pigs in each house is not specified, how many ways of accommodation are there? First of all, the first pig can live in any house at will, so there are five choices; Similarly, the second pig can live in these five houses at will, so there are five kinds; Similarly, the third pig has five choices. So the total * * * can form an arrangement of 5×5×5= 125.
Let's consider a more complicated situation. If only one pig and one * * * can live in a house, how many arrangements are there? First of all, the first pig can live in any house at will, so there are five choices; The second pig can only live in the remaining four houses, so there are four choices; Similarly, the third pig can only live in the remaining three houses, and there are only three choices. So the total * * * can form an arrangement of 5×4×3=60.
If we don't consider the arrangement order of piglets, but only consider the combination when they live in the house, how many ways are there? This situation, which we call combination, will be much less than arrangement. In other words, if you don't consider the arrangement mentioned above, that is, if the piglets are all the same, many of them are repetitive. So, how many kinds are repetitive? For the arrangement of three piglets, we can know that there are 1, 2, 3; 1,3,2; 2, 1,3; 2,3, 1; 3, 1,2; 3, 2,1* * six permutations, that is to say, for the above permutations, each case will be repeated six times to form 60 permutations. When we don't consider the arrangement order of piglets, the combination number becomes 60÷6= 10.
Permutations and combinations are really interesting, but they are also difficult to understand. I still need to learn more knowledge.
Diary of Mathematical Thinking 4 Reflections on Age
I have done a lot of research reports before, but I have not done an age report, so I want to do it this time.
My brother, little cousin, big cousin, first cousin, little cousin, big cousin, mother, father, grandparents, grandparents, grandpa.
10 13 6 7 9 17 14 15 38 43 74 65 82 82
The oldest is 82 years old and the youngest is 6 years old. The difference between the oldest and the youngest is 82-6=76 years old. The average age is the total age divided by 14 people (10+13+6+7+9+14+15+38+43+74+65+82+82). The difference between the highest age and the lowest age is 76 years, with an average age of 34 years. People in our family are quite young.