Example: 1: Unload several cases from the freighter with a total weight of 10 ton, and the weight of each case shall not exceed 1 ton. In order to ensure that these boxes can be transported at one time, how many cars with a load of 3 tons are needed at least?
[Analysis and Solution] Because the weight of each box is not more than 1 ton, the weight of each box that a car can transport is not less than 2 tons, otherwise another box can be put. So five cars are enough, but four cars may not be able to carry all the boxes away. For example, if there are 13 boxes, then each car can only transport 3 boxes, and 13 boxes cannot be transported by 4 cars at a time.
So in order to ensure that all the boxes can be transported at one time, at least five cars are needed.
Example 2: Intercept 100 short bamboo poles with a length of10 foot respectively. How many raw materials should be used at least? What is the most cost-effective cut?
[Analysis and solution] A bamboo pole with a length of 10 feet should be cut in three ways:
(1) 3 feet 2 and 4 feet 1, the most economical;
(2) three feet three, more than one foot;
(3) 4 feet 2, more than 2 feet.
In order to save materials, try to use the method of (1). With 50 raw materials, 100 3-foot bamboo poles and 50 4-foot bamboo poles can be cut. If 50 4-foot bamboo poles are short, it is best to choose method (3), which requires the least raw materials, only 25, and at least 75 raw materials.
Example 3: The lengths of the three sides of an acute triangle are two digits respectively, and they are three consecutive even numbers. The sum of their numbers is a multiple of 7. What is the longest circumference of this triangle?
[Analysis and Solution] Because the three sides of a triangle are three consecutive even numbers, their unit digits can only be 0, 2, 4, 6, 8, and their sum is even, and because the sum of their unit digits is a multiple of 7, it can only be 14, and the maximum value of the three sides of a triangle can be 86, 88, 90, so the longest circumference is 86+.
Example 4: Decomposition of 25 into the sum of several positive integers to maximize their products.
[Analysis and Solution] Start the experiment with a small number shape and find its law:
6 divided by 3+3, its product is 3×3=9.
Divide 7 by 3+2+2, and the product is 3×2×2= 12.
Divide 8 by 3+3+2, and the maximum product is 3× 3× 2 =18;
9 divided by 3+3+3, its product is 3×3×3=27. ……
That is to say, in order to maximize the product of decomposition numbers, 3 should appear as much as possible. When a natural number can be expressed as the sum of several 3' s and 1, one 3' s and 1 should be taken out and then decomposed into two 2' s, so that 25 can be decomposed into 3+3+3+3+2+2.
Example 5: A and B are going to explore the desert. They go deep into the desert for 20 kilometers every day. It is known that each person can carry one person's food and water for up to 24 days. If some food is not allowed to be stored halfway, how many kilometers can one of them go deep into the desert (the last two need to return to the starting point)? What if some food can be stored on the way back?
[Analysis and Solution] Suppose A goes back after X days, A leaves the food he needs when he goes back, and the rest is transferred to B. At this time, B*** has (48-3X) days of food, so X=8. For the remaining 24 days of food, B can only move forward for 8 days, leaving 16 days of food for him to return, so B can.
If the conditions are changed, the crux of the problem is the food left over in B24 days when A returns. Because 24 days of food can make B go deep into the desert alone 12 days, and the other 24 days of food will provide A and B with a walk back and forth, that is, 24÷4=6 days, so B can go deep into the desert 18 days, that is, one day.
Ex. 6: Every worker and every piece of equipment in two garment factories, A and B, can completely produce suits of the same specifications. Factory A produces tops, pants and only 900 suits a month. Factory B spends a lot of time producing tops and trousers, and just produces 65,438+0,200 suits a month. Now the two factories are jointly producing, trying to produce more suits. So how many more suits are produced each month than in the past?
[Analysis and solution] According to the known conditions, the time ratio of producing a pair of trousers and a coat in a factory is 2: 3; Therefore, the ratio of the number of shirts and trousers produced by a factory in a unit time is 2: 3; It can also be seen that the ratio of the number of shirts and trousers produced by factory B in unit time is 3: 4; Because of this, A factory is good at producing pants, and B factory is good at producing tops. The two factories jointly produce, give full play to their respective specialties, and arrange factory B to fully produce jackets. Since factory B produces 1 0,200 jackets a month, factory B can produce 1 0,200 ÷ = 210,000 jackets a month, and factory A is arranged to fully produce pants, so factory A can produce 900 pants a month.
In order to support production, a factory first fully produced 2 100 pairs of trousers, which required 2 100÷2250 = pieces a month, and then a factory independently produced 900×=60 suits a month, so now the joint production produces more suits every month than in the past.
(2100+60)-(900+1200) = 60 sets.
There are 65,438+0,400 Weiqi pieces today. Party A and Party B play the game of taking Go. Party A takes it first, and Party B takes it later. They take turns eating once. It is stipulated that only 7P(P is 1 or any prime number not exceeding 20) can be taken at a time. Who will win the game in the end? Ask both parties who has a winning strategy.
[Analysis] Because 1400=7×200, the original question can be translated as: There are 200 chess pieces, and both parties take p pieces in turn, and whoever takes the last one wins.
[Solution] B has a winning strategy.
Since 200=4×50, p is either 2 or can be expressed in the form of 4k+ 1 or 4k+3 (k is zero or a positive integer). The strategy adopted by B is: If A takes 2,4k+1 and 4k+3, then B takes 2,3,1,so the remaining pieces are still multiples of 4. So the last remaining number is a multiple of 4, not more than 20. At this time, A can't take it all, and B can take it all and win.
[Description] (1) In this question, B is the "late Mover", so the first Mover does not necessarily have a winning strategy. The key is to look at the "situation" they face;
(2) We can analyze the solution of this problem in this way, and divide all situations-the number of remaining pieces into two categories. The first category is a multiple of 4, and the second category is others. If someone encounters the second situation when playing chess, they can go 1 or 2 or 3, so the rest is the first situation. If he is faced with the first situation when playing chess, then the second situation must be left to another person who has finished playing chess. Therefore, whoever faces the second situation first will win, and this method can be used in most double-match problems.
There is a tour group of 80 people, including 50 men and 30 women. Their hotel has three room types: 1 1, 7 and 5 people. Men and women live in different rooms. How many rooms should they live in at least?
【 Analysis and Solution 】 In order to minimize the number of rooms, first arrange 1 1 rooms, so that 50 men will arrange 3 1 1 rooms, 2 5 rooms and 1 7 rooms. 30 women should be assigned to11room, two 7 rooms, 1 5 room, and * * * has 10 room.
Example 9 has a 3×3 chessboard and 9 cards of square size. Write a number on each card at will. Party A and Party B play games, choose a card in turn and put it in one of the nine squares. For Party A, calculate the sum of six numbers in the upper and lower columns, and for Party B, calculate the sum of six numbers in the left and right columns. The person with the larger amount wins. Proof: No matter what number is written on the card, if A goes first, there can always be a strategy to make B impossible to win.
There are three kinds of [certificate]:
(1) When A 1+A9 > A2+A8, A wins. A's strategy is: first choose a9 and put it in box A, and then choose as small as possible the second time.
If you put the numbers in B or D, the sum of the numbers in A and C is not less than a 1+a9, and the sum of the numbers in B and D is not more than a2+a8, then A wins.
(2) When A 1+A9 < A2+A8, A will also win. A takes a 1 and puts it into B grid for the first time, and a8 or a9 is put into A or C grid for the second time, so that the sum of the numbers of A and C grids is not less than a2+a8, and the sum of the numbers of B and D grids is not more than a 1+a9, and A wins.
(3) When a 1+a9 = a2+a8, if A wins or draws, A can adopt any of the above strategies.