Tisch
Xiao Li and Xiao Zhang start to make the same kind of parts at the same time, and each person can make 1 part per minute, but Xiao Li has to rest for 1 minute every three parts and Xiao Zhang has to rest for 1.5 minutes every four parts. Now they have to finish the task of manufacturing 300 parts together, which takes _ _ _ _ _ _ _ minutes. Analysis: Xiao Li's cycle is 4 minutes, and Xiao Zhang's cycle is 5.5 minutes. Their minimum time is 44 minutes.
Li Can Jr. completes three parts in a cycle, he can complete 33 parts in 44 minutes, Xiao Changcan completes 48=32 parts in 44 minutes, and they can complete 33+32=65 parts in 44 minutes, 300 divided by 65 = 4 ...
So after four 44 minutes, there are still 40 parts left, which means it takes 176 minutes to complete 260 parts.
We just need to calculate how long it will take two people to cooperate on 40 parts. Let's roughly estimate that if two people don't rest, it is 402=20 (minutes). Xiao Li only has 15 minutes to go to work in 20 minutes, so Xiao Zhang is not convenient to complete the calculation. Let's first calculate, Xiao Zhang completed 16 in four cycles, Xiao Li completed 15 in five cycles, and completed two in another two minutes, so two people * * * completed 16+ 17=33 parts in 22 minutes, and the remaining seven were the 23rd minute. So you can finish these 40 in 26 minutes.
So it takes 176+26=202 (minutes) to complete 300 parts.
Comments: This question is difficult to rest. This stop-and-go problem has a cycle for everyone. Let's find a common cycle and see how many cycles are needed to complete the work or walk the whole course. The rest will inevitably use enumeration. Usually the problem lies in the enumeration step. It is easy to find ideas as long as you touch these questions, but it is not easy to do them right. Why? The difficulty of this problem lies in the time required for these 40 parts. We estimate how many will be finished in 22 minutes. If you calculate the rest separately, you must be careful and patient. Ok, 150% and 90%. There may be something wrong with the last seven calculations. When listing, you must have clear thinking and clear standards. Issues irrelevant to the discussion need not be considered at all. So you can concentrate on solving the problem. Training enumeration method is very beneficial to learning classification discussion in the future. The problem of learning function in middle school is often discussed in categories, and the big situation often has small situations. At this time, the order of thinking is very important.
extreme
1,2, 3 ... How many numbers in 2006 included the number 6? Analysis: It is considered that 1 is indirect. We just need to look at how many numbers in one, two, three and four digits do not contain the number 6 to think.
In fact, if you look at a specific category here, you are actually arranging it. Look at the number first. Obviously, there are eight people besides six. In the second case, our goal is double digits. Ten digits are divisible by 0 and 6. Pay special attention to the fact that multiple numbers cannot be 0. One digit is 0-9, and there are 9 cases when 6 is removed. According to the principle of gradual multiplication, we know that there are 8 times 9=72. The number 6 does not contain three digits. There are 8 cases in hundreds, 9 cases in tens and singles, and * * * has 8 times 9 times 9=648 cases. Consider four more digits. The fourth big case is 1 and two small cases. When the number of thousands is 1, there are nine cases of hundreds, tens and units. So ***9 times 9 times 9=729 cases, and 6 cases in thousands are 2 ***735 cases. So those without the number 6 are 8+72+648+735= 1463, and those with the number 6 are 2006- 1463=543.
Idea 2: Just do it. The 6th bit is the number of 10 * * 2000 divided by 10=200 occurrence 1 times, and there are 2006 bits that are 6. So the number of the sixth place is ***20 1. The six digits are 10 *** and there are 200 per 100. Similarly, there are 200 people/kloc-6 out of 0/00. The numbers considered here are all 1- 1999, and the tens and hundreds above 2000 are not 6. According to the digits of 6 and their inclusion and exclusion, the digits are166,266 ...1966 * * 20, and the digits of 6 are 606,000. 1606,1616 ...1.696 * * 20, and the number of tens and hundreds is also 20. Only1666,6662 people have numbers of 6. According to the relationship between inclusion and exclusion, there is 20 1+200+200-20-20+2 = 543.
Tisso
The distance between AB is1000m. Party A and Party B will return to AB, and Party B will leave in 6 minutes. At this time, catch up with A at C, with a distance of 600 meters. B will return immediately after arriving at B, and A will return after resting for 1 minute. A speeds up when returning, and catches up with B at C. How soon will A return to A before B? Analysis: The most important thing about travel problem is to find invariants. At first, it takes six minutes to walk 600 meters to see B chasing A. Then 600 divided by 6= 100. It takes 100 meters to walk 1 minute. The back 400 meters can save 4 minutes. By the time A got to A, B had been back for four minutes. When A came back, B walked for five minutes. It means walking 400 meters at one speed takes 5 minutes less than B. It takes 1 minute to walk 80 meters. So in the last 600 meters, A is less than 5 times 1.5=7.5 minutes, or 600 divided by 80=7.5 minutes. Example 4 A and B start from the same point on the 400-meter circular runway and walk back to back. When we first met, A turned and ran back. When they meet again, B turns and runs back; After each encounter, A and B turn around alternately. The speed of the two keeps constant during the movement. A runs 3 meters per second, and B runs 5 meters per second. When they first met, the intersection of A and B happened to be the first starting point?
Analysis: We use the equivalent method to replace the ring with a square ABCD. Starting from a, each side 100 meters. Assuming that A starts clockwise and B starts counterclockwise, they first meet at the midpoint of BC. The second time, it becomes chasing B counterclockwise, and 400 divided by 2=200 seconds. A walked 600 meters and met at the midpoint of AD. The third B turn, A turn counterclockwise, meet at C, and the fourth A turn clockwise. B chase armor and walk 600 meters to meet at a.