2. If the sum of a four-digit number and a three-digit number is 1999, and the four-digit number and the three-digit number are composed of seven different numbers, then there are at most () such four digits?
A cart and a car both drive from A to B, and the speed of the cart is 80% of that of the car. It is known that the big car started 17 minutes earlier than the small car, but it stopped at the midpoint of the two places for 5 minutes before continuing to B. However, the small car did not stop halfway and went directly to B. Finally, the small car arrived at B 4 minutes earlier than the big car, and it is known that the big car was there. Then the car caught up with the bus in the morning.
4. The sum of1997 is 1+9+7+26. Please write down all four digits less than 2000 except 1997.
Some children line up, starting from the first person on the left, giving an apple to every two people, and giving an orange to every four people starting from the first person on the right. As a result, 65,438+00 children got all the apples and oranges, so there were at most () children.
6. Trams leave from the tram terminal at regular intervals. Party A and Party B walk in the same direction on the same street. Party A walks 82 meters every minute and meets an oncoming tram every 10 minute. B walks 60 meters every minute and meets an oncoming tram every 10 minute 15 seconds. So the tram terminal runs every () minutes?
7. Calculation:19971997+9971997+971997+91997+7 =
8. The numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 are combined into three digits and one digit, and the sum of these digits is 999. We require the largest three digits to be as small as possible, so these three digits are ().
Answers and analysis:
The first question:
Solution:1(114+1/20) =140/17 = 8+4/17 (.
It can also be understood that it takes 4/ 17 hours for Party A and Party B to complete the task after 8 hours each. Then in 4/ 17 hours, the workload completed by both parties is (4/17) × (114+1/20) = 1/35, so Party A needs to do it alone.
Therefore, the following results are obtained:
It took Party A and Party B 16( = 8+8) hours and 24 minutes to finish this manuscript.
The second question:
1abc，xyz
a+x=b+y=c+z=9
For x, there are three points (0, 1, 8) and seven choices that are not desirable.
For y, there are two values of 1, 8 and a, x, and six choices.
For z, 1, 8, a, x, b, y is not desirable.
4×6×7= 168
There are three digits of 168, corresponding to four digits of 168.
The third question:
V is big: V is small = 4: 5, and the time ratio is 5: 4.
When the bus arrives at B, the big car is behind the small car 17-5+4= 16 minutes. The time here refers to the time when everyone is driving.
It takes 80 points for the big car to complete the whole journey and 64 points for the small car. The starting time of the cart at the midpoint is 80/2+5=45 minutes, and the starting time of the car at the midpoint is 64/2+ 17=49 minutes.
At the midpoint, the cart starts 4 minutes earlier than the cart, which is 4*4= 16 minutes longer than the cart catching up with the cart.
So it took 49+ 16=65 minutes to catch up with the cart, and set off at 10 in the morning, and the time to catch up was 1 1: 05.
The fourth question:
Four digits less than 2000, the first digit is 1, and the sum of the other three digits is 25,3× 8 = 24, so one digit must be 9, and the other two digits are 9,7, or 8,8.
So besides 1997, there are 1988, 1979, 1898, 1889, 1799.
The fifth question:
Starting from the first person from the left, for every14 [= (1+2) * (4+1)-1] person, there will be one person who has both oranges and apples, so every * * 10 child has one.
The sixth question:
A 10 minute walking distance: 82 *10 = 820m;
B10.25min walk: 60 *10.25 = 615m;
Comparatively speaking, A walked 820-6 15 = 205 meters more than B, which means that the tram needs 10.25- 10 = 0.25 minutes.
Therefore, the tram speed is 205/0.25 = 820m per minute.
The distance interval of departure is: (82+820) *10 = 9020m;
The departure time interval is: 9020/820= 1 1 minute.
That is, there is a tram every 1 1 minute.
Question 7:
Answer 3099 1086
1997 1997+997 1997+97 1997+7 1997+ 1997+997+97+7
= 1997200+9972000+972000+72000+2000+ 1000+ 100+ 10-(8*3)
=3099 1086
Question 8:
If the hundredth digit of a three-digit number is 7 or 8 or 9, then the hundredth digits of the other two three-digit numbers can only be 1 and 2, so the sum of the three digits exceeds 999. Therefore, the hundredth place should be less than 7.
If the hundredth digit of a three-digit number is 6, the hundredth digits of the other two digits can only be 1 and 2, and the sum of the hundredth digits is 9, then their ten digits (including the carry of each digit) can only be 9 and cannot be carried (otherwise the hundredth digit will exceed 9). In this way, the ten digits of three numbers can only be 0, 3, 4 or 0, 3, 5, and the corresponding single digits are 5, 7, 8, 9 or 4, 7, 8, 9. After a group of single digits, the last digit is not 9, so it is excluded.
Considering the former group, we can form the following numbers that meet the requirements of the topic:105,237,649,8, and get the largest three-digit 649, which is what we want.
Give you a few more multiple-choice questions:
1. It is known that the function y = x2+1–x, and the point P(x, y) is on the image of this function. Then, the point P(x, y) should be on the rectangular coordinate plane ().
(a) first quadrant (b) second quadrant (c) third quadrant (d) fourth quadrant
2. There are m red balls, 10 white balls and n black balls in a box, and each ball is the same except the color. If you choose a ball from them, the probability of getting a white ball is the same as the probability of not getting a white ball, then the relationship between m and n is ().
(A)m+n = 10(B)m+n = 5(C)m = n = 10(D)m = 2，n = 3
3. Our province stipulates that a junior high school math contest will be held on the last Sunday of June every year 1 1, and the date of next year's junior high school math contest is ().
(a)165438+1October 26th (b)165438+/October 27th (c)165438+1October 29th.
4. There are two points A (–2,2), B (3 3,2) and C in the plane rectangular coordinate system. If △ABC is a right triangle, then the point C that meets the conditions is ().
1 (B)2 (C)4 (D)6。
5. A company ordered 22 lunches in a fast food restaurant at a cost of 140 yuan. There are three kinds of lunches: A, B and C, and the unit prices are 8 yuan, 5 yuan and 3 yuan respectively. Then the possible different sorting schemes are ().
1 (B)2 (C)3 (D)4。
6. it is known that a > 0, b>0 and a (a+4b) = 3b (a+2b). Then the value of a+6ab–8b2a–3ab+2b is ().
(A) 1(B)2(C) 19 1 1(D)2
Answer: BADDCB