Math problems that you want to solve in your head, math problems that open your mind?

For this kind of question, you can generally do it like this:

A and B should stop and rest for 1 minute every 200 meters, A approximates: 200 (200 60 + 1) = 600 13 meters, B approximates: 200 (200 50 + 1) = 600 15 meters, that is, every 600 meters, A takes 2 minutes less than B;

13 2=, we can look at the situation at 13*6=78 minutes:

A ran = 6 * 600 = 3600 meters, 7 laps and 100 meters, just after resting;

78 = 15 * 5 + 3, B ran = 5 * 600 + 3 * 50 = 3150 meters, 6 laps and 150 meters, and then 1 minute of rest after running;

A is still 50 meters away from catching up with B; Is it the first time or the second time? Let's go back a little further:

77 minutes: A ran 3600 meters, B ran 3100 meters, here A has just caught up with B for the first time, so A catches up with B for the first time 77 minutes from the start;

500 meters, because every 200 meters to stop for one minute to rest, A to run an extra 500 meters to stop two more times, 2 minutes. In these 2 minutes, B can run 50 2 = 100 meters. Therefore, it is equivalent to that A needs to catch up with B 500 + 100 = 600 meters to meet for the first time.

600 (60-50) = 60 minutes, 60 minutes A runs a total of 60 60 = 3600 meters, and needs to stop 3600 200-1 = 17 times, totaling 17 minutes. Therefore, A needs 60 + 17 = 77 points to catch up with B for the first time.

Check this out, maybe it will be useful! Although I haven't been able to figure out the answer, I hope it helps

Question: The circular track is 500 meters long, and A and B start at the same time, A is 120 meters long and B is 100 meters. Every 200 meters is rested for 1 minute, ask when A will catch up with B?

Answer: In fact, the speed of A is 200 meters per (200 120 + 1) = 8 3 minutes, which is 600 meters for 8 minutes.

B's speed is 200 meters every 3 minutes, then 600 meters takes 9 minutes.

The so-called when A catches up with B is when A runs one more lap than B, that is, 500 meters more. 500 (600 8) = minutes, and every 600 meters is saved by 1 minute, then, to save points, try to calculate 6 points, you need to run 3600 meters. At this time, B used 54 points, just ran 3600 and had already rested.

Let's look at A. It only takes 48 minutes for A to run 3600 meters, so how much more can A run in the remaining 54-48=6 points? 6 = 18 3, it is already known that the arrangement of these 18 3 points is the first two.

8 3 ran 400 meters at 3 o'clock, there were 2 3 minutes left, ran 2 3 * 120 = 80 meters, A ran 400 + 80 = 480 meters in 6 minutes, that is, A ran 4080 meters in these 54 minutes.

One minute after 54 minutes, that is, in the 55th minute, A just ran the remaining 200-80 = 120 meters of 200 meters, and at this time, A ran 4080 = 120 = 4200

And B ran 3600 + 100 = 3700 meters at this time.

It can be seen that at 55 minutes, A runs 500 meters more than B, and falls 1 lap of B, which is the so-called when A catches up with B.

When A catches up with B for the first time, the distance difference between A and B is 500 meters, and A rests 2 times more than B, so in fact, the distance A has chased is 500 + 50 * 2 minutes = 600 meters; So in fact, A's running time is 600 (60-50) = 60 minutes, and the number of breaks is 60*60 200=18 times, but the last time has caught up, that is to say, A rests for 17 minutes in the middle, and A spent a total of 77 minutes.

7+2+5＝143547

The first two digits are 7*2 14, the middle two digits are 7*5 35, and the last two digits are 7*(2+5) 2 47.

Can it be more complicated, the answer is 143547, and the math alone is 14

Xie Xi Tong: or Yantan Zao Yu x + 5 8 = 7 10

x=7/10-5/8

x=28/40-25/40

x=3/40

(1) Define a new operation, a b=a-b+2, try to calculate 2 3;(2) (3) Solution: a b=a-b+2 2 3=2-3+2=1 (2) (3)=-2-(-3)+2=1+2=3 (2) Knowing x=,y=,z=,find the value of the algebraic formula -x+y-z. Solution:

-x=,-z= -x+y-z=(-x)+y+(-z)= (3) if|a|=4，|b|=2，|c|=5, and |a+b|=a+b，|a+c|=-(a+c) to find the value of a-b-c. Solution: From|a+b|=a+b:

a+b 0 is caused by |a+c|=-(a+c) gets: a+c 0 -c a -b |a|=4，|b|=2，|c|=5 If c=5, the inequality does not hold So c=-5 To make a -b, then a=4, b=-2 or a=4, b=2 when a=4, b=2, c=-5 a-b-c =4-2-(-5) =2+5 =7 when a=4, b=-2, c=-5 a-b-c =4+2-(-5) =11

(1) a b=a-b+2, the formula has been given, the corresponding a is regarded as 2, b is regarded as 3, and substituting is: 2 3=2-3+2=1;（-2）△(3)=-2-（-3）+2=3

2) Directly substitute the values of x, y, and z to obtain: -x+y-z=(3)|a+b|=a+b, then a+b must be positive, and |a|=4，|b|=2, then there must be a=4, b=+2 or -2

a+c|=-(a+c), then a+c must be negative, and |a|=4，|c|=5

Then there must be c = -5 and a = +4 or -4

In summary, we can get a = 4, b = +2 or -2, c = -5

So a-b-c = 7 or 11 (7 when b = +2 and 11 when b is -2).

(1）2△3=2-3+2=1

2）△(3)=（-2）-（3）+2=3(2)-x+y-z=

3) According to the meaning of the question: a=4, b=2 or -2, c=-5, then a-b-c=4-2-(-5)=7 or a-b-c=4-(-2)-(5)=11

1.（1）2-3+2=1 （2）（-2）-（3）+2=33.Because |a|=4，|b|=2，|c|=5, and |a+b|=a+b, so a+b>0, a=4 b=2 or -2

And |a+c|=-(a+c), then a+c<0, c=-5 a=-4 or 4. Combined, the above results a=4 b=2 or -2 c=-5

So a-b-c = 4-2-(-5) = 7 or 4-(-2)-(5) = 11

Nobody divides the time for four hours, and then everyone adjusts the time, which is the fairest.

There are three shifts of 4 hours each. Class A: A and B; Class B: B C; Class C: C A. Each person works 8 hours a day.

One person can work 40 hours a week.

1c2a

3a4a5b6c

The answer is completely correct, please give hard points.

Please donor be self-reliant, thank you!

Regularity of two numerical converter junctions.

Converter 1: = A2 + 2 * A+1 = A*A+2*A+1

Converter 2: = a + 1) 2 = a + 1) *a + 1) = a * a + 2 * a + 1

Both are equal.

It's not clear, it's a bit eye-catching.