Solve 3 Primary School Math Olympiad Problems Urgently Needed!!!! ）

Draw a line diagram and see...

A and B met for the first time, and a total of 1 whole course.

Of them, line B is 60 meters.

A and B met for the second time, and a total of 3 whole journeys.

Among them, B traveled 1 full course plus 10 meters.

A and B have a total of 3 full journeys, and the time taken is 3 times that of 1 full journey.

B should be OK: 60 3 = 180 meters.

That is: 1 full journey plus 10 meters, equals 180 meters.

The whole distance (AB distance) is: 180-10=170 meters.

If you climb level 10, you can only go up from level 8 or level 9, so calculate how many possibilities there are to level 8 and level 9 respectively.

In the same way, if you climb level 8, you can only go up from level 6 or 7.

Ascend level 9 and only go up from level 7 or 8.

That is: different ways to climb level 10 = way to climb level 9 + way to climb level 8.

The way to climb level 9 = the way to climb level 8 + the way to climb level 7.

The way to climb level 8 = the way to climb level 7 + the way to climb level 6.

The way to climb level 3 = the way to climb level 2 + the way to climb level 1.

There is 1 type to climb level 1.

To ascend level 2, there are 1+1=2 types.

Ascend level 3, there are 1+2=3 types.

Ascend level 4, there are 2+3=5 types.

Ascend to level 5, there are 3+5=8 types.

Ascend level 6, there are 5 + 8 = 13 types.

Ascend level 7, there are 8 + 13 = 21 types.

Ascend level 8, there are 13 + 21 = 34 types.

Ascend level 9, there are 21 + 34 = 55 types.

Ascend level 10, there are 34 + 55 = 89 types.

Even factors, including factor 2 or factor 4

Containing only factors 3 and 5, there are:

2+1) (3+1) = 12.

Even factors are: 12 2 = 24.

1.A and B met for the first time, and they walked a total of 1 whole course, of which B traveled 60 meters.

A and B met for the second time, and a total of 3 full journeys were taken, of which B traveled 1 full course plus 10 meters.

A and B have a total of 3 full journeys, and the time taken is 3 times that of 1 full journey.

B should be OK: 60 3 = 180 meters.

That is: 1 full journey plus 10 meters, equals 180 meters.

So AB is 180-10 = 170 meters apart.

2.It's a combination problem.

Let a = the number of times you climb the first step, b = the number of ways to reach the top, and limit the number of times you can climb the first level to reach the top floor only after climbing two steps.

Eleven cases of a=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 were analyzed.

When a=0, the only way to reach the top is to climb two levels each time, and climb five times, that is, there is only one possibility for b, b=1;

When a=1,3,5,7,9, the rest of the climbers cannot reach the top, and b=0

When a=2, the combination problem is that two of the ten steps are not arranged to choose two poles as the position of ascending one pole, and the rest ascending to the top of the two poles, i.e., b=c(10 2)=45

When a = 4, the same as a = 2 analysis process, i.e. b = c (10 4) = 210

When a = 6, b = c (10 6) = 210

When a=8, b=c(10 8)=45

When a=10, the same as a=0 analysis, can only be level-by-level, b=1

Sum is sufficient: b=1+45+210+210+210+45+1=512

3.Decompose prime factors.

If there is no 2, there is no even factor.

If there is 2, ignore factor 2 and combine the prime factors of other odd numbers to get some results, and then combine it with factor 2.

If there are more than 2s, it is the result of combining the above odd prime factors 2, 2 2, 2 2 2....That's it. The answer is 24