There are 200 lights on, each controlled by a pull switch. 25

The first time the lights go out in multiples of 2 times there are 100 lights on.

The second time the lights were extinguished by multiples of 3, 66 should have been extinguished, and 34 were left.

But 2 and 3 have common multiples of 6, and the lights of multiples of 6 are on, i.e. 33.

The last time, 5 times the lights went out, 40 should have been extinguished, and 27 were left.

But 5 and 2 have a common multiple of 10, and the lights of a multiple of 10 are lit, i.e. 20.

Plus the common multiples of 5 and 3 15 light up i.e. 13.

So there should be a total of 60 lights up.

I did it myself, you see, it's not necessarily right.

I don't know if it's right or not, don't laugh if it's wrong, let's talk about it together, I think the correct answer is 100Why, that's how I analyze it.

There are a total of 200 lights, the multiple of 2 is just turned off and half turned on, so now the light and off are 100 each, the second step is still the same, the multiple of 3 is operated once, the number of multiples of 3 inside and out should be equal, the number within 100 is lit here and the corresponding number within 200 below will be extinguished, so that it is just offset, there are still 100 bright 100 off.

The third step is to operate a multiple of 5, or the same reason, the number of multiples of 5 inside the bright and the inside should be equal, and the last extinguished and bright ones should be 100 each. I don't know if the answer is right or not, but I did a simulation from 1-20 to verify that my idea is correct, and you can combine it with your own thinking.

Using the principle of repulsion, the multiple of 2 is 100

3 multipliers of 66.

Multiples of 5 40 pcs.

The common multiple of 235 is 6.

23 is a common multiple of 33.

The common multiple of 25 is 20.

The common multiple of 35 is 13.

So it's 100 + 66 + 40 - 33 - 20 - 13 + 2x6 = 155

Ratchet several times, the light is off, the lamp is on, the light is on.

10 is the number of even gods, and pulling the light 10 times is the cover friend who is on

So the answer is: on

Select 1002

First list 30 numbers, then pull multiples of 2 (a total of 15 bright), and then pull multiples of 3 (find 5 bright lights, extinguish 5 lights.) In the end, it is still 15 lights), and then pull a multiple of 5 (found that 3 lights are on, 3 lights are off, and finally 15 lights are on). Then 30 lights, pull 3 times and turn on 15 lights.

2000÷30=66...20 (lamp). The remaining 20 lamps can be listed from 1 to 20, and after pulling three times, it is found that there are still 12 lights left.

66 15 + 12 = 1002 (lamp).

Option A considers the number of pulls, the number of times all lights are pulled are times, where the last of the once pulled and the three times is off, and the numbering of the lights that are only pulled once and pulled 3 times is:

Just a multiple of 2: 1000-333-200+66

Just a multiple of 3: 666-333-133+66

Just a multiple of 5: 400-200-133+66

of common multiples, a total of 66 lamps.

The total number of lights turned off is: 1000-333-200+66+666-333-133+66+400-200-133+66+66=998

Then the total number of lights that were last lit is: 2000-998=1002

Tell me about the analysis process.

It is only a multiple of 2: the multiple of 2 is a total of 1000, subtract the multiple of 2 and 3, subtract the multiple of 2 and 5, where all the multiples of 2 and 3 contain the multiple of , and all the multiples of 2 and 5 also contain the multiple of , which is equivalent to the multiple of subtracting once, then a multiple of 66 should be added.

The same is true for multiples of 3 and 5. It is relatively simple to think in this way, and the essence is that the multiples are contained in the multiples of 2 and 3, the multiples of 2 and 5, and the multiples of 3 and 5, because all the lights are originally on, so it is much simpler to consider only the number of lights that are turned off than the number of lights that are on.

For the first time 1,000 lights were turned off, all numbered evenly;

The second pull was made on the 666 light switches. But the switch in the even-numbered position (i.e. a multiple of 6, which was turned off for the first time) is lit again. The lights that are on at this point are: 1000 - 333 + 333 = 1000

The third pull of the 400 light switches.

Of these, there are 66 lights in multiples of 30. Because it was lit up in the second operation, it was turned off again this time;

The lights in the multiples of 15 but not 30 multiples position have a total of 111 - 66 = 45 because they were turned off in the second operation and turned on again this time;

The lights in the multiples of 10 but not 30 multiples position have a total of 200 - 66 = 134, because they were turned off in the first operation and turned on again this time;

The rest is only a multiple of 5 but not a multiple of 15 The lights in the position have a total of 400 - 134 - 45 - 66 = 155 and are turned off in this operation.

So, the lights that are on at this time are: 1000 - 66 + 45 + 134 - 155 = 958

From 1 to 200, multiples of 3 are: 200 3 = 66 pcs....2；

Multiples of 5 are: 200 5 = 40;

3 5 = 15, then the common multiple of 3 and 5 is:

200 15 = 13 pcs....5；

Then the ones that don't light up after pulling are:

66 + 40-13 = 93 (lamp).

A: The number of lights that do not turn on after pulling is 93

So the answer is: 93

From 1 to 150, multiples of 3 are: 150 3 = 50 (pcs);

Multiples of 5 are: 150 5=30 (pcs);

3 5 = 15, then the common multiple of 3 and 5 is:

150 15=10 (pcs);

150-50-30+10,100-30+10,80 (lamp);

A: The number of lights that turn on after pulling is 80

So the answer is: 80

When pulling a multiple of 3, pull out 50 lights, and when pulling a multiple of 5, instead of pulling out 30 lights, pull out 20 lights, and pull out 10 lights, so it is 90 lights.

This answer is incorrect, it should be 90 lights on. The first step is to pull the multiple of 3, leaving 100 lights to be bright, and the second step is to pull the multiple of 5, if you don't count the multiple of the first pull of 3, you should extinguish 30 lights, but among the 30 lights, you should light 10 lights (common multiples of 3 and 5), then say, the total number of lights is 110, and then extinguish the non-common multiple of 20, so the number of lights is 90. The mistake in this question is:

I didn't pay attention to the switch twice (common multiples of 3 and 5) and the light would turn on again, so 80 should add 10 lights that are repeatedly turned on.

150 lamps Pull down a multiple of 3 Leave 100 150-150 3=100

150 pulls the second time The multiple of 5 should be 30, 150 5=30 but 3*5=15 150 15=10 10 out of 150 is a common multiple of 3 and 5.

There are 10 lamps that are pulled repeatedly, and these 10 are added for the second time, which means that the first time 150-50=100

The second time the 30 times the rope is drawn, 20 times is when the lights are out, and 10 times when the lights are on, 100-20+10=90

So, in the end, there are 90 lights on.

Hello dear, the light goes off after pulling it, and the light turns on after pulling it twice. The cable of a lamp that is a multiple of 3 or a multiple of 5 but not a multiple of 15 is pulled once, and the cable of a lamp that is a multiple of 15 is pulled twice. Multiples of 3 are:

150 3 = 50 (pcs) Multiples of 5 are: 150 5 = 30 (pcs) Multiples of 15 are: 150 15 = 10 (pcs) where is a multiple of 3 or a multiple of 5 but not a multiple of 15 has:

50 + 30-10-10 = 60 (pcs) so the lights on are: 150-60 = 90 (pcs) so the answer is: 90

Dear, this is the specific calculation step, can you see it, I hope it can help you.

From 1 to 150, multiples of 3 are: 150 3 = 50 (pcs);

Multiples of 5 are: 150 5=30 (pcs);

3 5 = 15, then the common multiple of 3 and 5 is:

150 15=10 (pcs);

150-50-30+10,100-30+10,80 (lamp);

A: The number of lights that turn on after pulling is 80

So the answer is: 80

The first pass is a multiple of 3, 150 3 = 50.

The second pass is a multiple of 5, 150 5 = 30.

Focus!!! The problem lies in the second pass, does the second pass really turn off 30 lights, no!!

Calculate that the common multiple of 3 and 5 is 150 (3 5) = 10, that is, the second time, there are 10 lights that are turned off the first time, and the second time they are pulled again, is it on again? So the second time I actually turned out 20 lights and turned on 10 lights at the same time, so how many lights were on in the end.

150-50-20+10=90.

The lights that have not been touched are on, and the lights that have been pulled twice are also on, 80 ten 10 = 90 lights,

This question mainly tests lights that are multiples of 3 and 5, and they are pulled twice and then turned on.

So it should be 150 3=50

The lights that are pulled out are 50 30 2 10 = 60

The light that is on is 150 60 = 90

Multiples of 3 within 100 have 100 3 = 33.

Multiples of 5 within 100 are 100 5 = 20.

Within 100 is a multiple of 3 and 5 at the same time, that is, the multiple of 15 has 100 15 = 6 because the switch that is repeatedly pulled is in the state of open when it is pulled for the first time, and it returns to the state of off when it is pulled again for the second time, so there are 33 * 2 = 66 lights on when it is pulled for the first time, and a total of 12 lights are turned off when it is pulled for the second time, and there are 66-12 = 54 lights that are lit when it is pulled for the first time, and there are (20-6) * 2 = 28 lights when it is pulled for the second time.

Since each switch turns on 2 lights, there are a total of 54 + 28 = 82 lights on.

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