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It can be simulated and inferred.
The light numbered 1 was pulled only 1 time. (The first person pulled).
The lamp numbered 2 was pulled 2 times. The lamp numbered 3 (pulled by the first and second people) was pulled 2 times. The lamp numbered 4 (pulled by the first person and the third person) was pulled 3 times.
The first, second, and fourth person pulled) the lamp numbered 5 pulled 2 times. It can be inferred that the number of times the lamp is switched on and off is the approximate number of the lamp number.
of the number of pieces. Any number (other than 1) has at least two divisors: 1 and itself, and if there are other divisors, they are generally in pairs, assuming that the divisor of any number x has a and b in addition to itself.
If a and b are not equal, then the divisor of this number is even, and if a=b, then the divisor of this number is odd, so x should be 1 squared, 2 squared to 10 squared.
Only the odd numbered lights are lit, numbered by:
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The number of times a certain number of bulbs are pulled is the approximate number of the number, and the light is only lit when the number of times is odd, that is, when the approximate number of the number is odd, the light is on.
There are only perfectly squared numbers that are odd: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
This is the number of the last light bulb to turn on.
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The first time you pull the switch of the light numbered as a multiple of 3, it will turn off 33 lights, and the second time you pull the switch of the light numbered as a multiple of 5, the light that is a multiple of 3 and 5 will be turned on again, a total of 6, but the light that is a multiple of 5 will be turned off, a total of 14, so the last light is n=100-33+6-14=59
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What does it mean to pull the switch --- once?
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1 light pulled 1 time.
The factor is odd and still on:
The factor of 1 is 1, which is lit.
The factor of 2 is 1, and 2 is extinguished.
The factor of 3 is 1, and 3 is extinguished.
The factor of 4 is 1, 2, 4, and it is lit.
The factor of 5 is 1, and 5 is extinguished.
The factor of 6 is 1, 2, 3, 6, extinguished.
The factor of 7 is 1, and 7 extinguishes ......
And so on ......
Because, the factors of a number can form this number in pairs, such as 6, 1x6=6, 2x3=6, so 6 is 4 factors.
That is to say, in general, the factors of a number are even (paired) unless the number is a perfectly squared number, such as 9, where a pair of 3x3 is only counted as one factor, so there can be an odd number of factors.
So, at the end, it's all perfectly squared, 1 4 9 16 25 36 49 64 81 100
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The light is turned off, the light that pulls even times is off, and the light that pulls the odd number of times is on, this problem can be understood as the number of factors from 1 to 100 is odd, that is, how many are there in the perfect square.
From 1 to 100, the perfect square number is 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
The lights are on.
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Considering the number of factors of each positive integer, as long as the number of factors is an odd number, the lamp is odd several times, and the lamp is on; If the number of factors is even, then the lamp is pulled even several times, and the lamp is off.
Since only the number of factors of the square number is an odd number, and the number of the factors of the other numbers is an even number, the last light is the lamp numbered by the square number, that is, the numbered lamp is on.
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Approximate number.
The number is even, and it is lit and it is finished.
Full square number 1 4 9 16 25 36 4964 81 100
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1) According to the title, the number of times the bulb is pressed is equal to the number of all the factors of its number;
2) The start state is off, and then it is on, indicating that the number of times it has been pressed is an odd number;
3) The number of all factors is an odd number, and the natural number of the sullen Tong is the only one that is completely squared.
To sum up, the bulb numbered is a full square number and is lit at the end.
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Hello, the pass rate is 92 100 = 92%, processing 10000 pieces is about 10000 92% = 9200 qualified.
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