There is a 100-100 off lamp in the room

There is a 100 to 100 100 off lamp in the room, and there are 100 people from 1 to outside the door. Everyone wants to enter the room once, pull the lamp ropes that correspond to their serial numbers and are multiples of their serial numbers (for example, to pull all the lamp ropes on the first and to pull 2, 4, and 6 on the second, and 100 only need to pull the 100 light rope) Q: When 100 people come out, how many lights are on in the room?


In this case, we all know that the answer is the number of shards 100, 9 ....

OK, first of all, you must think that it must be opened only when the light is pulled an odd number of times...
Think about it. If the number is N, it must be that the I that pulled n before will be able to divide N, so I must be the factor of N...
So the question is converted to N, and the number of factors of N is an odd number.

1. The factor of the number of distinct is an odd number, because for example, if n = A * B and A and B are different, they are all paired. Only when a *, there is only one factor, so it must be the number of shards...

2. Proof of a high-end atmosphere:
By the unique decomposition theorem, p is the prime number, corresponding to a is the exponent of P, and then the unique decomposition is:

So how many factors does n have? That is: S = (A1 + 1) * (A2 + 2 )*... * (An + 1), which combines mathematics. It is composed of 0, 1, and an Pn.

The problem is converted to making the number S of the factor an odd number... I think about it here. It must be an odd number * odd number = odd number. Therefore, AI + 1 is an odd number, so AI is an even number.

Since all AI is an even number, for example, we can not extract all AI numbers from a 2? n = (P1 ^ (A1/2) * P2 ^ (A2/2) *... * PN ^ (AN/2) ^ 2, then n is a full shard number ~~~

Proof complete!

There is a 100-100 off lamp in the room