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Solve a simple math problem.
There is a better way. In my impression, this method existed more than 20 years ago.

First of all, remember the value of

There are twelve months in a year, and each month corresponds to a number. I call it the monthly value, that is, the values from 1 month to1February are (033 6 14 625 035) respectively.

Two. Constant value. In 2008, the constant value was "2"

Third, the calculation method:

For example:

1, May 2008 1 (monthly value is 1, constant is 2), then there are:

1+1+2 = Thursday 4/7

2. On May 2, 2008 (the monthly value is 1 and the constant is 2), there are:

2+ 1+2=5 5/7 Friday

3. On May 3, 2008, (the monthly value is 1 and the constant is 2), there are:

3+ 1+2=6 6/7 Saturday.

4. In September 2008 16, (monthly value is 5, constant is 2), there are:

16+5+2=23 23/7 on Tuesday, 2nd.

5. On September 7, 2008, (the monthly value is 5, and the constant is 2), there are:

17+5+2=24 24/7 Yu 3 Wednesday

6.2008165438+1October 20th (monthly value is 3, constant is 2), there are:

20+3+2=25 25/7 Thursday

etc

Seeing here, it is not difficult for us to figure out what day it is today. Some people may question that it's all from 2008, and there is a calendar, so it doesn't count. What about other years? Let me say:

1, law of constants: constants can only be one of seven numbers (0, 1, 2, 3, 4, 5, 6), and so on. The constant value changes from 0 to 6 for a period, and returns to 0 when it changes again. )

2. Changing principles,

Generally, there is a value corresponding to a fixed value in a year. If it is a running year, the year bounded by February 29th has two values.

That is to say, the constant value is 1 increase 1 in the month of 65438+ 10 every new year (for example, February 2008, April 320 10 in 2009) and also increases 1 every February 29th.

Example: On February 23, 2009, (the monthly value is 3, and the constant is 3), there are:

23+3+3=25 29/7 on 1 Monday