Math Cycle Problems, Primary Math Cycle Problems

The period is 4 algorithms:

In addition, x=x+1 is brought into the known equation to obtain: f(x+2)=[1+f(x+1)] [1-f(x+1)].1)

It is also known that f(x+1) [1+f(x)] [1-f(x)].2) Bring in (1) to simplify: f(x+2)=1 [-f(x)].3) From (3) and x=x+2 into (3) formula:

f(x+4)=1/[-f(x+2)].4) Bring Eq. (3) into Eq. (4) to get:

f(x+4)=f(x)

The period definition of the function can be obtained: the minimum positive period of f(x) is 4 if the standard answer is not this.

Then I can't help it.

This method is the dead way to solve all cycle problems!

Doesn't the landlord give extra points?

The first reply was garbage.

I actually wrote like that, fainted!

1) Add on 7 and 1, the cycle knot is 7654321, a total of 7 digits, 300-2 = 298;298/7=42……4

Then it's the 4th, it's the 4th

2) If it is 9 and 1, then 100 9 = 11 ......1, is 9, and if it is 8 and 1, then (100-1) 8=12......3, is 6, if it is 7 and 1, then (100-2) 7=14, then is 1, if it is 6 and 1, then (100-3) 6=16......1, is 6, if 5 and 1, then (100-4) 5=17......1, then is 5 If it is 4 and 1, then (100-5) 4=23......3, then is 2 If it is 3 and 1, then (100-6) 3=31......1, then is 3, if it is 2 and 1, then (100-7) 2=46......1, then 2 is only 5 and 1 fit the topic.

Analysis: It can be known from the meaning of the title:

Arranged in the order of four 1 cents, three 2 cents, and two 5 cents, then there are 9 coins in a cycle.

And 111 9 = 12....3, which means that the 111th coin is the 3rd coin after 12 cycles, then the 111th coin in his row is (1) cent.

Because there are 4 1 + 3 2 + 2 5 = 20 cents in a cycle, then these 111 coins have a total of 20 12 + 3 1 = 243 cents = yuan.

If the capacity of the pool is less than 2 3 for 4 cycles, the water will not overflow.

The question asks when the water overflows, and then we calculate that after the end of the 5th cycle, the capacity of the pool is 3 4, and the water will not overflow.

The capacity of the pool is still 1 4, and the water inlet rate per hour is 1 3, and 1 3 is greater than 1 4, indicating that the water will overflow during the water inlet process of pipe A, and the overflow time point is 1 3 (1 4) = 3 4

So the duration is 5 cycles plus 3 4

The cycle is 4

Algorithm: Another x=x+1 is brought into the known equation to obtain: f(x+2)=[1+f(x+1)] [1-f(x+1)].1)

It is also known that f(x+1) [1+f(x)] [1-f(x)].2)

2) Bring in (1) to simplify: f(x+2)=1 [-f(x)].3)

From Eq. (3) and x=x+2 into Eq. (3), we obtain:

f(x+4)=1/[-f(x+2)].4)

Bringing Eq. (3) into Eq. (4) gets:

f(x+4)=f(x)

The period definition with a function gives that the minimum positive period of f(x) is 4

If that's not the standard answer.

Then I can't help it.

This method is the dead way to solve all cycle problems!

Doesn't the landlord give extra points?

The first reply was garbage.

I actually wrote like that, fainted!

1) Add on 7 and 1, the cycle knot is 7654321, a total of 7 digits, 300-2 = 298;298/7=42……4

Then it's the 4th, it's the 4th

2) If it is 9 and 1, then 100 9 = 11 ......1, is 9, and if it is 8 and 1, then (100-1) 8=12......3, is 6, if it is 7 and 1, then (100-2) 7=14, then is 1, if it is 6 and 1, then (100-3) 6=16......1, is 6, if 5 and 1, then (100-4) 5=17......1, then is 5 If it is 4 and 1, then (100-5) 4=23......3, then is 2 If it is 3 and 1, then (100-6) 3=31......1, then is 3, if it is 2 and 1, then (100-7) 2=46......1, then it is 2 and only 5.

1. Fit the topic.

The last two digits of the product of 1991 1990 multiplied are 0, and only the last two digits of the product of 1990 1991 multiplied are considered. The last two digits of 1 1991 are 91, the last two digits of 2 1991 are 81, and the last two digits of the product of 3 1991 are 71....4 10 is 61, 51, 41, 31, 21, 11, the last two digits of the results of 1991 are 91, it can be seen that 10 cycles, that is, the period is 10, 1990 10 = 199, so the last two digits of the product obtained by multiplying 1990 1991 are 01.

The answer is 01

The answer is found in the library, which also has several periodic questions with detailed explanations.

The same can be said. Number remainder.

12 0...Travel teasing.

It can be seen that the remainder is 12 cycles.

2005 12 remainder is 1

Therefore, the answer is that the remainder of the rubber cavity is 3

This disturbance is actually a problem of summing the first n terms of the number series, one-seventh decimal place is so decimal point after the decimal point is six digits a cycle, talking about the sum of these six numbers is 1 + 4 + 2 + 8 + 5 + 7 = 27, six a cycle, so there are 300 6 = 50 in the 300 bits, there are 50 periods, there are 50 cycles, the sum of 1 cycle is 27, then there are 500 cycles, and the sum is 500 * 27 = 13500. The cyclical problem will definitely give a relatively large book, so it is certainly not possible to calculate one by one, but to find out the relationship within a cycle and find out the relationship between the period and the total number. Relieve hunger.

Divide by 5 and make the remainder of the circle, just look at the single digit.

The first n digits are.

1,1,2,3,5,8,3,1,4,5,9,4,3,7 ,0,7,0,7,0,7，。。Troublesome.

Add it all up, and get 56+7*(2002-14) 2=7014, and the remainder is the liquid cavity friend 4