If there is 10 5 yuan, then there is only 1 species.

If there are 9 in 5 yuan, 5 yuan's method will be classified according to the number of 2 yuan, and 2 yuan will use 2 at most, or 1, or it may not be used, so there will be 3 kinds.

If there are 8 cards in 5 yuan, the way to collect 10 yuan is to classify them according to the number of cards in 2 yuan. 2 yuan can use at most 5 cards, or 4 cards, 3 cards, 2 cards, 1 card, or it can be omitted.

There are six kinds of such meetings.

If there are 7 pieces in 5 yuan, add 15 yuan and there will be 8 kinds (for the same reason as above, according to the number of pieces used in 2 yuan).

…………

I always thought that 5 yuan didn't need one, so I used at most 25 2 yuan ones, at least not 2 yuan ones. There will be 26 kinds.

Total:1+3+6+8+1+13+16+18+21+23+26 =/kloc.

2. The natural number from 1 to 1999 is divisible by multiples of 2 and 5.

The multiple of 2 is 999, and the multiple of 5 is 399. Yes 199 is a multiple of 2 and 5.

So 999+399-199 =1199 is a multiple of 2 or a multiple of 5.

Other natural numbers are1999-1199 = 800. 800 simplest truth points can be formed.

3. The worst case is to take all 1 as 49, and then take 49 numbers from 50 to 100. A * * * took over:

(1+2+3+...+49)+49 * 51= 3724. At this time, there are no 50 duplicate numbers. One more, that's 3725, and one more.

It takes 720 meters to return.

It takes 720 /75=9.6 minutes for A to return to its original position, and 720/90=8 minutes for B. ..

So in 48 minutes, A walked five times and B walked six times, which was the first time for two people to return to the starting point.

A walked 75*48=3600 meters.

This score is written in the simple form of ABC/999, and the sum of numerator and denominator is 58.

Analyzing the prime factor of 999, 999=3*3*3*37. Therefore, a three-digit ABC must have three factors of 3 and 999*** (trial calculation by yourself). Go to three threes.

Then the score will be 2 1/37. The score before irreducibility is 567/999. The cyclic decimal number is 0.567(567 is the cyclic node).

6. If it is the sum of 1 to 9999, it can be calculated as follows.

1 appears in the unit, 10, 100 and 1000 respectively 1000 times, and * * * appears 4000 times.

2, 3, 4 ...9 Each number appears 4000 times.

So the sum of the numbers is (1+2+3+) ...+9) * 4000 =180000.

Sum of numbers from 9996 to 9999: 9* 13+6+7+8= 138.

The final total is:180000-138 =179862.

7. The value range of x is 1 to 6, and the first cycle segment of the obtained decimal is as follows: (In any case, the sum of the numbers of each cycle segment is 27).

0. 142857

0.2857 14

0.42857 1

0.57 1428

0.7 14285

0.857 142

When the sum of the nth bit is 20 10, there must be 2010/27 = 74 ... 12 (74 cycles, the following figures need to be added to12).

Through experiments, we can get 12 by adding the first two digits of 0.57 1428. Answer 1: When x is 4, N=74*6+2=446.

Add the first three digits of 0.7 14285 to get 12. Answer 2: When x is 5, N=74*6+3=447.