After 1 round, the remaining 1, 3, 5, 7, ... 79;
After two rounds, the remaining 1, 5, 9, ...;
After three rounds, the remaining 1, 9, 17, ...;
After four rounds, the remaining 1, 17, 33, 49, 65,
In the future, it will be 65 1 33;
33,65,
33,
So the last one is 33,
It is my habit to analyze a topic, especially a typical topic. We can discuss the above topics:
1. If it is an even number and is 2 to the power of n, the result is 1.
2. If it is in the form of j * 2 n, there is one more series after n rounds, 1 is the first number, and the tolerance is 2 n, * * * j. 。
3. If it is odd, 1 is even, and the mantissa is the first number, followed by 1.
4. For the results of several simple odd numbers,
3- on 3,
5- on 3,
7- on 7,
9- on 3,
1 1- at 7,
13- Yu 1 1,
15- Yu 15,
2 n- 1- remainder 2n-1,
With these results, you can use them flexibly, so you can solve a lot of screening.
Continue the discussion:
The above game can be transformed into the following form: arrange 1 to n into a circle, start with 1, and continue the cycle until the end, and find out which number is left.
1. Look at 2 n first. After the first round is removed, what remains is 2 (n- 1). After removing them one by one, 1 remains.
2. Look at the situation between 9 and 15:
9- remaining 3,
10- on 5,
1 1- at 7,
12- Yu 9,
13- Yu 1 1,
14- Yu 13
15- Yu 15,
When n increases 1, the result increases by 2.
3. Choose any number k, (K+ 1 is not in the form of 2 n) and his result is a,
Then, we can compare k with K+ 1 after removing one round, and we can find that the result is the same as that of K+ 1, that is to say, the number of results is +2, that is, 2* 1.
Based on the above situation and the principle of mathematical induction, we can draw the following conclusions.
a = 2 *(N-K)+ 1；
A-the last remaining figures;
N-the total number of starting cards;
K-the maximum number less than or equal to n and divisible only by 2,
For example, N=80, K=64,
a = 2(80-64)+ 1 = 2 * 16+ 1 = 33；
A scholar in Suzhou. July 24, 2005.
Thank you for your happiness. I will recommend it.