10,000 people, report the number in turn, report the singular number, and so on, and ask what is the

The title is not very clear. I understand that 10,000 people form a circle, and then report the number in a loop, and report the singular number to leave the circle, until there is only one person left in the circle, and that person reports the number.

Number of rounds Total number of people The number of people who have reported doubles.

5 625 312 (the last person to report the order).

9 39 20 (the last person reported a double).

12 5 2 (the last man to report).

After a total of 13 rounds, there was finally only one person left. The last person to report the double is the sum of all the numbers in the column of the total number of people above -1, and the last person to report the double is the sum of the numbers in the column of the total number of people (that is, the sum of the numbers in the column of the total number of people).

And the last man was the seventh of the 10,000 people.

One person reports, 10,000 people are 10,000, and the last number is 10,000.

After the first count, the one called 2 stood first. There are 50 people left. After the second count, the first time to call 4 stood first.

There are 25 people left. After the third count, the first time he called 8 stood in the first place. There are 12 people left.

After the fourth count, the first call for 16 stood first. There are 6 people left. After the fifth count, the first time he called 32 stood first.

There are 3 people left. After the sixth count, the first time he called 64 stood first. There is 1 person left.

So, the last remainder reported 64 in the first report. Complete.

Summary. The number reported by the first 5 people is 1,4,7,10,13, and the number reported by the 6th person is 11 for 35, and the number reported by the last 10 people can be regarded as the first item is 14, and the tolerance is 3 in the equal difference series (14+14+9 3) 5=55 5=275275+35+11=321, so the last 16 people report the number together equals 321

There are 16 people lined up in a bend of the old line in turn to report the number, the first person reported 1, and then the number of each person reported is to add the number of the previous person to 3, the sixth person reported the wrong number in the process of reporting, and the number of the previous person reported by the buried rose minus 2 reported out, and the number of the last 16 people reported by the last 16 people added up to what is equal to?

There are 16 people lined up in a line to report the number in turn, the first person reported 1, and the number of each person reported in the future is to add 2 to the number reported by the previous person to report the number of the 6th person in the process of returning to the ear, and the number of the previous buried people reported minus 2 reported out, and the last 16 people reported the number of what is equal to?

I'm sorry, but the first question is right.

The number reported by the first 5 people is 1,4,7,10,13, and the number reported by the 6th person is 11 for 35, and the number reported by the last 10 people can be regarded as the first item is 14, and the tolerance is 3 in the equal difference series (14 + 14 + 9 3) 5 = 55 5 = 275275 + 35 + 11 = 321 buried by the last 16 people reported the number of Yuxiang added up to Qingqiao Bo is equal to 321

Each ordinal number divided by 2 remains, leaving 1.

The first time, 1, 3, 5 ,......99

The second time, 1, 5, 9 ,......97

The third time, 1,9 ,......97

Fourth, 1, 17 ,......97

5th, 1, 33 ,......97, 6th, 1, 65th, 7th, 1 is 1.

After the first report, after the odd number is listed.

The rest is 2n, n = 1, 2 ,..25

Then 2(2k-1), k=1,2 ,..25, out of the ranks.

The rest is. 4k,k=1,2,..25

And then will. 4（2k-1),k=1,2,..13 out of the ranks.

Leftover. 8k,k=1,2,..12

Again will. 8（2k-1),k=1,2,..6. Out of the queue.

Leftover. 16k,k=1,2,3,4,5,6

Again will. 16 (2k-1), k = 1, 2, 3 out of the column.

Leftover. 32k,k=1,2,3

Again will. 32 (2k-1), k = 1, 2 out of column.

64k is left, k=1, so the last one left is 64.

Because the even number remains, the longer the number n has the more factor 2, 64=2 6 If m=128 then m=2 7 will last to be 128, if m=127 then 64 has the most factor 2 in 127, so the maximum value of m is 127

These 2018 people are numbered 1 2018 in order from the top to the bottom of the row. Obviously, the first registration with an even number is reported as an even number 2. When registering for the second time, among the ten people in 2018 and 2009, because the first one was reported by the even number 2018, the first 5 people registered, the people with the even number reported the odd number, and the people with the odd number reported the even number, and the last 5 people reported from the 2013 number, then the people with the even number reported to the even number, and the people with the odd number reported to the odd number.

The same is true for the next 2008 1999 ten, i.e. only 2 out of 10 people are reported twice and the number is even. 2018/10=201……8. From No. 9, 2018, there are 201 * 2 = 402 people who reported an even number twice. In the final 8 1 number, there is no one in the first five, and only one in the last three.

Then the number of people who report both times is 402+1=403

The first time the rest is a multiple of 2.

The rest of the second time is a multiple of 2 = 4.

For the third time, the remainder of Tubira is a multiple of 2 = 8.

The nth time is left, which is the number of times the front of 2 n.

The last remaining is 64, then m should be less than 64 2=128, and the maximum is 128-1=127

If m 128, then the last thing left is 128).

7*71=497

Full loop 71 times.

Each time the number and 28

The sum of the numbers of the complete cycle is 71*28=1988

The last incomplete loop: 1, 2, 3 and 6

The sum of all numbers.

500/7=71.。。3

1 to 7 cycles 71 times 71 * 28 = 1988 3 more times 1 + 2 + 3 = 6

And is 1988 + 6 = 1994