32 sheep, killed in 7 days, killing single but not double. 20

Topic: Thirty-two sheep, killed in seven days, killing single but not double every day, how to kill?

Analysis: The focus is on the principle of "killing singles and not killing doubles", but this principle does not limit how to kill, nor does it explain that "single and double" must refer to sheep.

Answer: 1. Thirty-two sheep, which can be understood as 3+10+2 15, 15 can be divided into seven odd numbers.

2. It is to number the sheep, kill the sheep on the first day, and then divide the remaining 16 in 6 days to divide it very well, (3 + 3 + 3 + 3 + 1 + 3 16).

3. Kill in seven days, and you can order the order number (week) according to the day of the week.

One, three, five, seven) kill, can also be by date (1st, 3rd......In short, as long as it is killed in seven days, it is also in line with the principle of killing a single and not a double.

4. It is to kill sheep across time, in any two days of seven days, in the early morning across zero o'clock to kill a sheep, this can be counted before zero o'clock to kill, can also be counted after zero o'clock killing, 3 + 3 + 3 + 3 + 3 + 8 (+1) + 8, this "1" and the previous "8" add up is singular, and the back of the "8" add is also singular.

7+ 7 +7+ (six and a half don't count as a kill) + one and a half doesn't count as a kill) +1 +1 = 32

Kill 31 on the first day, rest for 6 days, and kill 1 on the 7th day.

32 divided by 7 On average, one sheep is killed every day

You didn't finish this question, do you kill at least one every day?

Otherwise, 31 1 will be killed in two days.

Kill 2 in the first point, 5 in the next 6 days! ~

0 does not belong to the double bar.

But I think Article 4 on the first floor is okay.

One day, if you don't kill it, you can't do it.

If there is a time when you don't kill.

Buy one more sheep and make 33.

28 sheep, 7 days to kill, only allowed to kill single, not to kill double, how to kill in 7 days?

Hello, there is no answer to this, you can use a simple proof question: according to the question, assume that the sum of 7 singular numbers is 28, as we know, the addition of two singular numbers is an even number, and the addition of 3 singular numbers is a singular number. Whether n singular numbers are added to be even or singular mainly depends on whether this n is singular or even.

If n is singular, then the value is singular, and vice versa, the value is even. 7 is an odd number, and 28 is an even number. Therefore, it is impossible for 7 singular numbers to add up and still be an even number.

Therefore, the proposition is not true and unsolvable.