The two of them split three piles of candies, each with 3, 5, and 7 candies.

1。First of all, the three piles only consider the quantity, not the order. (a,b,c) is used to represent the remaining quantity of the three piles, which is easier to verify (3,2,1); (n,n,0) When these two are combined, whoever takes it will lose.

2。Combining the above two "must losers" can be recursively obtained (5,4,1); (7,6,1)..If you take it first, you will lose.

3.The question starts with (3,5,7), and the result after taking A must be (2,5,7),(3,4,7),(3,5,6) Otherwise, B only needs one step to make A reach the above "defeat point", and A will naturally not do this.

4.Taking (2,5,7) as an example, combined with the above points of defeat, B can only be divided into (2,4,7),(2,5,6), and then A can be divided into (2,4,6) no matter what.

5。In the same way, (3,4,7) and (3,5,6) have the same result. Therefore, the question A will win. The first step is to pick one of any of the piles.

And take 1-n of the n candies left in the pile???

There's nothing I can do!!

I used to play chess before, but it's actually very simple.

No matter which pile, take an even number for the first time, and then add it to an even number according to the number taken by the other party and yourself.

The last one must be the one who takes it first to win.

After thinking about it, it is indeed who takes first and who wins first.

The other party has no chance. The specific description is too complicated. Because the first person to take it can calculate how many piles of candy there are, and calculate that 2 people need to take it at least many times in total, and you can control this number of times to an odd number.

If it's an odd number, you're guaranteed to win.

I don't know if the landlord will give points.

What does it mean to take 1-n of the remaining n candies? Get it in your pocket and still have candy???

Whoever takes first wins. The first person to take it, calculate that 2 people need to take it at least many times in total, and you can control this number of times to an odd number. If it's an odd number, you're guaranteed to win.

First of all, I'm going to keep it with three piles of sugar, and whoever makes it two piles first loses.

And take 1-n of the n candies left in the pile???

And take 1-n of the n candies left in the pile??? "I don't know what this sentence means?

It's okay to have an odd number of candies to get yourself.

It's clear that the first to take advantage of it!

There's only one left in any pile.

One by one, the remaining one is smashed into foam.

I'm sorry I don't know. I wish you a speedy resolution of the issue.

Just don't have two piles of the same amount left

Take 3 from the pile of 7.

In the end, 3,4,5 remained

Whoever is left with 3, 3 or 4, 4 or 5 5 wins.

Summary. Hello dear<>

The answer you're looking for: This is a combinatorial problem that can be solved using exhaustive methods. We can first determine the distribution scheme of one of the candies, then calculate the distribution scheme of the other candy, and finally multiply the two schemes to get the total number of allocations.

Taking the distribution of 9 candies as an example, it can be divided into 5 schemes: 2-2-2-2-1, 3-2-2-1-1, 3-3-1-1-1-1, 4-2-1-1-1-1, and 4-1-1-1-1-1-1. For the distribution of 16 candies, it can be assigned as 2-2-2-2-2-2-2, 3-3-2-2-2-2-2-2-2, 3-3-3-2-2-2-2-1, 4-3-3-2-2-1-1 and 4-4-3-2-1-1-1, a total of 5 schemes. Therefore, the total number of distribution schemes is 5 5 = 25.

There are 9 and 16 kinds of candies respectively, and they should be given to 7 people in full, and the total number of candies each person gets is at least 2 and up to 5, and there are many kinds of candies in total.

Hello dear<>

The answer you're looking for: This is a combinatorial problem that can be solved using exhaustive methods. We can first determine the distribution scheme of one of the candies, then calculate the distribution scheme of the other candies, and finally multiply the two schemes to get which is the largest of the total distribution of celery.

Taking the distribution of 9 candies as an example, it can be divided into 5 schemes: 2-2-2-2-1, 3-2-2-1-1, 3-3-1-1-1-1, 4-2-1-1-1-1, and 4-1-1-1-1-1-1. For the distribution of 16 candies, it can be divided into 5 types of squares: 2-2-2-2-2-2-2, 3-3-2-2-2-2-2-2, 3-3-3-2-2-2-2-1-1, 4-3-3-2-2-1-1-1, and 4-4-3-2-1-1-1. Therefore, the total number of distribution schemes is 5 5 = 25.

There are 25 kinds.

This means that if you add 2 more, it will be exactly a multiple of 7 and 8.

The trouser cracks are n*56-2; (n is a positive integer)

There are 56-2=54 to the burial raids.

A: ...

7x+5=8x+6

8x-7x=6-4

x=27x+5=7*2+5=19

Answer calendar: There are 19 closed-posture pants in this pile of sugar.

Summary. First of all, we can calculate the total number of candies in this pile: 152 + 153 + 154 + 155 + 156 + 157 + 158 = 1085.

We then divide the total by the number of shares per person, i.e.: 1085 6 = 180 and 5. Finally, we know that there are 5 candies that cannot be divided equally, so an extra pack needs to be added, so the total number of packs is 25.

It should be noted that when the remainder of the division is not 0, we need to round up, i.e. add one more packet to make sure everyone gets the same amount of candy.

A pile of candies with a quantity of between 152 158 and 6 pieces per person, then the pile of candies can be divided into several packs.

A pile of candies is between 152 158 in packs of 6 pieces, then the pile of candies can be divided into several packs.

The answer is 26, and this pile of candies can be divided into 25 packs.

First, we can calculate the total number of candies in the pile: 152 + 153 + 154 + 155 + 156 + 157 + 158 = 1085. We then divide the total by the amount each person gets a share, i.e.:

1085 6 = 180 surplus 5. Finally, we know that there are 5 candies that cannot be divided equally, so we need to add an extra pack, so the total number of packs is 25. It should be noted that when the remainder of the division is not 0, we need to round up, i.e. add one pack to make sure everyone gets the same amount of candy.

The answer is wrong, isn't it?

A pile of candies is between 152 158 in packs of 6 pieces, then the pile of candies can be divided into several packs.

The answer is correct.

The answer is 26 why.

Because all numbers between 152 and 158 are divisible by 6, and the numbers divisible by 6 must be in this range. Then we only need to divide these numbers into groups of 6 according to the order of the big open, and we can get 26 groups, that is, 26 bags of candy.

Think of it like this:

First, roll up 9 pieces of fruit candy and divide them into 3 piles of 3 pieces each.

Then divide the remaining 3 pieces into 1-3 piles according to the number of combinations, and see if there are several cases:

Add 3 pieces to each pile.

So there are 3 divisions.

If you are aware of the caution and the full of respect, welcome my answer, thank you.

If you have any questions, please feel free to ask.

Summary. The mathematical method is, 164-7+5=162 162 (1+3+2)=27 2 27-5=49, <>

There are 164 candies in the three piles, of which the number of candies in the first pile is 3 times that of the second pile, 7 pieces, and the number of candies in the third pile is 5 pieces less than twice that of the second pile, and how many candies are in the third pile.

Hello! Dear, I can solve the equation laughing and disturbing the second pile of Qiaosheng laugh is x, then the first pile is, 3x+7, the third pile is, 2x-5, and the column equation is x+3x+7+2x-5=1646x+2=1646x=164-26x=162 x=162 6x=27<>

The doll has not yet learned to solve equations and equations.

The first pile is, 3 27 + 7 = 88 pieces, and the third pile is, 2 27-5 = 49 <>

What grade is your child? Pro-<>

Four. Why is it different?

Oooh, the solved x is the second pile.

The third pile is 2 27-5 = 49 seeds, pro.

Can you write me out every step of the way?

The mathematical method is, 164-7+5=162 162 (1+3+2)=27 2 27-5=49, <>

Summary. Dear, there are at least 46 pieces of these sugars, equation 6 7 + 4 = 46, there is a bunch of candies divided evenly between 6 people, two less, and three less for 7 people, how many candies are there in this pile at least.

Dear, there are at least 46 pieces of these sugars, and the formula 6 7 + 4 = 46 First of all, we remove 4 of these sugars, so that they can be divided equally among 6 people, and also among 7 people.

The least common multiple of 6 and 7 is 6 7 = 42, and then add the 4 sugars subtracted.

So in the end it's 46 pieces of sugar.

Why? There are two fewer people who are evenly divided into 6 people, is there 4 left, because 4 plus two equals 6, so that the average score can be given to 7 people less than three, and there are still 4 left, Lao Tong can be evenly divided because 4 plus three equals 7.

So remove the 4 in your hand first.

Still don't understand???

You don't understand, tell me.

I'll explain it to you.

Summary. The common multiple of two or more integers is called their common multiple, and the smallest common multiple other than 0 is called the least common multiple of these integers. The least common multiple of the integers a,b is denoted as [a,b], and similarly, the least common multiple of a,b,c is denoted as [a,b,c], and the least common multiple of multiple integers is also denoted by the same notation.

The concept corresponding to the least common multiple is the greatest common divisor, and the greatest common divisor of a, b is denoted as (a, b). Regarding the least common multiple and the greatest common divisor, we have this theorem: (a,b)x[a,b]=ab(a,b are integers).

A pile of candies, 8 8 points less than three, 6 6 points more than three, how many candies are in total.

Try to hurry. Hello kiss a pile of candies, 8 8 points less three, 6 6 points more than three, this pile of candy a total of 24. The minimum number of candies in this pile is required, that is, to find the least common multiple of 6 and 8, that is, the least common multiple is the product of the prime factor of the Beidou Gong Jiling of 6 and 8 and the unique prime factor 6=2 3,8=2 2 2, so the least common multiple of 6 and 8 is:

2 2 2 3 = 24, the common multiple of two or more integers is called their common multiple, and the smallest common multiple other than 0 is called the least common multiple of these integers. The least common multiple of the integers a,b is denoted as [a,b], and similarly, the least common multiple of the scattered a,b,c is denoted as [a,b,c], and the least common multiple of the dispersion of multiple integers is also denoted with the same notation. The concept corresponding to the least common multiple is the greatest common divisor, and the greatest common divisor of a, b is denoted as (a, b).

Regarding the minimum common divisor and the greatest common divisor, we have the following theorem: (a,b)x[a,b]=ab(a,b are integers).

Summary. The number of tanggu is a common multiple of three and five, and it cannot exceed 40, so it is a multiple of 3 and 5 cm 3 5 = 1515 2 = 3015 3 = 45

There is a pile of candies with less than 40 pieces, and 3 or 5 people can just divide the maximum number of pieces of candy in this pile.

The specific calculation process is as follows:

The number of Tanggu is a common multiple of three and five, and it can't touch Chi Chao's laughing model Li Guo 40, so it is a multiple of 3 and 5 centimeters 3 5 = 1515 2 = 3015 3 = 45

So up to 30 pieces.

So the final answer is 30 bucks at most? Yes.