Write any 5 natural numbers so that the difference between any two of them is not a multiple of 4.

1) First of all, suppose there is a natural number a1 a2 a3, if the difference between a2-a1 divided by the remainder of 4 is equal to the remainder of the difference between a3-a1 divided by 4, then the difference between a3-a2 must be a multiple of 4.

Proof: Let a2-a1=4k1+n, a3-a1=4k2+n, k1, k2, and n are all integers. (i.e. the difference between two numbers divided by the remainder of 4 is the same as n).

Then a3-a2=a3-a2-a1+a1

a3-a1）-（a2-a1）

4k2+n-（4k1+n）

4k2+n-4k1-n

4k2-4k1

4 (k2-k1), that is, the difference between a3 and a2 is a multiple of 4.

2) Then, not a multiple of 4, that is, dividing by 4 will leave 1 or 2 or 3.

Subtracting 5 numbers in pairs yields 4 sets of differences, which are set to n1, n2, n3, and n4 respectively, and the remainder of these four groups of differences divided by 4 must have at least two of the same or multiples of 4.

Proof : Suppose n1 4 is more than 1, n2 4 is more than 2, n3 4 is more than 3, then the remainder of n4 4 must be one of or 3, the case.

1. If the remainder is 0, i.e. n4 itself is a multiple of 4;

Circumstance. 2. If the remainder of n4 4 is either 1 or 2 or 3, then the same remainder is obtained from n1 or n2 or n3, and the same remainder is the difference between the subtraction of the two numbers is a multiple of 4, that is, the part (1) that has been proved above.

So, in conclusion, there are no 5 natural numbers that meet the conditions.

There are four numbers for 4a, 4b c d+3, and then write that the fifth number must be one of these four numbers, this problem is unsolvable.

The four numbers are a1, a2, a3, a4, and a5

Remember that the remainders of them except for the four are b1, b2, b3, b4, b5, then bi belongs to 0, 1, 2, 3 (i=1, 2, 3, 4, 5), according to the principle of the branch old line drawer, it can be known that there must be two bs equal.

The remainder of the corresponding two aces divided by four is equal, i.e., the difference is a multiple of four.

The difference between two different natural numbers is divided by 4, and the remainder can be Huizhou, so the difference between 5 different natural numbers, of which at least two natural numbers is a multiple of 4.

There are two cases of dividing a natural number by 4: one is that the remainder is 0 divisible, and the other is that there is a remainderIf the remainder of 2 natural numbers divided by 4 is the same, then the difference between the two natural numbers is a multiple of 4.

If you think of these four cases as 4 drawers, and 5 different natural numbers as 5 apples, there must be at least 2 numbers in one drawer, and the remainder of these two numbers is the same, and their difference must be a multiple of 4. So for any 5 natural numbers that are not identical, at least two of them are multiples of 4.

Divide 1 number by 5, the remainder includes 0, 1, 2, 3, 4 in the 5 case, and then choose any 1 number will have the same remainder, the difference of this number is a multiple of 5, so at least choose any 6 numbers, to ensure that at least two numbers of difference is a multiple of 5

Answer: Choose at least 6 numbers, which can ensure that the difference between at least two numbers is a multiple of 5

This is where the drawer principle comes into play (without going into details).

Any 5 natural numbers, according to the remainder divided by 4, can be divided into four categories.

That is, the remainder, the remainder 1, the remainder 2, and the remainder 3.

Subtracting the same class of numbers, the difference must be a multiple of 4.

If there are only 4 natural numbers, then the four may be evenly distributed in the four classes, in which case they will not be multiples of 4.

However, if a number is being added, then the number added must be a number of the above class, so the difference from the number in that class must be a multiple of 4.

So, at least 2 numbers are multiples of 4.

The remainder of any natural number divided by 4 may be 0,1,2,3According to the drawer principle, there must be 2 natural numbers in any 5 natural numbers, and the remainder divided by 4 is the same, and the difference between the two natural numbers is exactly a multiple of 4.

Proof: Copy.

Arbitrary. However, the remainder of the number is divided by 5, and there are only 5 cases of 0 and bai.

The structure of the du is 5 draws

Drawer: [0], dao[1], [2], [3], [4].

When there are 6 different natural numbers, divide each of these 6 different natural numbers by 5, and there must be at least 2 numbers with the same remainder, and the remainder is the same, that is, the remainder is subtracted to 0.

So, write 6 different natural numbers arbitrarily, and at least one of the two numbers is a multiple of 5.

bai arbitrarily writes 6 different natural numbers, which du

There is at least a difference of at least two numbers in is.

multiple of DAO5.

Prove that any natural number can be divided by 5 remainders There are only these 5 cases, which may be constructed into 5 drawers:

When there are 6 different natural numbers, divide these 6 different natural numbers by 5, there must be at least 2 numbers with the same remainder, the remainder is the same, that is, the remainder is subtracted to 0, so if you write out 6 different natural numbers arbitrarily, there is at least a group of two numbers where the difference between the two numbers is a multiple of 5.

The drawer principle proves that the natural number of any bai is divided by 5 and the remainder du is only 0 and zhi These 5 situations may be constructed as 5 drawers respectively

When there are 6 different volumes.

Natural numbers, divide these 6 different natural numbers by 5, there must be at least 2 numbers with the same remainder, the remainder is the same, that is, the remainder is subtracted to 0, so, arbitrarily write 6 different natural numbers, at least a group of two numbers The difference between the two numbers is a multiple of 5.

Oh, oh, 7 heads.

This conclusion.

That's right. The remainder of a natural number divided by 4 may be , so consider these 4 cases as 4 drawers, and any 5 different natural numbers as 5 elements, and then according to the drawer principle, there must be at least 2 numbers in a drawer, and the remainder of these two numbers is the same, and their difference must be a multiple of 4. So, for any 5 different natural numbers, at least two of them have a difference of a multiple of 4.

If the remainder of the two integers a and b divided by the natural number m is the same, then their difference a-b is a multiple of m. According to this property, this problem only needs to prove that 2 of these 5 natural numbers have the same remainder divided by 4. We can divide all natural numbers into four categories by four different remainders divided by four.

That's 4 drawers. Take any 5 natural numbers, according to the drawer principle, there must be two numbers in the same drawer, that is, they divide by 4 and the remainder is the same, so the difference between these two numbers must be a multiple of 5.