Proof Given any of the 5 integers, at least 2 numbers are multiples of 4 with a difference of 4.

This is a question based on the principle of the drawer:

The drawer principle, also known as the Dirichlet principle, is an important and fundamental mathematical principle.

Drawer principle (1): If more than n elements are divided into n sets in any definite way, then there must be at least one set containing at least two elements.

Drawer principle (2): If you put more than m n elements in n drawers, then there must be a drawer with m 1 or n 1 or more elements.

Use it to solve this problem:

Any integer can be represented in four ways: 4n, 4n, n n, 3 (n 0).

So let's give 5 integers, at least two of which are in one of the above sets, which we denote by 4m a and 4n a (m, n 0, a=0, 1, 2, 3).

They are subtracted, assuming that m is greater than n, then the result is 4(m n), which is definitely a multiple of 4.

Solution: Suppose these 5 numbers are.

Because 5-1 = 4

So 4 is a multiple of 4.

In the same way: Suppose these 5 numbers are.

Because 6-2 = 4

So 4 is a multiple of 4.

Solution: Let these five numbers be a a+1 a+2 a+3 a+4

then there is a+4-a=4

Suppose these 5 numbers are.

Because 5-1 = 4

So 4 is a multiple of 4.

Tongheng Reason: Suppose this is to sell 5 numbers.

Because 6-2 = 4

So 4 is a multiple of 4.

So for 5 integers, at least 2 of them are multiples of 4.

There are only three remainders for any integer divided by 3: 0, 1, 2

If any 4 numbers are given, there must be two numbers divided by 3 and the remainder will be the same.

Then the difference between these two numbers can be removed by 3 whole slag and dismantled Zen, that is, their difference is a multiple of 3.

Because the remainder of any integer divided by 4 is only 0, 1, 2, and 3;

Therefore, using the drawer principle, it can be seen that any five integers divided by 4 must have two congruent (the remainder is the same);

then take these two numbers with the same surplus, and the difference they contain must be a multiple of 4 (the remainder is the same, and the remainder is subtracted as soon as the remainder is subtracted);

Answer]: Let a be the set of these 52 integers, |a|=52。Remembering ai, then a0a1 and a50 constitute the 51 "pigeon nests" of a, so that there is ak to make |ak≥2。Let a, b ak

Then a and b are divided by 100, and the rest of the numbers are either the same or their sum is 100, i.e. yes.

a=100m+k b=100n+k

Either a=100m+100-k, b=100n+100-k, or a=100m+k, b=100n+100-k, in either case, a-b or a+b can be covered and the deficit is divisible by 100.

6 pcs. 1) 5 is not good.

For example: 1 2 3 4 5

6) 6 pcs are OK.

Because the remainder divided by 5 is only 0, 1, 2, 3, 4 a total of 5 kinds, by the drawer principle book liquid, its feast has at least 2 divided by 5 remainder is the same, so the difference between these two is a multiple of 5 state mascot.

The last digit of the integer (0 9) is divided into 6 categories: in the given 7 integers, if there are two numbers with the same last digit, the difference is a multiple of 10;

If the last digit of these 7 digits is different, then two of them must belong to one of the 6 categories above, and their sum is a multiple of 10

Therefore there must be two row-thick integers, and the sum or difference of their rows is a multiple of 10

Divide each number in 1-24 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 the answer is 6, so the answer is: 6

You've probably learned the concept of "remainder".

There are 9 types of remainders for dividing any number by 9.

Yu so according to the principle of drawers.

10 numbers are placed in a drawer made up of 9 remainders.

There must be two of them in the same drawer

Therefore, the above two numbers are about the remainder of 9.

So the difference between these two numbers is a multiple-denier number of 9. Certification.