The product of four consecutive natural numbers must be a multiple of 12, why?

Let's first explain why "the product of two consecutive natural numbers must be divisible by 2":

For all natural numbers, they can be divided into 2 categories, which are those divided by 2 by 0 and 2 by 1, that is, even and odd, and the two adjacent numbers must be 1 odd and 1 even, which belong to these two categories respectively. In other words, one of the two adjacent numbers must be divided by 2 to remainder 0, that is, divisible by 2, which is a multiple of 2. Therefore, the product of these two numbers must be divisible by 2.

For three consecutive natural numbers, from the above analysis, we know that one must be a multiple of 3, and their product is divisible by 3; And just look at the first 2 numbers, it becomes 2 adjacent natural numbers, and there must be 1 of them is a multiple of 2, so their product is divisible by 2; It is also known that 2 and 3 are coprime, so their product must be divisible by 6 (2x3).

For 4 consecutive natural numbers, again, there must be 1 number divisible by 4, and their product is also divisible by 4. Of the remaining 3 numbers, it is clear from the above analysis that there must be a number divided by 4 and the remainder 2, that is, divisible by 2, so that their product is divisible by 8; At the same time, as in the case of 3 numbers, we can know that at least 1 of these 4 numbers is divisible by 3, and thus their product is divisible by 3; And 8 is coprime with 3, so their product must be divisible by 24. Then their product must be divisible by 24, which is a multiple of 12.

The product of four consecutive natural numbers must be a multiple of 12, why?

One of the three consecutive integers must be divisible by 3, and at least two of the four consecutive natural numbers are even, i.e., divisible by 4.

3x4=12

Of the four contiguous natural numbers, one must be divisible by two, one divisible by three, and one divisible by four.

So multiply four consecutive natural numbers, and their product must be a multiple of 12.

It certainly contains two even numbers, a multiple of three 2*2*3=12

Four successive natural numbers must be 1 divisible by 2, 1 divisible by 3, and 1 divisible by 4.

Sowei Li multiplies four consecutive natural numbers, and their product must be a multiple of 12.

Because the difference between two adjacent numbers of continuous natural numbers is '1', from 10 to 2, the difference is 10-2=8.

This is a 'difference problem'.

Difference = 10-2 = 8, Multiplier = 1 and 4 9 = 13 9 2nd Number = Difference (Multiple-1).

It can be seen that these 13 consecutive natural numbers start from 17 to 29.

Their middle number is 23, so their sum.

There are 13 consecutive natural numbers, and the 10th number is 1 and 4/9 times the 2nd number, so what is the sum.

Let the 7th number be a, then the 13 consecutive natural numbers are: a-6, a-5, a-4, a-3

a-2, a-1, a, a+1, a+2, a+3, a+4, a+5, a+6(a+3) (a-5)=13 9,,a=23 So, the 13 consecutive natural numbers are:

The sum of the sum of 5 consecutive natural old numbers must be a multiple of (5).

Proof: Let the integers of the five splitting continuations be: a, a+1, a+2, a+3, a+4.

a+ (a+1)+(a+2)+(a+3)+(a+4)5a+105(a+2)

Therefore, the sum of 5 consecutive natural numbers must be a multiple of (5).

It's a formula.

1²+2²+3²+…n = n n 1 2n 1 6, so the product of three consecutive natural numbers must be a multiple of six, because this is just a simple formula.

2，3，6

Because at least one of the three consecutive natural numbers must be even, and one must be a multiple of 3.

So their product is both divisible by 2 and 3 and 6.

1003=17×59

So the sum of these two natural numbers is:

Because of three. Natural number.

Multiplication, for example: a, b, c represent three natural standby judgments, and the product is imitation of the excavation car ABC, which is A.

bc times, the same goes on, so it must be a multiple of which numbers, hope.

There must be a multiple of 2 and a multiple of 3 in three consecutive natural numbers, so bei must be a multiple of 6.