1 hours before bedtime-divisible by Math series

Everyday Math.

Divisible.

Division is the most basic concept in mathematics. Division. But. This stuff is different from the usual compulsory education division.

Definition: Set A is a non-zero integer. B is also an integer. If there is an integer p, make b=a*p; then we'll call it

B can be divisible by a "divisible"!. Remember as a|b.

So b is called a multiple. A, the approximate name of B, also called the factor.

Lift the chestnuts. 3|12 21|63 5|15 is like this.

Divisible. There are also many properties.

1, if a|b && b|c so a| c a little transitivity flavor.

Chestnuts: 5|15 && 15| 250 so 5|250.

2 if a|b && a|c satisfies any integer x and Y has this a| (b*x+c*y)

For a chestnut: 3|15 && 3 |6 3| (15*2+6*3)

Calculate Oh, (15*2+6*3) = = 48 very interesting 3|48

3, if M is not equal to 0 m belonging to an integer, when a|b (a*m) | (b*m)

These two are the necessary foundation. Of course there are other properties, but the others are very simple.

4, some of the nature of the proof: to raise a chestnut:

Set integer x and y to satisfy the following formula: A*x+b*y=1, and a|n, B | n So (a*b) |n

Prove:

Because A|n and b|n

According to Nature 3: (a*b) | (n*b) and (a*b) | (a*n)

By Nature 4: (a*b) | (n*b*y+n*a*x)

Then: (a*b) | N (b*y+a*x)

Because:a*x+b*y==1;

So: (a*b) | N

5, there are a lot of elementary school what to learn.

An integer A

The last one can be divisible by 2, 2|a

The last two potentials are divisible by 4, 4|a

The last three bits can be divisible by 8, 8|a.

If 3 can divide the sum of all the numbers on a, 3|a

If 9 can divide the sum of all the numbers on a, 9|a

At the same time, the characteristic of the number that can be 7,11,13, divisible: If the difference between the last three bits of a and the last three bits is divisible by 7,11,13

then 7|a && 11|a && 13|a

There's a new math problem later: Church church. and password strongbox.

1 hours before bedtime-divisible by Math series