This problem can be solved with the knowledge of primary school.

First of all, understand that the number divisible by 3 has a characteristic, that is, your sum is divisible by 3.

So this five-digit number can be divisible by 3, divided by a 6, leaving the sum of four digits.

Now we're going to branch:

1。 6 is at the top of five digits. The "first bit" of the remaining four bits can be 0. So there are (9999-0)/3+ 1=3334 numbers in the total number of 0-9999 (it is easy to ignore here).

2。 When 6 is not in the first place. Then the "first place" of the remaining four digits cannot be 0. So this four-digit number is the number divisible by 3 in 1000-9999. * * * (9999- 1002)/3+ 1 = 3000.

Insert 6 again. 6 can't be in the first place, so there can only be four positions: thousands, hundreds, tens and units. So a * * * has 3000*4= 12000.

The two possibilities add up to 12000+3334= 15334.

The second question.

We should also divide the situation.

1。 When a 1 >; When b 1

Because A is increasing and B is decreasing.

So all a values are greater than b values.

Numbers whose values are n+ 1 to 2n.

The value of b is 1 to n.

The formula = a1-b1+a2-B2+...+an-bn.

=(n+ 1 + n+2 +...+ 2n)-( 1+2+...+n)

=n^2

2。 When a 1

Because A is increasing and B is decreasing.

So all b values are greater than a values.

The value of b is a number from n+ 1 to 2n.

A number with a value of 1 to n.

Equation = b1-a1+B2-A2+...+BN-An

=(n+ 1 + n+2 +...+ 2n)-( 1+2+...+n)

=n^2

3。 When a 1

Because A is increasing and B is decreasing.

So there must be AK & GTBKA (k-1) < B(k- 1)- brackets indicate angle marks.

Equation = b1-a1+B2-a2+…+b (k-1)-a (k-1)+AK-bk+…+an-bn.

Make all the minuets into a new group C.

All subtractions form a new group d.

c 1=b(k- 1) c2=b(k-2)...c(k- 1)=b 1 ck=ak...cn=an

d 1=a(k- 1) d2=a(k-2)...d(k- 1)=a 1 dk=bk...dn=bn

Because c1= b (k-1) >; a(k- 1)=d 1

A is increasing, so c 1 is greater than a(k- 1) and the previous number, here is both d(k- 1) and the previous number.

Because c 1=b(k- 1) b is decreasing.

So c 1 is greater than bk and later, and here is dk and later.

So c 1 is greater than all numbers of d.

C 1 to c(k- 1) are all increasing, so they are all greater than d.

It can also be proved that ck is greater than all d.

Ck to cn is also increasing.

So all c's are bigger than d's.

Go back to the first situation.

The result is still n 2

It is more troublesome to elaborate in words. I hope you can understand.

You can draw when you tell your son.

The abscissa indicates the size of the corner mark.

The ordinate represents a numerical value.

All point connections obtained:

Series is an increasing line.

B is the decreasing line.

They intersect (mainly for the third case, the first two don't make sense, and the intersection is not in 1 quadrant).

The result of the formula in the problem is always the number above the intersection minus the number below.

So we always add up the big ones and subtract the small ones.

The result is always n 2