The number 3 is arithmetic progression, set as a 1, a2, a3.

Then in the number 3, the difference between a3 and a 1 must be an integer multiple of 2.

That is, a 1 and a3 must both be odd or even.

Choose a 1 and a3, and the middle number a2 will be determined.

So, if a 1 and a3 are odd numbers.

From 1 to 19.

* * * Choose two odd numbers from 10, namely a 1 and a3. Note that the tolerance of arithmetic progression can be negative, that is, a 1 can be greater than or less than a3. So there is an arrangement order.

So there are A(2 10)= 90 methods.

Here, A(2, 10) indicates the permutation in the permutation combination, that is, two of 10 are selected in sequence.

Similarly, if a 1 and a3 are even numbers.

From 2 to 20

*** 10 even number, just choose two numbers. There are also 90 ways to take A (2, 10).

To sum up, * * has 2 * a (210) = 2 * 90 =180 methods.

2.

Obviously, if you choose all dimes, you won't get a dollar, and if you choose all two coins, you won't get a hundred.

Therefore, the same currency will not be formed in different ways.

There are three dimes,

We can choose four methods: none, 1, 2, 3, * * *.

There are six two-dollar coins,

We can choose seven methods: don't choose, choose 1, choose 2, choose 3, choose 4, choose 5, choose 6, * * *.

There are four hundred-dollar coins,

We can choose five methods: don't choose, choose 1, choose 2, choose 3, choose 4, * * *.

Therefore, according to the principle of multiplication, there are

4*7*5= 140 methods.

Therefore, it corresponds to 140 currencies.

But there is one thing to be removed here, that is, nothing is selected. At this time, the currency value is 0, so it should be removed.

Therefore, 140- 1= 139 different currencies can finally be formed.

3.

( 1)

1 digit (1-9) cannot contain 0.

(2)

There are 9 cases of 2-digit (10-99).

(3)

3 digits (100-999)

At this moment,

There are 1 to 9 in a hundred miles, and there are 9 methods.

There are two ways to put one digit or zero in ten digits.

The remaining digits cannot be 0. You can fill in 1 to 9. There are 9 ways.

Therefore, * * * has 9*2*9= 162 cases.

(4)

4 digits (1000- 1999)

At this time, it can only be 1.

There are three ways to fill in the company, the tenth and the hundredth, which can be 0.

The remaining two digits can be filled in 1 to 9, so there is a 9*9 method.

Therefore * * * Hu 3*9*9=243 cases.

Synthesis (1), (2), (3) and (4)

* * * There are 0+9+ 162+243=4 14 methods.