I've got a very hard math problem?

Question

If you have a four figure number and add the numbers together it should get 2004. What number can do this (except the example)?

For example:

1983 + (1+9+8+3) = 2004

Answer 1

To begin, take the example:

1983 = 1000·1 + 100·9 + 10·8 + 1·3

Let ABCD represent the digits in a four-digit number.

ABCD = 1000·A + 100·B + 10·C + 1·D

----

Thus we have:

2004 = (1000·A + 100·B + 10·C + 1·D ) + (A + B + C + D)

2004 = 1001·A + 101·B + 11·C + 2·D

----

Since it has to be a in the 2000s, A (the thousands place) must be at least 1. However, if it was 1, then that would provide the previous answer. Note how 1983 is close to 2000 as it has to be to be over 2000 when its individual digits are added to it. That's the only solution using 1 in the thousands place.

----

However, the thousands place c also be 2 without going over:

2004 = 1001·(2) + 101·B + 11·C + 2·D

2004 - 2002 = 101·B + 11·C + 2·D

2 = 101·B + 11·C + 2·D

B and C can clearly not be bigger than 0 here (and they can't be negative since they are just digits of a bigger number):

2 = 101·(0) + 11·(0) + 2·D

2 = 2·D

1 = D

----

The answer is 2001.

2001 + ( 2 + 0 + 0 + 1 ) = 2004