How many 4 digit numbers have their thousands's digit equal to the sum of the first three digits?

Vivekanand Vellanki (last edited 4 years ago) (Expert, Founder @Qalaxia)

Re-stating the question, given

Th, H, T, O where Th = H + T + O and Th > 0, how many such combinations exist? Also, all numbers are > 0 and < 10.

In how many ways can we find H, T, O such that their sum is < 10?

I see three cases:

Case AD: All H, T and O are different. One such combination will give 6 numbers.
Case 2S: Two of the numbers are same, while the third one is different. Each such combination gives 3 numbers (the number of places where the different number can be placed)
Case AS: All the numbers are same. Each such combination gives 1 number

The total number of such numbers:

6 * AD + 3 * 2S + AS

Estimating AS: This is easy. Since H, T, and O are same. The only possible choices and 1,1,1; 2,2,2 and 3,3,3. 4+4+4 > 10 and 0+0+0 = 0 (not > 0).

AS = 3.

Estimating 2S:

Lets start by picking the number that is different and finding out the number of choices.

After picking 0, there are 4 choices 1, 2, 3, and 4. The largest sum being 4*2 + 0 = 8.

After picking 1, there are 4 choices 0, 2, 3, and 4 (9)

After picking 2, there are 3 choices 0, 1, and 3 (8)

After picking 3, there are 3 choices 0, 1, and 2 (7)

After picking 4, there are 3 choices 0, 1, and 2 (8)

After picking 5, there are 3 choices: 0, 1 and 2 (9)

After picking 6, there are 2 choices: 0 and 1 (8)

After picking 7, there are 2 choices: 0 and 1 (9)

After picking 8, there is 1 choice: 0 (8)

After picking 9, there is 1 choice: 0 (9)

2S = 2*4 + 4*3 + 2*2 + 2*1 = 8 + 12 + 4 + 2 = 26

Estimating AD:

In this case, all the numbers have to be different:

The way we will pick numbers is that we will pick numbers in increasing order, that is, if 1 is picked, 0 or 1 cannot be picked after 1. This ensures that the same combination is not picked twice.

After picking 0 and 1, there are 7 choices (2, 3, 4, 5, 6, 7, and 8)

After picking 0 and 2, there are 5 choices (3, 4, 5, 6, and 7)

After picking 0 and 3, there are 3 choices (4, 5, and 6)

After picking 0 and 4, there is 1 choice (5)

Picking 0, there are 16 choices.

After picking 1 and 2, there are 4 choices (3, 4, 5 and 6)

After picking 1 and 3, there are 2 choices (4, and 5)

Picking 1, there are 6 choices

After picking 2 and 3, there is one choice (4)

Picking 2, there is 1 choice

If 3 is the first number picked, the smallest sum is 12. Hence, for 3 and above, there are no options.

AD = 16 + 6 + 1 = 23

Total possibilities = 6 * 23 + 3 * 26 + 3 = 219.