Question

Problem for Mathletes:

Consider the positive integers that such that its digits are either \(1, 2,\) or \( 3\). Define a function \( f \) such that \( f(N) = \) the number of positive integers consisting of only \(1, 2 ,\) and \(3\) whose digits sum to \(N\). find \( f(10) \)

Answer 1

cool. trying to understand what the problem wants

Answer 2

NO HINTS PLEASE!

Answer 3

274

Answer 4

Not cool, zarkon! Don't explain your answer yet!

Answer 5

This problem is giving me a headache, I'm still missing some cases.

Answer 6

Especially the 7 digits case!

Answer 7

I was able to get 194 only :(

Answer 8

\[1+9+42+77+30+31+4\]
@Zarkon, if you could tell us how you worked it out, that would be great. Thank you!

Answer 9

Lets discuss it together. The maximum number of possible digits is 10, obviously:
# of digits # of combinations
10: 1,1,1,1,1,1,1,1,1,1 1
9: 1,1,1,1,1,1,1,1,2 9


5: 3,3,2,1,1 OR 2,2,2,2,2 30+1
4: 3,3,3,1 4
3: - -
These are probably the easy cases. I need to count for the 8, 7 and 6 digits.

Answer 10

use a generating function

Answer 11

Or you can use recurrence relation.

Answer 12

I used this method:
10x1 = 10
8x1 + 1x2 = C(9,1) = 9
7x1 + 1x3 = C(8,1) = 8
6x1 + 2x2 = C(8,2) = 28
5x1 + 1x3 + 1x2 = C(7,2) * C(2,1) = 21 * 2 = 42
4x1 + 2x3 = C(6,2) = 15
4x1 + 3x2 = C(7,3) = 35
3x1 + 1x3 + 2x2 = C(6,3) * C(3,1) = 20 * 3 = 60
2x1 + 2x3 + 1x2 = C(5,3) * C(3,1) = 10 * 3 = 30
2x1 + 4x2 = C(6,4) = 15
1x1 + 3x3 = C(4,3) = 4
1x1 + 1x3 + 3x2 = C(5,4) * C(4,1) = 5 * 4 = 20
2x3 + 2x2 = C(4,2) = 6
5x2 = 1

Total = 274