# Analytical solution

• Mar 3rd 2010, 07:53 AM
Pinsky
Analytical solution
Hy!

My younger brother got an assignment and asked me for help. The problem goes:

Find all the four digit numbers whose sum of all digits lies on the interval <12,16>.

It's hardly a problem to solve it numerically, but does anybody know an analytical solution?
• Mar 3rd 2010, 06:21 PM
Soroban
Hello, Pinsky!

What idiot assigned this problem?

Quote:

My younger brother got an assignment and asked me for help.

Find all the four-digit numbers whose sum of all digits lies on the interval <12,16>.

It's hardly a problem to solve it numerically, but does anybody know an analytical solution?

This is a truly stupid problem at any level of mathematics!

It does not ask "how many" . . . it says Find.
. . It wants a list of the numbers!

For a digital sum of 12, there are over 300 such numbers.
Does the teacher want to see all of them?

• Mar 3rd 2010, 10:17 PM
Pinsky
It's a practice test for a math competition on a regional level for the fourth grade.
My first opinion was that it's kind of lame, but i thought that there is an analytical solution but i just don't see it.

If we would change the "find" with a "how many". What would the solution be then?
How to combine the condition that says (let's just take one number instead of an intevral)

$\displaystyle a+b+c+d=12$

with the permutation formula (this is for all the possible permutations) :

$\displaystyle 9\cdot10\cdot10\cdot10$

(this is just for my curiosity :) )

Tnx
• Mar 3rd 2010, 10:27 PM
Bacterius
Hello,
I do not know about an analytical solution, but I can give you the exact amount of such numbers :

For a sum of 12, there are 415 such numbers.
For a sum of 13, there are 480 such numbers.
For a sum of 14, there are 540 such numbers.
For a sum of 15, there are 592 such numbers.
For a sum of 16, there are 633 such numbers.

Which gives a total of 2660 numbers to list. Boring, and useless ...

First (perhaps) step to an analytical solution, note that the number of solutions increases with the sum (which is normal since there are more possible combinations).

PS : what kind of assignment is that o_O'