### Sample Problem

In the beginning, there are three positive natural numbers on the blackboard. Then we replace one of them with a number 1 less than the sum of the other two. Thus we proceed for certain rounds and the final numbers are 32, 45 and 76. Find out the original numbers with the least sum.

#### Solution

We work backwards by “undoing” the largest numbers until we reach the smallest sum:

1.      32, 45 and 76

First we start from the 76, the largest number, and replace it with a smaller number a:

Since 32 + a-1 = 45, a = 14;

2.      32, 45 and 14

Then we replace 45 with a smaller number:

14 + a-1 = 32, a = 19

3.      32, 19 and 14

Then we replace 32:

14 + a-1 = 19, a = 6

4.      6, 19 and 14

We replace 19 with 14 + 1-6 = 9

5.      6, 9 and 14

We replace 14 with 9 + 1-6 = 4

6.      6, 9 and 4

We replace 9 with 6 + 1-4 = 3

7.      6, 3 and 4

We replace 6 with 4 + 1-3 = 2

8.      2, 3 and 4

We replace 4 with 3 + 1-2 = 2

9.      2, 3 and 2

We replace 3 with 2 + 1-2 = 1

10. 2, 1 and 2

Since 2 + 1-1 = 2, one of the 2s can be replaced with a 1, the least positive natural number.

Thus the original numbers should be 1, 1 and 2, giving a sum of 4.