1991 OIM Problems/Problem 4
Problem
Find a number of five different and non-zero digits, which is equal to the sum of all the numbers of three different digits that can be formed with five digits of
.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Let or in a better format:
The total number of combinations is given the following way:
For the first digit of any three-digit number we have 5 numbers to chose from.
For the second digit we have 4 numbers to chose from.
For the third digit we have 3 numbers to chose from.
Total numbers of three digit numbers is (5)(4)(3)=60.
Now we need to find their sum.
From all 60 ways, in the first digit we will have each digit of N showing with (4)(3)=12 configurations.
From all 60 ways, in the second digit we will have each digit of N showing with (4)(3)=12 configurations.
From all 60 ways, in the last digit we will have each digit of N showing with (4)(3)=12 configurations.
Therefore the sum, since each digit of is shown in each position 12 times, then
Since , then
The only case above is which will give all different digits and the sum of the digits equal to the sum of all of the three-digit numbers.
Therefore, and
- Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. At the competition, I messed up one of my calculations and multiplied all the numbers by 1331 instead of 1332. So, I got partial points.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.