Difference between revisions of "1991 OIM Problems/Problem 4"
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<math>\begin{cases}(15)(1332)=19980; 1+9+9+8+0=27; 27 \ne 15; \text{NO} | <math>\begin{cases}(15)(1332)=19980; 1+9+9+8+0=27; 27 \ne 15; \text{NO} | ||
| − | + | \end{cases}</math> | |
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 22:31, 13 December 2023
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 
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
