Difference between revisions of "1991 OIM Problems/Problem 4"
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(35)(1332)=46620\text{; } 4+6+6+2+0=18\text{; } 18\ne 35; & \text{NO} | (35)(1332)=46620\text{; } 4+6+6+2+0=18\text{; } 18\ne 35; & \text{NO} | ||
\end{cases}</math> | \end{cases}</math> | ||
+ | |||
+ | Therefore, <math>\sum_{i=1}^{5}d_i=27</math> and <math>N=35964</math> | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 23:36, 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
Therefore, and
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.