Difference between revisions of "1991 OIM Problems/Problem 4"
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Since <math>12345\le N \le 98765</math>, then <math>15 \le \sum_{i=1}^{5}d_i \le 35</math> | Since <math>12345\le N \le 98765</math>, then <math>15 \le \sum_{i=1}^{5}d_i \le 35</math> | ||
− | < | + | <math>\begin{cases} (15)(1332)=19980; & 1+9+9+8+0=27; & 27\ne 15; & \text{NO}\ |
(15)(1332)=19980; & 1+9+9+8+0=27; & 27\ne 15; & \text{NO}\ | (15)(1332)=19980; & 1+9+9+8+0=27; & 27\ne 15; & \text{NO}\ | ||
(16)(1332)=21312; & 2+1+3+1+2=9; & 9\ne 16; & \text{NO}\ | (16)(1332)=21312; & 2+1+3+1+2=9; & 9\ne 16; & \text{NO}\ | ||
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(33)(1332)=43956; & 4+3+9+5+6=27; & 27\ne 33; & \text{NO}\ | (33)(1332)=43956; & 4+3+9+5+6=27; & 27\ne 33; & \text{NO}\ | ||
(34)(1332)=45288; & 4+5+2+8+8=27; & 27\ne 34; & \text{NO}\ | (34)(1332)=45288; & 4+5+2+8+8=27; & 27\ne 34; & \text{NO}\ | ||
− | (35)(1332)=46620; & 4+6+6+2+0=18; & 18\ne 35; & \text{NO} | + | (35)(1332)=46620; & 4+6+6+2+0=18; & 18\ne 35; & \text{NO} |
− | \end{cases}</ | + | \end{cases}</math> |
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 22:21, 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.