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> | ||
| − | \begin{cases} (15)(1332)=19980; & 1+9+9+8+0=27; & 27\ne 15; & \text{NO}\\ | + | <cmath>\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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(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}</cmath> |
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 22:20, 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 
\[\begin{cases} (15)(1332)=19980; & 1+9+9+8+0=27; & 27\ne 15; & \text{NO}\\
(15)(1332)=19980; & 1+9+9+8+0=27; & 27\ne 15; & \text{NO}\\
(16)(1332)=21312; & 2+1+3+1+2=9; & 9\ne 16; & \text{NO}\\
(17)(1332)=22644; & 2+2+6+4+4=18; & 18\ne 17; & \text{NO}\\
(18)(1332)=23976; & 2+3+9+7+6=27; & 27\ne 18; & \text{NO}\\
(19)(1332)=25308; & 2+5+3+0+8=18; & 18\ne 19; & \text{NO}\\
(20)(1332)=26640; & 2+6+6+4+0=18; & 18\ne 20; & \text{NO}\\
(21)(1332)=27972; & 2+7+9+7+2=27; & 27\ne 21; & \text{NO}\\
(22)(1332)=29304; & 2+9+3+0+4=18; & 18\ne 22; & \text{NO}\\
(23)(1332)=30636; & 3+0+6+3+6=18; & 18\ne 23; & \text{NO}\\
(24)(1332)=31968; & 3+1+9+6+8=27; & 27\ne 24; & \text{NO}\\
(25)(1332)=33300; & 3+3+3+0+0=9; & 9\ne 25; & \text{NO}\\
(26)(1332)=34632; & 3+4+6+3+2=18; & 18\ne 26; & \text{NO}\\
(27)(1332)=35964; & 3+5+9+6+4=27; & 27=27; & \textbf{YES}\\
(28)(1332)=37296; & 3+7+2+9+6=27; & 27\ne 28; & \text{NO}\\
(29)(1332)=38628; & 3+8+6+2+8=27; & 27\ne 29; & \text{NO}\\
(30)(1332)=39960; & 3+9+9+6+0=27; & 27\ne 30; & \text{NO}\\
(31)(1332)=41292; & 4+1+2+9+2=18; & 18\ne 31; & \text{NO}\\
(32)(1332)=42624; & 4+2+6+2+4=18; & 18\ne 32; & \text{NO}\\
(33)(1332)=43956; & 4+3+9+5+6=27; & 27\ne 33; & \text{NO}\\
(34)(1332)=45288; & 4+5+2+8+8=27; & 27\ne 34; & \text{NO}\\
(35)(1332)=46620; & 4+6+6+2+0=18; & 18\ne 35; & \text{NO}\\
\end{cases}\] (Error compiling LaTeX. Unknown error_msg)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
