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Difference between revisions of "1991 OIM Problems/Problem 4"
Line 34: Line 34:
 
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}\\
+
<math>\begin{cases} (15)(1332)=19980; & 1+9+9+8+0=27; & 27\ne 15; & \text{NO}\\(35)(1332)=46620; & 4+6+6+2+0=18; & 18\ne 35; & \text{NO}\end{cases}</math>
−
(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}</math>
 
  
  

Revision as of 22:23, 13 December 2023

Problem

Find a number $N$ 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 $N$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Let $N=d_1d_2d_3d_4d_5$ or in a better format: $N=10000d_1+1000d_2+100d_3+10d_4+d_5$

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 $N$ is shown in each position 12 times, then

$S=\left( 12\sum_{i=1}^{5}d_i \right)100+\left( 12\sum_{i=1}^{5}d_i \right)10+\left( 12\sum_{i=1}^{5}d_i \right)$

$S=1332 \sum_{i=1}^{5}d_i =N$

Since $12345\le N \le 98765$, then $15 \le \sum_{i=1}^{5}d_i \le 35$

$\begin{cases} (15)(1332)=19980; & 1+9+9+8+0=27; & 27\ne 15; & \text{NO}\\(35)(1332)=46620; & 4+6+6+2+0=18; & 18\ne 35; & \text{NO}\end{cases}$ (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.

See also

https://www.oma.org.ar/enunciados/ibe6.htm

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