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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>
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The only case above is <math>(27)(1332)</math> which will give all different digits and the sum of the digits equal to the sum of all of the three-digit numbers.
  
 
Therefore, <math>\sum_{i=1}^{5}d_i=27</math> and <math>N=35964</math>
 
Therefore, <math>\sum_{i=1}^{5}d_i=27</math> and <math>N=35964</math>
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 +
* Note.  I actually competed at this event in Argentina when I was in High School representing Puerto Rico.  At the competition, I messed up one of my calculations and multiplied all the numbers by 1331 instead of 1332.  So, I got partial points.
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~Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}
  
 
== See also ==
 
== See also ==
 +
 +
[[OIM Problems and Solutions]]
 +
 
https://www.oma.org.ar/enunciados/ibe6.htm
 
https://www.oma.org.ar/enunciados/ibe6.htm

Latest revision as of 08:40, 23 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 $15 \le \sum_{i=1}^{5}d_i \le 35$

$\begin{cases} (15)(1332)=19980\text{; } 1+9+9+8+0=27\text{; } 27\ne 15; & \text{NO}\\ (16)(1332)=21312\text{; } 2+1+3+1+2=9\text{; } 9\ne 16; & \text{NO}\\ (17)(1332)=22644\text{; } 2+2+6+4+4=18\text{; } 18\ne 17; & \text{NO}\\ (18)(1332)=23976\text{; } 2+3+9+7+6=27\text{; } 27\ne 18; & \text{NO}\\ (19)(1332)=25308\text{; } 2+5+3+0+8=18\text{; } 18\ne 19; & \text{NO}\\ (20)(1332)=26640\text{; } 2+6+6+4+0=18\text{; } 18\ne 20; & \text{NO}\\ (21)(1332)=27972\text{; } 2+7+9+7+2=27\text{; } 27\ne 21; & \text{NO}\\ (22)(1332)=29304\text{; } 2+9+3+0+4=18\text{; } 18\ne 22; & \text{NO}\\ (23)(1332)=30636\text{; } 3+0+6+3+6=18\text{; } 18\ne 23; & \text{NO}\\ (24)(1332)=31968\text{; } 3+1+9+6+8=27\text{; } 27\ne 24; & \text{NO}\\ (25)(1332)=33300\text{; } 3+3+3+0+0=9\text{; } 9\ne 25; & \text{NO}\\ (26)(1332)=34632\text{; } 3+4+6+3+2=18\text{; } 18\ne 26; & \text{NO}\\ (27)(1332)=35964\text{; } 3+5+9+6+4=27\text{; } 27=27; & \textbf{YES}\\ (28)(1332)=37296\text{; } 3+7+2+9+6=27\text{; } 27\ne 28; & \text{NO}\\ (29)(1332)=38628\text{; } 3+8+6+2+8=27\text{; } 27\ne 29; & \text{NO}\\ (30)(1332)=39960\text{; } 3+9+9+6+0=27\text{; } 27\ne 30; & \text{NO}\\ (31)(1332)=41292\text{; } 4+1+2+9+2=18\text{; } 18\ne 31; & \text{NO}\\ (32)(1332)=42624\text{; } 4+2+6+2+4=18\text{; } 18\ne 32; & \text{NO}\\ (33)(1332)=43956\text{; } 4+3+9+5+6=27\text{; } 27\ne 33; & \text{NO}\\ (34)(1332)=45288\text{; } 4+5+2+8+8=27\text{; } 27\ne 34; & \text{NO}\\ (35)(1332)=46620\text{; } 4+6+6+2+0=18\text{; } 18\ne 35; & \text{NO} \end{cases}$

The only case above is $(27)(1332)$ which will give all different digits and the sum of the digits equal to the sum of all of the three-digit numbers.

Therefore, $\sum_{i=1}^{5}d_i=27$ and $N=35964$

  • Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. At the competition, I messed up one of my calculations and multiplied all the numbers by 1331 instead of 1332. So, I got partial points.

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

OIM Problems and Solutions

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

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