Difference between revisions of "1991 OIM Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | {{ | + | |
| + | Let <math>N=d_1d_2d_3d_4d_5</math> or in a better format: <math>N=10000d_1+1000d_2+100d_3+10d_4+d_5</math> | ||
| + | |||
| + | 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 <math>N</math> is shown in each position 12 times, then | ||
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| + | <math>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)</math> | ||
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| + | <math>S=1332 \sum_{i=1}^{5}d_i =N</math> | ||
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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> | ||
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| + | <math>\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}</math> | ||
| + | |||
| + | 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> | ||
| + | |||
| + | * 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}} | ||
== 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 
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 
The only case above is 
which will give all different digits and the sum of the digits equal to the sum of all of the three-digit numbers.
Therefore, 
and 
- 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.
