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; | + | <math>\begin{cases} |
− | + | (15)(1332)=19980\text{; } 1+9+9+8+0=27\text{; } 27\ne 15; & \text{NO}\\ | |
− | (16)(1332)=21312; | + | (16)(1332)=21312\text{; } 2+1+3+1+2=9\text{; } 9\ne 16; & \text{NO}\\ |
− | (17)(1332)=22644; | + | (17)(1332)=22644\text{; } 2+2+6+4+4=18\text{; } 18\ne 17; & \text{NO}\\ |
− | (18)(1332)=23976; | + | (18)(1332)=23976\text{; } 2+3+9+7+6=27\text{; } 27\ne 18; & \text{NO}\\ |
− | (19)(1332)=25308; | + | (19)(1332)=25308\text{; } 2+5+3+0+8=18\text{; } 18\ne 19; & \text{NO}\\ |
− | (20)(1332)=26640; | + | (20)(1332)=26640\text{; } 2+6+6+4+0=18\text{; } 18\ne 20; & \text{NO}\\ |
− | (21)(1332)=27972; | + | (21)(1332)=27972\text{; } 2+7+9+7+2=27\text{; } 27\ne 21; & \text{NO}\\ |
− | (22)(1332)=29304; | + | (22)(1332)=29304\text{; } 2+9+3+0+4=18\text{; } 18\ne 22; & \text{NO}\\ |
− | (23)(1332)=30636; | + | (23)(1332)=30636\text{; } 3+0+6+3+6=18\text{; } 18\ne 23; & \text{NO}\\ |
− | (24)(1332)=31968; | + | (24)(1332)=31968\text{; } 3+1+9+6+8=27\text{; } 27\ne 24; & \text{NO}\\ |
− | (25)(1332)=33300; | + | (25)(1332)=33300\text{; } 3+3+3+0+0=9\text{; } 9\ne 25; & \text{NO}\\ |
− | (26)(1332)=34632; | + | (26)(1332)=34632\text{; } 3+4+6+3+2=18\text{; } 18\ne 26; & \text{NO}\\ |
− | (27)(1332)=35964; | + | (27)(1332)=35964\text{; } 3+5+9+6+4=27\text{; } 27=27; & \textbf{YES}\\ |
− | (28)(1332)=37296; | + | (28)(1332)=37296\text{; } 3+7+2+9+6=27\text{; } 27\ne 28; & \text{NO}\\ |
− | (29)(1332)=38628; | + | (29)(1332)=38628\text{; } 3+8+6+2+8=27\text{; } 27\ne 29; & \text{NO}\\ |
− | (30)(1332)=39960; | + | (30)(1332)=39960\text{; } 3+9+9+6+0=27\text{; } 27\ne 30; & \text{NO}\\ |
− | (31)(1332)=41292; | + | (31)(1332)=41292\text{; } 4+1+2+9+2=18\text{; } 18\ne 31; & \text{NO}\\ |
− | (32)(1332)=42624; | + | (32)(1332)=42624\text{; } 4+2+6+2+4=18\text{; } 18\ne 32; & \text{NO}\\ |
− | (33)(1332)=43956; | + | (33)(1332)=43956\text{; } 4+3+9+5+6=27\text{; } 27\ne 33; & \text{NO}\\ |
− | (34)(1332)=45288; | + | (34)(1332)=45288\text{; } 4+5+2+8+8=27\text{; } 27\ne 34; & \text{NO}\\ |
− | (35)(1332)=46620; | + | (35)(1332)=46620\text{; } 4+6+6+2+0=18\text{; } 18\ne 35; & \text{NO} |
\end{cases}</math> | \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}} | {{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 09:40, 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.