Difference between revisions of "1991 OIM Problems/Problem 4"
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== Solution == | == Solution == | ||
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| + | 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> | ||
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| + | The total number of combinations is given the following way: | ||
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| + | For the first digit of any three-digit number we have 5 numbers to chose from. | ||
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| + | For the second digit we have 4 numbers to chose from. | ||
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| + | For the third digit we have 3 numbers to chose from. | ||
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| + | Total numbers of three digit numbers is (5)(4)(3)=60. | ||
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| + | Now we need to find their sum. | ||
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| + | From all 60 ways, in the first digit we will have each digit of N showing with (4)(3)=12 configurations. | ||
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| + | From all 60 ways, in the second digit we will have each digit of N showing with (4)(3)=12 configurations. | ||
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| + | From all 60 ways, in the last digit we will have each digit of N showing with (4)(3)=12 configurations. | ||
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| + | 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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| + | {{alternate solutions}} | ||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe6.htm | https://www.oma.org.ar/enunciados/ibe6.htm | ||
Revision as of 23:02, 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 
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
