Question

Find the sum of all numbers greater than 10000 formed by using digits 1, 3, 5, 7, 9, no digit being repeated in any number.

Answer 1

Hint: First of all find the total number of numbers satisfying the given conditions. Now fix 4 out of 5 digits each time and divide total numbers into 5 categories and then consider the place value of each digit of 5 categories to find the total sum of numbers.

Complete step by step answer: -
In this, we have to find the sum of all numbers greater than 10000 formed by using digits 1, 3, 5, 7, 9, no digit being repeated in any number. We know that any number greater than 10000 would be of 5 digits or more than 5 digits. Also, since we have only 1, 3, 5, 7, 9, and repetition is not allowed, we can make only 5 digit numbers and any number formed by these five digits would be greater than 10000.

Any of the five digits can occupy ten thousand places. Since repetition is not allowed any of the remaining 4 digits can occupy a thousand’s place. Similarly, any of the remaining 3 digits can occupy a hundred’s place. Now, ten’s and unit’s place can be occupied by any of the remaining two digits and last remaining digit respectively.
\[\dfrac{5}{{{10}^{4}}}\times \dfrac{4}{{{10}^{3}}}\times \dfrac{3}{{{10}^{2}}}\times \dfrac{2}{{{10}^{1}}}\times \dfrac{1}{{{10}^{0}}}=120\]
Therefore, the total numbers formed = 120.
Now, the number of ways for a fixed digit at the place of \[{{10}^{4}}\] is
\[\dfrac{{}}{{{10}^{4}}}\times \dfrac{4}{{{10}^{3}}}\times \dfrac{3}{{{10}^{2}}}\times \dfrac{2}{{{10}^{1}}}\times \dfrac{1}{{{10}^{0}}}=24\text{ ways}\]
The number of ways for a fixed digit at the place of \[{{10}^{3}}\] is
\[\dfrac{4}{{{10}^{4}}}\times \dfrac{{}}{{{10}^{3}}}\times \dfrac{3}{{{10}^{2}}}\times \dfrac{2}{{{10}^{1}}}\times \dfrac{1}{{{10}^{0}}}=24\text{ ways}\]
The number of ways for a fixed digit at the place of \[\text{1}{{\text{0}}^{2}}\] is
\[\dfrac{4}{{{10}^{4}}}\times \dfrac{3}{{{10}^{3}}}\times \dfrac{{}}{{{10}^{2}}}\times \dfrac{2}{{{10}^{1}}}\times \dfrac{1}{{{10}^{0}}}=24\text{ ways}\]
The number of ways for a fixed digit at the place of \[\text{1}{{\text{0}}^{1}}\] is
\[\dfrac{4}{{{10}^{4}}}\times \dfrac{3}{{{10}^{3}}}\times \dfrac{2}{{{10}^{2}}}\times \dfrac{{}}{{{10}^{1}}}\times \dfrac{1}{{{10}^{0}}}=24\text{ ways}\]
The number of ways for a fixed digit at the place of \[\text{1}{{\text{0}}^{0}}\] is
\[\dfrac{4}{{{10}^{4}}}\times \dfrac{3}{{{10}^{3}}}\times \dfrac{2}{{{10}^{2}}}\times \dfrac{1}{{{10}^{1}}}\times \dfrac{{}}{{}}\text{ }=24\text{ ways}\]
As shown, every fixed digit placed at the place of \[{{10}^{4}}\], we can fill other digits at other places in 24 ways without repetition.
Hence, the sum of the place values of \[{{10}^{4}}=\left( 1+3+5+7+9 \right)\times 24\times {{10}^{4}}....\left( i \right)\]
Similarly, the sum of the place values of \[{{10}^{3}}=\left( 1+3+5+7+9 \right)\times 24\times {{10}^{3}}....\left( ii \right)\]
Similarly, the sum of the place values of \[{{10}^{2}}=\left( 1+3+5+7+9 \right)\times 24\times {{10}^{2}}....\left( iii \right)\]
Similarly, the sum of the place values of \[{{10}^{1}}=\left( 1+3+5+7+9 \right)\times 24\times {{10}^{1}}....\left( iv \right)\]
Similarly, the sum of the place values of \[{{10}^{0}}=\left( 1+3+5+7+9 \right)\times 24\times {{10}^{0}}....\left( v \right)\]
Sum of all numbers = (i) + (ii) + (iii) + (iv) + (v)
\[=25\times 24\times {{10}^{4}}+25\times 24\times {{10}^{3}}+25\times 24\times {{10}^{2}}+25\times 24\times {{10}^{1}}+25\times 24\times {{10}^{0}}\]
\[=\left( 25\times 24 \right)\left[ {{10}^{4}}+{{10}^{3}}+{{10}^{2}}+{{10}^{1}}+{{10}^{0}} \right]\]
\[=\left( 25\times 24 \right)\times \left( 11111 \right)\]
\[=6666600\]
Therefore, we get the sum of total numbers greater than 10000 formed by using digits 1, 3, 5, 7, 9 without repetition is 6666600.

Note: Here, some students randomly select the numbers and try to add them which is very lengthy as there are 120 numbers and can lead to mistakes. Also, some students do not consider the place value but in questions such as addition, subtraction, etc. of numbers, place value method is most promising.