## Puzzle #124: Add the Numbers

This is a gem of a puzzle that my nephew, Karan, gave to me yesterday. He and I came up with different ways to come to the answers, and there are probably many more ways.

For all the possible 5 digit numbers with the digits 1, 3, 5, 7 and 9. Each of the digits can appear only once in a number and all of them need to appear in every number (therefore 11379 is not a valid numbers). What is the addition of all these numbers?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy counting!

This entry was posted in Puzzles and tagged , . Bookmark the permalink.

### 8 Responses to Puzzle #124: Add the Numbers

1. Batman says:

6666600

2. Sandeep says:

On first blush, Does every number get repeated 24 times in each addition column.Total of each vertical column is 25*24 ??

• Alok Goyal says:

Absolutely correct!

3. aditya says:

There are 120 (5!) possible numbers with each digit (1, 3, 5, 7, 9) repeating 24 times (120/5) in every position. So 24*25 (sum of all digits) * sum of place value of each position. Just add up and you get 6666600

• Alok Goyal says:

Perfect!

4. riteshbanglani says:

If you fix a number for one decimal position, the other numbers can be arranged in 4! ways.
The sum of all those numbers is 4! * (1+3+5+7+9)
If we do this for every decimal place, we have to add (10^4+10^3+10^2+10^1+10^0)
So the total sum is 4! * (1+3+5+7+9) * (10^4+10^3+10^2+10^1+10^0) = 6666600

• Alok Goyal says:

Awesome!

5. Pratik Poddar (@pratikpoddar) says:

Number of numbers with x at i_th position is 4! = 24 for all valid x and i

Summation of all numbers at i_th position = (1+3+5+7+9)*24*10^i

So the summation is (1+3+5+7+9)*24*(1+10+100+1000+10000) = 6666600