 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Further Descending Integers (Posted on 2006-10-19) Consider all possible positive whole numbers less than 10,000,000 not containing the digit 2. These numbers are now arranged in descending order of magnitude.

What would be the 2,222,222nd number?

Now consider all possible positive whole numbers less than 10,000,000 not containing the digits 1 and 2. These numbers are now arranged in descending order of magnitude.

What would be the 1,212,121st number?

 See The Solution Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 2
We simply consider the set of numbers in base 9. Instead of the ten digits 0123456789 we have nine digits 013456789. So we only need to convert 2,222,222 into base nine (with digits 012345678), take the difference between that and 10 million base 9, and then upgrade all digits greater or equal to 2 up by one. Setting up powers:
1, 9, 81, 729, 6,561, 59,049, 531,441 (that's enough)

2,222,222 base 10 = 4,156,275 base 9

Therefore the number in base 9 is 10000000-4156275=
4732614

After upgrading the digits greater than/equal to 2 our number is:
5,843,715

Similarly, the 1,212,121st number of the decending order set of numbers less than 10 million not containing the digits 1 or 2 will be solved by converting 1,212,121 to base eight and upgrading each digit greater than or equal to one by two.

1,212,121 base 10 = 4,477,331 base 8

10,000,000 base 8 - 4,477,331 base 8  is:
3,300,447 base 8

After upgrading digits greater than or equal to 1 by two we have:
5,500,669

 Posted by Eric on 2006-10-19 16:12:58 Please log in:

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