20141108, 11:17  #1 
"Matthew Anderson"
Dec 2010
Oregon, USA
2^{4}·3^{2}·5 Posts 
10 digit number puzzle
Hi fun math people,
I got this one from www.mathisfun.com :) I'm sure it will be too easy for some of you. Find a 10digit number where the first digit is how many zeros in the number, the second digit is how many 1s in the number etc. until the tenth digit which is how many 9s in the number. Best of luck, Matt 
20141108, 14:09  #2 
Oct 2014
Bari, Italy
3·13 Posts 
6210001000?
It should be nice to find the general case. Last fiddled with by Luis on 20141108 at 14:30 
20141108, 14:17  #3 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
9,000,000,000
8,100,000,000 probably more than these I'm just not able to see them easily. 
20141108, 14:27  #4 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6,143 Posts 
This is an old puzzle.
But there is a simple function that can generate such numbers for any arbitrary integer base except 2, 3, 6 and 7. I think it would be more interesting to try and find the general function first and then get the answer for the OPs question in base 10. Last fiddled with by retina on 20141108 at 14:28 
20141108, 14:29  #5 
Oct 2014
Bari, Italy
3×13 Posts 

20141108, 15:45  #6  
Oct 2014
Bari, Italy
3·13 Posts 
Quote:
Quote:
digit_i, i=0, ... base1 digit_0 = base4 (it's not a number if base > 10) digit_1 = 2 digit_2 = 1 digit_(base4) = 1 digit_i = 0 otherwise A base 4 solution is 1210, but I can't find solutions for base 5... Last fiddled with by Luis on 20141108 at 15:48 

20141108, 16:03  #7  
"Forget I exist"
Jul 2009
Dumbassville
8384_{10} Posts 
Quote:
edit: doh forgot to change the first digit to 6 instead fo 9 so change one of the 1's, so 6,210,001,000 Last fiddled with by science_man_88 on 20141108 at 16:07 

20141108, 16:54  #8  
Oct 2014
Bari, Italy
3·13 Posts 
Quote:
9,000,000,000 wrong because you have 1 'nine', so the tenth digit should be 'one'. 9,000,000,001 wrong because there are not 9 'zero's, but only 8. 8,000,000,010 wrong, because you have 1 'one', so the second digit should be 'one'. If the second digit is 'one', you have 2 'one's. And so on... 6,210,001,000 is the only solution I found for base 10. I don't know if there are another ones. 

20141108, 19:21  #9 
May 2013
East. Always East.
11×157 Posts 
I wrote a fairly neat little Excel Spreadsheet for this one.
It's an iterative process where each cell represents one digit and the formula for the cell is the count of each digit. It started with 9,000,000,000 and iterated to 8,000,000,001 and eventually to 6,210,001,000. I next added a bit of code which changed the starting values to random numbers between 0 and 9 if it reached 6,210,010,000. The intention is to "reset" the game to a new start point automatically. The solution should "freeze" at another valid solution if it finds a new one since the "count" of digits will not change from iteration to iteration, but it will also not reset since it didn't reach 6,210,001,000. I have attached it for your enjoyment. You can notice the iterative process and then the change when it reaches the currently known solution. Excel 2010. It "SHOULD" default to manual calculation mode AND allowing iterative calculations, which means you must simply hold down the F9 button (unless you've changed your manual calculation button). Last fiddled with by TheMawn on 20141108 at 19:21 
20141108, 19:34  #10 
May 2013
East. Always East.
11×157 Posts 
For 9 digits: 521,001,000
For 8 digits: 42,101,000 For 7 digits: 3,211,000 For 4 digits: 1,210 There's a clear pattern in going from 9 digits down but once the pattern is impossible to fit there appear to be no solutions. I can't find any for 6 digits or 5. Four digits appears to be a special case in this regard. 
20141108, 19:39  #11 
May 2013
East. Always East.
1727_{10} Posts 
And sure enough:
B21,000,000,001,000 A2,100,000,001,000 9,210,000,001,000 821,000,001,000 72,100,001,000 Essentially, there's as many 0's as you want, a first digit indicating that number of zeros, and then a 1 for that digit. Then, you trigger an extra 1 for the number of 1's, giving you a 2 for the number of 1's, giving you a 1 for the number of 2's (which conveniently replaces the 1 for the number of 1's you had earlier. Last fiddled with by TheMawn on 20141108 at 19:44 
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