write a number that has ten digits
the first digit must tell u how many 0's there are in the number
the second digit must tell u how many 1's there are in the number
and so on, until the tenth tells u how many 9's there are in the number
2006-07-31 07:58:20 · 10 answers · asked by locomexican89 3 in Science & Mathematics ➔ Mathematics
ok that was quick, now a little tougher
9 digits, start with 1 and stop with 9
same rules
2006-07-31 08:09:04 · update #1
never mind all the answers are wrong so far, 900000000000 would need a 1 in the last digit for the nine at the beginning
2006-07-31 08:10:46 · update #2
6210001000
6 0's
2 1's
1 2
1 6
2006-07-31 08:40:35 · update #3
This is impossible.
Proof:
Starting at the 10th digit (the # of 9's place).
If you put anything in this spot greater than 0, you don't have enough spaces left in the number to accommodate it. Let's use 1 for example. If you put 1 in the 10th spot, you need a 9 somewhere else in the number. The one, plus the 9 is two digits, leaving only 8 digits left - not enough spaces left to place 9 of any digit. So the only possible number in the 9 spot is 0.
Moving on to the 8's spot.
Placing a 1 in the 8's spot, would mean an 8 would have to appear somewhere in the number, but there are only 7 digits left (we have a 0 in the 9's spot, 1 in the 8's spot, and an 8 somewhere else). Again, not enough room, so a 0 is the only option.
It continues like this until you reach the 0 spot, where 0 is the only option in any of the digits. You have 9 0's at that point, so 9 is the only option in the 0's spot - making the number 9000000000. But, as has been pointed out, this is wrong since a one would be needed in the last place for this to work.
No solution is therefore possible, QED.
2006-07-31 08:44:39 · answer #1 · answered by Will 6 · 1⤊ 0⤋
"Additional Details"
"ok that was quick, now a little tougher
9 digits, start with 1 and stop with 9
same rules"
For the 9-digit problem, wouldn't 100,000,000 work? There's one 1 and zero of each of the digits 2-9.
For the 10-digit problem, I see 6,210,001,000, but I haven't found any others, yet. I'm curious if there are any more.
2006-07-31 10:17:13 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Interesting, these self-describing numbers.
With nine digits from 1 through 9 no solution is possible. You can see that in different ways:
* because every digit is at least one, every digit should at least occur once. Therefore there must be at least one nine, but that would mean that there are nine the same digits in the number!
* there are 9 digits in total so the sum of the digits should be 9, which is only possible if they are all ones, but that is no solution either.
2006-07-31 09:49:23 · answer #3 · answered by dutch_prof 4 · 0⤊ 0⤋
9000000000
There can be only one 9 because if there were to be more than one the number would have more than 10 digits. The only other number that can come 9 times is 0. The 0s tell us that no other digit is present.
Hence the number is 9000000000.
2006-07-31 08:10:52 · answer #4 · answered by tuhinrao 3 · 0⤊ 0⤋
ok that was quick, now a little tougher
9 digits, start with 1 and stop with 9
same rules
100000000
A digit no where the 1st digit tells u how many 1's u have, the 0's tell u that there are no 2,3...9
2006-07-31 15:54:16 · answer #5 · answered by peaceharris 2 · 0⤊ 0⤋
9 digits start with 1 and end with 9
123456789
Number: 000000000
2006-07-31 11:23:42 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Nope, I give up on this one. What's the solution?
Really cool loco!!!
2006-07-31 08:04:18 · answer #7 · answered by Anonymous · 0⤊ 0⤋
6200001001
2006-07-31 08:10:39 · answer #8 · answered by RandomNormality 3 · 0⤊ 0⤋
ah... when you find out, tell me
2006-07-31 08:09:13 · answer #9 · answered by mommy_mommy_crappypants 4 · 0⤊ 0⤋
2601111110 : -)
2006-07-31 08:08:57 · answer #10 · answered by Anonymous · 0⤊ 0⤋