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December 5, 2008

Find the Number

There is a 10 digit number whose first digit is the no:
of occurence of 0's in the number. Second number is
no: of occurence of 1's in the number. Third number
is the no: of occurence of 2's in the number. 4'th no:
is the number of occurence of 3's in the number. and so
on up to 10th digit is no of 9's in the number.
what is the number?

4 comments:

Unknown said...

6210001000

December 7, 2008 at 1:06 AM
Anirudh Tiwari said...

I guess there is no single answer to this as 9000000000 also satisfies the anser.
8100000000 also satisfies it . So u can make hell lot of numbers by this rule.

December 11, 2008 at 11:49 AM
Anonymous said...

6210001000 is the only answer.

9000000000 has 1 '9' so there should be a '1' in the last position.

Similarly, 8100000000 has an '8' so there should be a '1' in the 2nd last position.

December 11, 2008 at 9:05 PM
Anonymous said...

if it is a single 10 digit number then no solution.. if all the occurrences are happening in different number so it ,easy, happening

October 27, 2011 at 6:41 PM

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