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?
6210001000
ReplyDeleteI guess there is no single answer to this as 9000000000 also satisfies the anser.
ReplyDelete8100000000 also satisfies it . So u can make hell lot of numbers by this rule.
6210001000 is the only answer.
ReplyDelete9000000000 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.
if it is a single 10 digit number then no solution.. if all the occurrences are happening in different number so it ,easy, happening
ReplyDelete