Assume that we are allowed to collect one number each day.
On the first day we are bound to get a number we do not already have.
On the second day we are almost certain; in fact there is a 9 in 10 chance.
So we allow 10/9 of a day for the second number.
Then on the third day, the chance is 8 in 10, so we allow 10/8 of a day and so on, until for the last number we need 10 days, giving the solution:
1*1+1*10/9+1*10/8+1*10/7+1*10/6+1*10/5+1*10/4+1*10/3+1*10/2+1*10/1, or about 29 days, i.e. digits.
Mathematically: Sigma (1 to n) X/n where X is the number of items being collected (e.g. here, 10) and n=X, giving n* Hn, or n times the nth harmonic number.
Edited on July 7, 2012, 2:56 am
Posted by broll
on 2012-07-06 11:34:42