2001 IMO Shortlist Problems/N1
Problem
Prove that there is no positive integer 
such that, for 
, the leftmost digit (in decimal notation) of 
equals 
.
Solution
Suppose that there is such a number . Let 
be the number of digits of 
, and let 
be the number of digits of 
. Then we have:
and
Combining these inequalities via multiplication we get:
From this inequality, it can be seen that 
has at least 
and at most 
digits. However, if it has digits then the first digit is less than 
, which contradicts with how is defined. Thus, must have digits. Expressing this as an inequality, we get:
Combining these values with the earlier bounds for 
via division, we get:
Suppose that 
. Then 
is between 
and 
. Thus, one of the values 
must be . Call that value 
. This means that the digits of 
are just the digits of 
, followed by 
zeroes. Thus the first digit of the two numbers are equal, which contradicts with how is defined.
This means that:
Let 
be the number of digits of 
. Then we have:
Combining these with the bounds for 
, 
and 
via multiplication we get:
So 
has 
digits, but is smaller than the smallest digit number with the first digit 4, which means that its first digit can't be 4. This contradicts with how is defined, thus no such value of exists.
