and
have
different first digits. Let n > 3 and first digits of
and
coincide
and are a. ThenSolution :
At n = 1, 2, 3 the numbers
and
have
different first digits. Let n > 3 and first digits of
and
coincide
and are a. Then
for some natural p and q. The multiplication of these inequalities yields
Since 
,
it follows that 
.
Therefore, 
is
the only possibility. Since 
and
,
.