Three-fourths the square of a number
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1. Start the search among 2-digit numbers. A dropdown number (DDN) with 2 digits is a number such that
or
but both of these solutions violate the requirement that the digits are distinct, so there are no 2-digit DDNs.
A 3-digit DDN is such that
or if the average is
, or
if the average is
. The smallest possible value for
is 1 since we require 3 digits. Then
, and we can pick
and
to get the smallest DDN, 102.
2. In a 4-digit DDN , we have
or or
or
.
We're free to fix and
to try to get the smallest DDN. This leaves us with
or
or
or
.
The first two cases are impossible - the only choices for such that
are 1 and 2, and the sum of two positive integers must be positive. The smallest possible value of
is 2; this leaves us with
or
, but the latter case is impossible because 3 does not divide 2. So
, and the <em>smallest</em> 4-digit DDN is 1025.
To find the largest DDN, start with the largest possible values for and
. Let
and
. Then
or
or
or
. At most, we can have
with 7 and 6, so the first two cases are impossible. If we maximize
, then either
or
(which we don't want). So the <em>largest</em> 3-digit DDN is 9874.
3. I don't have an analytical solution to this, but using brute force (program) the total count is 112.
4. It is possible; consider 1249 and 1250, with digital averages
which happens to be the smallest pair. (Also found with brute force.)