Question

All digits in a dropdown number are different, and one of its digits is the average of all its digits. It has at least two digits. For example, 5021 is a dropdown number, but neither 4389 nor 6033 is a dropdown number.

Questions:
1. Find the smallest dropdown number.
2. What are the smallest and largest 4-digit dropdown numbers?
3. How many 3-digit dropdown numbers are there?
4. Is it possible for a pair of consecutive numbers to both be dropdown numbers? If not, explain why not. If it is possible, find the smallest such pair.

Answer 1

It is 9817 is the answer hope this help c3822

Answer 2

1. Start the search among 2-digit numbers. A dropdown number (DDN) with 2 digits is a number
`ab` such that

`\frac{a+b}2 = a \implies a + b = 2a \implies b = a`

or

`\frac{a+b}2 = b \implies a+b = 2b \implies a=b`

but both of these solutions violate the requirement that the digits are distinct, so there are no 2-digit DDNs.

A 3-digit DDN
`abc` is such that

`\frac{a+b+c}3 = a \implies a+b+c = 3a \implies b+c = 2a`

or
`a+c=2b` if the average is
`b`, or
`a+b=2c` if the average is
`c`. The smallest possible value for
`a` is 1 since we require 3 digits. Then
`b+c=2`, and we can pick
`b=0` and
`c=2` to get the smallest DDN, 102.

2. In a 4-digit DDN
`abcd`, we have

`\frac{a+b+c+d}4 = a \implies a + b + c + d = 4a \implies b+c+d=3a`

or
`a+c+d=3b` or
`a+b+d=3c` or
`a+b+c=3d`.

We're free to fix
`a=1` and
`b=0` to try to get the smallest DDN. This leaves us with
`c+d=3` or
`c+d=-1` or
`1+d=3c` or
`c=3d`.

The first two cases are impossible - the only choices for
`c,d` such that
`c+d=3` are 1 and 2, and the sum of two positive integers must be positive. The smallest possible value of
`c` is 2; this leaves us with
`1+d=6` or
`2=3d`, but the latter case is impossible because 3 does not divide 2. So
`d=5`, and the smallest 4-digit DDN is 1025.

To find the largest DDN, start with the largest possible values for
`a` and
`b`. Let
`a=9` and
`b=8`. Then
`c+d=19` or
`c+d=15` or
`17+d=3c` or
`17+c=3d`. At most, we can have
`c+d=13` with 7 and 6, so the first two cases are impossible. If we maximize
`c=7`, then either
`17+d=21\implies d=4` or
`24=3d\implies d=8` (which we don't want). So the largest 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

`\frac{1+2+4+9}4=4 \text{ and } \frac{1+2+5+0}4=2`

which happens to be the smallest pair. (Also found with brute force.)