Numbers that are averages of their digits

Question

A number is called special if its decimal value^ is the average of its digits. How many special numbers are there? Bonus: How many special numbers do not begin with the digit 0? Good luck!

^ You are allowed to append zeroes at the end. For example "1" is a special number, while "1.0" is not.

Answer 1

For the bonus, single-digit integers are trivial.

Rational numbers greater than 1, with trailing zeros, include:

1.1250000 = 9/8
1.500 = 6/4
1.687500000000000 = 27/16
1.8000 = 9/5
2.250 = 9/4
3.750 = 15/4

And this one is very special, having no trailing zeros.

4.5 = 9/2

Answer 2

There

are an infinite number of special numbers,

because

0.1...0 (total 10 digits), 0.01...0 (total 100 digits), and so on all work.

For the bonus part, I found

1, 2, ..., 9, and...

If you allow infinite decimals, we have 8.999... recurring.

Answer 3

There are

an infinite number of special numbers.

For example,

0, 0.0, 0.00, ...

Answer 4

I wanted to try my hand at it from a principled way, by creating an equation for it.

The simple case of single digits is obvious, so let's start our equations with 2 digits of interest: ie, we are looking for (a, b, n) such that a.b0... = (a + b) / n.

a + b / 10 = (a + b) / n
<=> n . a + n . b / 10 = a + b
<=> (n - 1) . a = (1 - n / 10) . b
<=> a = b . (1 - n / 10) / (n - 1)

Now, we can attempt to solve the equation for various values of n:

We restrict ourselves to a and b in [1, 2, 3, 4, 5, 6, 7, 8, 9], as a = 0 or b = 0 are trivial cases.

- For n = 2 we get a = b . 4 / 5, and thus a = 4 and b = 5, that is: 4.5 = 9 / 2.
- For n = 3 we get a = b . 7 / 20, and no value of b will make up that 20.
- For n = 4 we get a = b / 5, and thus a = 1 and b = 5, that is: 1.500 = 6 / 4.
- For n = 5 we get a = b / 8, and thus a = 1 and b = 8, that is: 1.8000 = 9 / 5.
- For n = 6 we get a = b . 2 / 25, and no value of b will make up that 25.
- For n = 7 we get a = b / 20, and no value of b will make up that 20.
- For n = 8 we get a = b / 35, and no value of b will make up that 35.

Given the fact that the denominator will only grow from there, and that the quotient b / a cannot exceed 9, there's no point in looking any further.

As a recap, we have gathered the following solutions:

4.5 = 9 / 2, 1.500 = 6 / 4, and 1.8000 = 9 / 5.

We can similarly proceed with (a, b, c, n) such at a.bc0... = (a + b + c) / n.

a + b / 10 + c / 100 = (a + b + c) / n
<=> n . a + n . b / 10 + n . c / 100 = a + b + c
<=> (n - 1) . a = (1 - n / 10) . b + (1 - n / 100) . c
<=> a = (b . (1 - n / 10) + c . (1 - n / 100)) / (n - 1)

Which we can once again attempt to solve for various values of n:

We restrict ourselves to: - a and c being non-zero digits, otherwise we fall back to the previous study. - while b can be any digit. And from there: - For n = 3 we get a = (b . 70 + c . 97) / 200, and just looking at the digits of the numerator tells us there's no solution with c non-zero. - For n = 4 we get a = (b . 15 + c . 24) / 75, once again, looking at the digits of the numerator, we notice that there's no solution with c != 5. With c = 5, we find a = 2 and b = 2, that is: 2.250 = 9 / 4, as well as a = 3 and b = 7, that is 3.750 = 15 / 4.
- For n = 5 we get a = (b . 10 + c . 19) / 80, and just looking at the digits of the numerator tells us there's no solution with c non-zero.
- For n = 6 we get a = (b . 20 + c . 47) / 250, nope.
- For n = 7 we get a = (b . 10 + c . 31) / 200, nope.
- For n = 8 we get a = (b . 5 + c . 23) / 175, no solution with c != 5, and the denominator is too large for b to make it up when c = 5.
- For n = 9 we get a = (b . 10 + c . 91) / 800, nope. - For n = 10 we get a = c / 10, and the denominator has grown too large for c to make up.

As a recap, we have gathered the following solutions:

2.250 = 9 / 4, and 3.750 = 15 / 4.

Observation:

It seems that in a multi-digits solution, there cannot be a solution in which the smallest non-zero digit is not 5, as otherwise it doesn't add-up cleanly with the previous smallest digit.

Automating the search with Python does not yield any more results beyond:

- For 4 digits, we find for n = 8, a = 1, b = 1, c = 2, and d = 5, or 1.1250000 = 9 / 8.
- For 5 digits, we find 1.6875 = 27 / 16.

And I do wish I had a more principled explanation for this.

---

Program:

    from fractions import Fraction


    def solve_rec(n, n_digits):
        def rec(n, den, fac, digits, acc):
            if den == fac:
                denominator = den * (n - 1)


                if acc <= 0 or acc % denominator != 0:
                    return


                a = acc // denominator


                digits = [a] + digits


                print("n = ", n, ", list = ", digits, ", fraction =", sum(digits), "/", n)
                return


            numerator = den - fac * n


            for digit in range(0, 10):
                new_acc = acc + digit * numerator


                if new_acc % (fac * 10) != 0:
                    continue


                rec(n, den, fac * 10, [digit] + digits, new_acc)


        last_numerator = 10 ** n_digits - n


        for last in range(1, 10):
            accumulator = last * last_numerator


            if accumulator % 10 != 0:
                continue


            rec(n, 10 ** n_digits, 10, [last], accumulator)




    if name == 'main':
        import sys


        limit = int(sys.argv[1])


        for n_digits in range(1, 10):
            for n in range(n_digits + 1, limit + 1):
                solve_rec(n, n_digits)