Chapter 177

Bollier Numbers

Riddler Classic

How many decimal numbers are there whose value equals the average of their digits? The digits of $2.3$ , for example, are $2$ and $3$ , with average $2.5$ , so it does not work. Two technical rules: no trailing zeros (so $2.250$ does not count, since it can be written $2.25$ ), and if the average is a repeating decimal it may be truncated to the original length, with standard rounding (so $5.66$ does not work because $(5+6+6)/3 = 5.6\overline{6}$ rounds up to $5.67$ ).

The Riddler, FiveThirtyEight, March 16, 2018(original post)

Following the puzzle setter, we call such numbers Bollier numbers, after the submitter Sam Bollier. The Riddler Express that week was an image-only shape-arithmetic puzzle that is not recoverable from the archived page, and is deferred.

Solution

There are infinitely many Bollier numbers. The cleanest demonstration combines two facts: a small explicit family with one digit, and an infinite family with arbitrarily many digits built from prime denominators.

Single-digit Bollier numbers. For any digit $d \in \{0, 1, \ldots, 9\}$ the number $d$ has one digit, whose average is $d$ . So all ten single-digit integers are Bollier numbers. That gives ten.

The repunit observation. Look at $4.5$ : digits are $4$ and $5$ , average $(4+5)/2 = 4.5$ . Bingo. Look at $3.8\overline{3}$ truncated to six digits, $3.83333$ : digit sum is $3 + 8 + 3 + 3 + 3 + 3 = 23$ , average $23/6 = 3.8\overline{3}$ , which truncates to $3.83333$ under the puzzle’s rule. Bingo again. So the family of Bollier numbers is not a curiosity of single digits; it contains many decimals.

The infinite family. Let $p > 5$ be a prime such that the decimal expansion of $1/p$ has a repetend of even length. Define $N_p \;=\; \tfrac{9}{2} - \tfrac{1}{2p} \;=\; \tfrac{9p - 1}{2p},$ truncated, with standard rounding, to $p$ decimal digits in total ( $1$ digit before the point, $p - 1$ after). We claim $N_p$ is a Bollier number.

Why. The number $9/2 - 1/(2p) = (9p - 1)/(2p)$ lies strictly between $4$ and $4.5$ , so its integer part is $4$ . Its expansion past the decimal point is eventually periodic with period equal to the period of $1/(2p)$ , which divides the period of $1/p$ (and equals it when $p$ is odd). Write the $p$ -digit truncated value as $4.d_1 d_2 \cdots d_{p-1}$ . We need $\frac{4 + d_1 + d_2 + \cdots + d_{p-1}}{p} \;=\; 4.d_1 d_2 \cdots d_{p-1}$ under the truncation rule. Equivalently, the average of all $p$ digits equals the displayed value to $p$ digits with standard rounding.

The reason this works for any prime $p > 5$ with even-length repetend is that $9p - 1$ is then exactly divisible by $2$ (since $p$ is odd, $9p - 1$ is even), and the average of the $p$ digits of $4. \underbrace{d_1 \cdots d_{p-1}}_{}$ is $(9p - 1)/(2p)$ by direct computation. Verifying that the truncation to $p$ digits with proper rounding agrees with the displayed value requires the periodic structure: even-length repetend ensures the rounding at digit $p$ is downward, leaving the displayed value unchanged.

Worked example, $p = 7$ . Then $N_7 = (9 \cdot 7 - 1)/(2 \cdot 7) = 62/14 = 31/7 = 4.\overline{428571}$ , truncated to $7$ digits as $4.428571$ . Digit sum $4 + 4 + 2 + 8 + 5 + 7 + 1 = 31$ ; average $31/7 = 4.428571\overline{428571}$ , which truncates to $4.428571$ under the rounding rule (the next digit is $4 < 5$ ). It matches.

Worked example, $p = 11$ . $N_{11} = (9 \cdot 11 - 1)/22 = 98/22 = 49/11 = 4.\overline{45}$ , truncated to $11$ digits as $4.4545454545$ . Digit sum $4 + 4+5+4+5+4+5+4+5+4+5 = 49$ ; average $49/11 = 4.4545\ldots$ , truncates to $4.4545454545$ . Matches.

Why "infinitely many". The construction produces a distinct Bollier number for each suitable prime $p$ . A classical result of Hasse (1965) on the distribution of orders of $10$ in $(\mathbb{Z}/p\mathbb{Z})^*$ gives that the density of primes $p$ for which the order of $10 \pmod p$ is even is $17/24 > 0$ . In particular there are infinitely many such primes, hence infinitely many Bollier numbers from this family alone. So, $\boxed{\,\text{There are infinitely many Bollier numbers.}\,}$

The computation

Verify the construction directly for the first several primes with even-length repetend. For each $p$ , compute $N_p = (9p - 1)/(2p)$ exactly as a fraction, truncate to $p$ digits with standard rounding, sum the digits, and check that their average rounds back to the same truncated value.

For odd primes $p > 5$ , compute the multiplicative order of $10$ modulo $p$ .
Keep only those with even order (even-length repetend of $1/p$ ).
Form $N_p$ as an exact Fraction, render to $p$ digits using Decimal with ROUND_HALF_UP.
Compute digit sum and average; compare the average’s $p$ -digit rounded form to $N_p$ .

from fractions import Fraction
from decimal import Decimal, getcontext, ROUND_HALF_UP

def order10(p):
    cur, k = 10 % p, 1
    while cur != 1:
        cur = (cur * 10) % p
        k += 1
    return k

def truncate(x, n):
    getcontext().prec = n + 40
    d = Decimal(x.numerator) / Decimal(x.denominator)
    q = Decimal(10) ** -(n - 1)
    return d.quantize(q, rounding=ROUND_HALF_UP)

def is_prime(n):
    if n < 2: return False
    for k in range(2, int(n**0.5) + 1):
        if n % k == 0: return False
    return True

count = 0
for p in range(7, 60):
    if not is_prime(p) or p in (2, 5): continue
    if order10(p) % 2: continue
    Np = Fraction(9 * p - 1, 2 * p)
    val = truncate(Np, p)
    digits = [int(c) for c in str(val) if c.isdigit()]
    avg = Fraction(sum(digits), p)
    avg_disp = truncate(avg, p)
    ok = (val == avg_disp)
    count += int(ok)
    print(f"p={p:3d} N_p={val} sumd={sum(digits):3d} match={ok}")
print(f"Bollier numbers produced so far: {count}")

The script prints "match=True" for $p = 7, 11, 13, 17, 19, 23, 29, 47, 59$ , the first nine primes in $[7, 60]$ with even-length repetend of $1/p$ . Each row is a freshly minted Bollier number. The same recipe applied to larger such primes produces arbitrarily many more, so the count is unbounded.