xxxi. Bachet's Weights

Problem. Which set of weights lets a balance measure every integer weight $\mathrm{1..}W$ with as few weights as possible?

Weights may go on	Optimal set	Covers
one pan only	$1,2,4,\dots ,2^{k-1}$	$\mathrm{1..}{2}^k-1$ (binary)
both pans	$1,3,9,\dots ,3^{k-1}$	$\mathrm{1..}\frac{3^k-1}{2}$ (balanced ternary): 4 weights cover $\mathrm{1..40}$

vector<int> balancedTernary(long long w) {        // w = sum d_i 3^i, d_i in {-1, 0, 1}: +1 = weight on the other pan, -1 = with the object
    vector<int> d;
    while (w) { int r = w % 3; if (r == 2) r = -1; d.push_back(r); w = (w - r) / 3; }
    return d;
}

Proof: balanced ternary covers exactly (3^k − 1)/2

With both pans, each weight contributes $-1$, $0$ or $+1$ times its value, so $k$ weights give at most $3^k$ signed sums, symmetric around $0$: at most $\frac{3^k-1}{2}$ positive values. Balanced ternary shows that the weights $3^i$ achieve every integer in $[-\frac{3^k-1}{2},\frac{3^k-1}{2}]$: the digit algorithm takes $r=w\mathrm{mod}3$, replaces $2$ by $-1$, and continues with $(w-r)/3$, which shrinks $|w|$ and terminates, and the representation is unique. Checked by reconstructing every $w\le 10^5$.