Tax deduction

[This feels like a very crude approach; hopefully someone else can post a more elegant way to find the answer.]

We are looking for the lowest cost, call it $x$ , such that when we calculate 6%, 6.1% and 6.2% of $x$ , the amounts rounded to the nearest cent would all be different, and since we are trying to minimize the cost, it stands to reason that the rounded amounts of cents would be consecutive whole numbers.

To simplify the calculations a little, we shall assume that $x$ is the cost in cents rather than in dollars. Then we are looking for $x, n \in \mathbb{N}$ such that Round $(0.06x) = n-1$ , Round $(0.061x) = n$ and Round $(0.062x) = n+1$ .

The three unrounded amounts are evenly spaced, separated by $0.001x$ ; to minimize $x$ , we need to minimize these gaps. This can be achieved by finding $x$ such that $0.061x$ is "very" close to $n$ , while $0.06x$ and $0.062x$ are "slightly" below $n - 0.5$ and above $n + 0.5$ respectively, as shown below.

So we need $\ 0.001x > 0.5 \$ , i.e. $\ x > 500$ . Since $0.061(500) = 30.5$ , the tax amount $n$ must be at least $31$ cents. The cost that would give that exact tax amount is $\approx 508.20$ cents, so we check what happens at $x = 508$ cents:

$\begin{array}{rlcrl} 6 \%: & 508(0.06) = 30.48 & \implies & & n = 30 \text{cents} \\ 6.1 \%: & 508(0.061) = 30.988 & \implies & & n = 31 \text{cents} \\ 6.2 \%: & 508(0.062) = 31.496 & \implies & & n = 31 \text{cents} \end{array}$

The two higher tax rates give the same rounded tax amount, so this is not our solution. We try $n = 32$ cents, but the cost for this tax amount is $\approx 524.59$ cents which is not a good sign as it's further from a whole number cost than our previous attempt; and indeed this again does not yield three different rounded tax amounts. However, when $n = 33$ cents, we find that the cost that would give that exact tax amount is $\approx 540.98$ cents, which is promising as it's very close to a whole amount; we try $x = 541$ cents:

$\begin{array}{rlcrl} 6 \%: & 541(0.06) = 32.46 & \implies & & n = 32 \text{cents} \\ 6.1 \%: & 541(0.061) = 33.001 & \implies & & n = 33 \text{cents} \\ 6.2 \%: & 541(0.062) = 33.542 & \implies & & n = 34 \text{cents} \end{array}$

which gives us the required three different rounded tax amounts. So the cost we're looking for is $541$ cents, i.e. $\boxed{\$5.41}$

In order to be able to determine a tax rate of $6.1\%$ assuming we know it has the form $\overline{a.b}\%$ , we simply need to be able to differentiate it from $6.0\%$ and $6.2\%$ . Writing the item's cost $C$ in cents and the tax $T$ also in cents, we have $0.06C < T - 0.5 \leq 0.061C < T + 0.5 \leq 0.062C$ and solving for $C$ yields $\max\left\{\frac{1000T-500}{61}, \frac{1000T+500}{62}\right\} \leq C < \min\left\{\frac{1000T-500}{60}, \frac{1000T+500}{61}\right\}$

For $T \leq 60$ , this inequality becomes $\frac{1000T+500}{62} \leq C < \frac{1000T-500}{60}$ which implies $\frac{1000T+500}{62} < \frac{1000T-500}{60} \iff T > 30.5 \iff T \geq 31,$ so we will define $\Delta = T-31$ and rewrite the inequality in terms of $\Delta$ using $\frac{1000(\Delta+31) + 500}{62} = 16\Delta + 508 + \frac{4\Delta + 2}{31} \\ \frac{1000(\Delta+31) - 500}{60} = 16\Delta + 508 + \frac{2\Delta+1}{3}.$ Therefore, noting $M := C - 16\Delta - 508$ is an integer, we have for $0 \leq \Delta \leq 29$ , $\frac{4\Delta + 2}{31} \leq M < \frac{2\Delta+1}{3}$

Now, minimizing $\Delta$ will minimize $T$ and therefore minimize $C$ , so we want to find the smallest integer $\Delta$ such that $0 \leq \Delta \leq 29$ and $\left[\frac{4\Delta+2}{31}, \frac{2\Delta+1}{3}\right) \cap \mathbb{Z} \neq \emptyset$

By direct observation, the smallest such solution is $\Delta = 2$ , for which $M=1$ . Therefore $C = 16\Delta + 508 + M = 16(2) + 508 + 1 = 541$ which in dollars gives an answer of $\boxed{5.41}$ .