You have how much left?

If the item costs between $N/2$ and $N$ dollars, then you can buy only one item. These extremes will produce remainders of $N/2$ and $0$ , respectively. The average amount of money left over in this region, which has length $N(1 - 1/2) = N/2$ , is therefore $N/4.$

Likewise, if the item costs between $N/3$ and $N/2$ dollars, then you can buy only two items. These extremes will produce remainders of $N/3$ and $0$ , respectively. The average amount of money left over in this region, which has length $N(1/2 - 1/3) = N/6$ , is therefore $N/6$ .

Continuing in this manner, we see that if the item costs between $N/(n + 1)$ and $N/n$ , then you can buy only $n$ items. These extremes will produce remainders of $N/(n+ 1)$ and $0$ , respectively. The average amount of money left over in this region, which has length $N(1/n - 1/(n + 1)) = N/n(n + 1)$ , is therefore $N/2(n + 1)$ .

The expected amount, $M$ , of money left over is therefore:

$M = \displaystyle\sum_{n=0}^\infty((1/n - 1/(n+1)) \times (N/(2(n+1))))$

which equals $N \times (1 - (\pi^2/12)) = 0.18N$

The cost of the item is a random variable $X$ that is uniformly distributed over $(0,N)$ , and we want the expectation of the change random variable $C \; = \; N - \left\lfloor \tfrac{N}{X}\right\rfloor X$ Thus $\begin{aligned} \mathbb{E}[C] & = \frac{1}{N}\int_0^N \left(N - \left\lfloor \tfrac{N}{x}\right\rfloor x\right)\,dx \; = \; N - \tfrac{1}{N} \sum_{j=1}^\infty \int_{\frac{N}{j+1}}^{\frac{N}{j}} \left\lfloor \tfrac{N}{x}\right\rfloor x\,dx \\ & = N - \frac{1}{N} \sum_{j=1}^\infty j \int_{\frac{N}{j+1}}^{\frac{N}{j}} x\,dx \; = \; N - \frac{1}{2N} \sum_{j=1}^\infty j\left(\frac{N^2}{j^2} - \frac{N^2}{(j+1)^2}\right) \\ & = N - \frac{N}{2}\sum_{j=1}^\infty\left(\frac{1}{j} - \frac{1}{j+1} + \frac{1}{(j+1)^2}\right) \; = \; N - \tfrac12N\big(1 + (\zeta(2)-1)\big) \\ & = \big(1- \frac{1}{12}\pi^2\big)N \; \approx \; 0.1775N \end{aligned}$ making the answer $\boxed{0.178N}$ .