10. One day, Xiao Ming went shopping with 100 yuan. At the first store, he bought several items of $\mathrm{A}$, at the second store, he bought several items of $\mathrm{B}$, at the third store, he bought several items of $\mathrm{C}$, at the fourth store, he bought several items of $\mathrm{D}$, at the fifth store, he bought several items of $\mathbf{E}$, and at the sixth store, he bought several items of $\mathbf{F}$. The prices of the six items are all different and are whole numbers of yuan, and Xiao Ming spent the same amount of money in each of the six stores. Then, Xiao Ming has $\qquad$ yuan left.
【Analysis】Assuming that $\mathrm{x}$ yuan was spent in each store, according to the problem, $\mathrm{x}$ must have at least 6 factors.
Since $6=(5+1)=(2+1) \times(1+1)$, the smallest $\mathrm{x}$ is $2^{2} \times 3=12$,
and other cases where more money is spent would exceed the 100 yuan limit, so they do not need to be considered. Therefore, the remaining amount is $100-12 \times 6=28$ (yuan)