Problem (Number Theory):
A student computed the repeating decimal expansion of $\frac{1}{N}$ for some integer $N$, but inserted six extra digits into the repetend to get $.0 \overline{0231846597}$. Compute the value of $N$.

Solution:
Because the given repetend has ten digits, the original had four digits. If $\frac{1}{N}=.0 \underline{A} \underline{B} \underline{C} \underline{D}=$ $\frac{\underline{A} \underline{B} \underline{C} \underline{D}}{99990}$, then the numerator must divide $99990=10 \cdot 99 \cdot 101=2 \cdot 3^{2} \cdot 5 \cdot 11 \cdot 101$.

Note that all 3- and 4-digit multiples of 101 contain at least one digit which appears twice. Because the 10-digit string under the vinculum (i.e., 0231846597) contains no repeated digits, $\underline{A} \underline{B} \underline{C} \underline{D}$ cannot be a multiple of 101 . So $\underline{A} \underline{B} \underline{C} \underline{D}$ divides $2 \cdot 3^{2} \cdot 5 \cdot 11=990$. The only divisor of 990 that can be formed from four of the given digits (taken in order) is 0165, that is, 165. Hence $\frac{1}{N}=\frac{165}{99990}=\frac{1}{606} \Rightarrow N=\mathbf{6 0 6}$.

Final answer: 606