Suppose $656_7=3ab_{10}$, where $a$ and $b$ represent base-10 digits. Find $\frac{a\cdot b}{15}$.
I recall that a number in base 7 means that each digit represents a power of 7, starting from the rightmost digit as the zeroth power. So, $656_7$ means $6\cdot 7^2 + 5\cdot 7^1 + 6\cdot 7^0$. I can simplify this by multiplying out the powers of 7: $6\cdot 49 + 5\cdot 7 + 6\cdot 1$. This equals $294 + 35 + 6 = 335$. So, $656_7 = 335_{10}$. This means that $3ab_{10} = 335_{10}$. The leftmost digit is 3, so that matches. The middle digit is $a$, so $a$ must be 3 as well. The rightmost digit is $b$, so $b$ must be 5. I have found $a$ and $b$: $a=3$ and $b=5$. I can substitute the values of $a$ and $b$: $\frac{3\cdot 5}{15}$. This simplifies to $\frac{15}{15}$, which is 1.

