The product of the base seven numbers $24_7$ and $30_7$ is expressed in base seven. What is the base seven sum of the digits of this product?
We can ignore the $0$ digit for now, and find the product of $24_7 \times 3_7$. First, we need to multiply the units digit: $4_7 \times 3_7 = 12_{10} = 15_7$. Hence, we write down a $5$ and carry-over the $1$. Evaluating the next digit, we need to multiply $2_7 \times 3_7 + 1_7 = 7_{10} = 10_{7}$. Thus, the next digit is a $0$ and $1$ is carried over. Writing this out:  $$\begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c}
& & & & & \stackrel{1}{2} & \stackrel{}{4}_7 \\
& & & \times & & & 3_7 \\
\cline{4-7} & & & & 1 & 0 & 5_7 \\
\end{array}$$ We can ignore the $0$ in $30_7$, since it does not contribute to the sum. Thus, the answer is $1+0+5 = \boxed{6}$.

Notice that the base seven sum of the digits of a number leaves the same remainder upon division by $6$ as the number itself.
6
Level 4
Number Theory
Problem: The product of the base seven numbers $24_7$ and $30_7$ is expressed in base seven. What is the base seven sum of the digits of this product?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the product of the base seven numbers $24_7$ and $30_7$, first convert these numbers to base ten. 

For $24_7$: 
2 in the seven's place contributes $2 \times 7^1 = 14$.
4 in the unit's place contributes $4 \times 7^0 = 4$.
Thus, $24_7 = 14 + 4 = 18_{10}$.

For $30_7$: 
3 in the seven's place contributes $3 \times 7^1 = 21$.
0 in the unit's place contributes $0 \times 7^0 = 0$.
Thus, $30_7 = 21 + 0 = 21_{10}$.

Now, calculate the product in base ten:
$18_{10} \times 21_{10} = 378_{10}$.

Next, convert $378_{10}$ back to base seven. 
Determine the largest power of 7 less than 378:
$7^0 = 1$, $7^1 = 7$, $7^2 = 49$, $7^3 = 343$, $7^4 = 2401$ (too large).
The largest is $7^3 = 343$.

Calculate how many times $343$ fits into $378$:
$378 \div 343 = 1$ (remainder $35$).
So, the coefficient for $7^3$ is 1.

Next, consider $7^2 = 49$:
$35 \div 49 = 0$ (remainder $35$).
So, the coefficient for $7^2$ is 0.

Next, consider $7^1 = 7$:
$35 \div 7 = 5$ (remainder $0$).
So, the coefficient for $7^1$ is 5.

Finally, for $7^0 = 1$:
$0 \div 1 = 0$.
So, the coefficient for $7^0$ is 0.

Thus, $378_{10}$ converts to $1050_7$.

Now, find the sum of the digits of $1050_7$:
1 + 0 + 5 + 0 = 6.

The final answer is $6$.<|END_OF_LATENT|> The final answer is $6$.