Express $213_{8}-142_{7}$ as a base 10 integer.
We have that \begin{align*} 213_{8} &= 2(8^2)+ 1(8^1) +3(8^0) \\
&= 2(64)+1(8)+3(1)\\
&= 128 + 8 + 3\\
&= 139\\
142_{7} &= 1(7^2)+ 4(7^1) +2(7^0) \\
&= 1(49)+4(7)+2(1)\\
&= 49 + 28 + 2\\
&= 79
\end{align*}So, $213_{8}-142_{7}=139-79=\boxed{60}$.
60
Level 3
Number Theory
Problem: Express $213_{8}-142_{7}$ as a base 10 integer.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, first convert both numbers from their respective bases to base 10. 

For $213_{8}$, calculate its base 10 equivalent:
- The digit 2 in the 8's place contributes $2 \times 8^2 = 2 \times 64 = 128$.
- The digit 1 in the 8's place contributes $1 \times 8^1 = 1 \times 8 = 8$.
- The digit 3 in the 8's place contributes $3 \times 8^0 = 3 \times 1 = 3$.
- Summing these values gives $128 + 8 + 3 = 139$.

Next, convert $142_{7}$ to base 10:
- The digit 1 in the 7's place contributes $1 \times 7^2 = 1 \times 49 = 49$.
- The digit 4 in the 7's place contributes $4 \times 7^1 = 4 \times 7 = 28$.
- The digit 2 in the 7's place contributes $2 \times 7^0 = 2 \times 1 = 2$.
- Summing these values gives $49 + 28 + 2 = 79$.

Now, perform the subtraction in base 10:
- Calculate $139 - 79$.
- This results in $60$.

Thus, the final answer is $60$.<|END_OF_LATENT|> The final answer is $60$.