Write $(-5)^5\div5^3+3^{4}-6^{1}$ as an integer.
Recall that $(-a)^n=-a^n$ when $n$ is odd so we can rewrite $(-5)^5$ as $-5^5$. Because $a^m\div a^n=a^{m-n}$ for positive integers $m>n$, we get $$-5^5\div5^3=-5^{5-3} =-5^2.$$ Now we can rewrite the expression to get  \begin{align*}
(-5)^5\div5^3+3^{4}-6^{1}&=-5^2+3^4-6\\
&=-25+81-6\\
&=\boxed{50}.
\end{align*}
50
Level 2
Prealgebra
Problem: Write $(-5)^5\div5^3+3^{4}-6^{1}$ as an integer.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the expression, first evaluate each term separately. Start with $(-5)^5$. This is calculated as $(-5) \times (-5) \times (-5) \times (-5) \times (-5)$. The product of the first two $(-5)$ gives $25$, then multiplying by the next $(-5)$ gives $-125$. Continuing this process results in $-3125$. 

Next, calculate $5^3$. This is $5 \times 5 \times 5$, which equals $125$. 

Now, perform the division: $-3125 \div 125$. This simplifies to $-25$ because $-3125$ divided by $125$ equals $-25$. 

Next, evaluate $3^4$. This is $3 \times 3 \times 3 \times 3$, which equals $81$. 

Then, calculate $6^1$. This is simply $6$. 

Now, combine the results: $-25 + 81 - 6$. Start with $-25 + 81$, which equals $56$. Then subtract $6$ from $56$, resulting in $50$. 

Thus, the final answer is $50$.<|END_OF_LATENT|> The final answer is $50$.