In a certain colony of bacteria, the number of bacteria doubles every day. The colony starts with 3 bacteria, and has 6 at the end of day 1, 12 at the end of day 2, and so on. What is the number of the first day which ends with the colony having more than 100 bacteria?
The number of bacteria is multiplied by 2 at the end of each day, so the number of bacteria at the end of day $n$ is $3\cdot2^n$. We want $3\cdot2^n > 100$, or $2^n > 33\frac{1}{3}$. The smallest $n$ for which this happens is $n = \boxed{6}$.
6
Level 2
Algebra
Problem: In a certain colony of bacteria, the number of bacteria doubles every day. The colony starts with 3 bacteria, and has 6 at the end of day 1, 12 at the end of day 2, and so on. What is the number of the first day which ends with the colony having more than 100 bacteria?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the first day when the number of bacteria exceeds 100, we need to analyze the growth pattern of the bacteria. The colony starts with 3 bacteria and doubles each day. 

On day 0, the number of bacteria is 3. 
On day 1, the number of bacteria is 3 * 2^1 = 6. 
On day 2, the number of bacteria is 3 * 2^2 = 12. 
On day 3, the number of bacteria is 3 * 2^3 = 24. 
On day 4, the number of bacteria is 3 * 2^4 = 48. 
On day 5, the number of bacteria is 3 * 2^5 = 96. 
On day 6, the number of bacteria is 3 * 2^6 = 192. 

We observe that on day 5, the total is 96, which is still less than 100. However, on day 6, the total reaches 192, which exceeds 100. 

Thus, the first day that ends with the colony having more than 100 bacteria is day 6.<|END_OF_LATENT|> The final answer is $6$.