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Home » Después de 3 horas: \(200 \times 2 = 400\) bacterias. - AIKO, infinite ways to autonomy.

Después de 3 horas: \(200 \times 2 = 400\) bacterias. - AIKO, infinite ways to autonomy.

📅 May 5, 2026 👤 scraface

May 05, 2026

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📌 $y = \pm2$: $x^2 + 4(4) = 16 \Rightarrow x^2 = 0 \Rightarrow x = 0$ (2 points).

📌 Total lattice points: $2 + 2 = \boxed{4}$.

📌 Question: What two-digit number is three more than a multiple of 7 and five more than a multiple of 6?

📌 Solution: Let the number be $n$. Then $n \equiv 3 \pmod{7}$ and $n \equiv 5 \pmod{6}$. Solve the system:

📌 From $n \equiv 5 \pmod{6}$, $n = 6k + 5$. Substitute into the first congruence: $6k + 5 \equiv 3 \pmod{7} \Rightarrow 6k \equiv -2 \pmod{7} \Rightarrow 6k \equiv 5 \pmod{7}$. Multiply both sides by the modular inverse of 6 mod 7 (which is 6, since $6 \times 6 = 36 \equiv 1 \pmod{7}$): $k \equiv 5 \times 6 \equiv 30 \equiv 2 \pmod{7}$. Thus, $k = 7m + 2$, and $n = 6(7m + 2) + 5 = 42m + 17$. The two-digit solutions are when $m = 0$: $n = 17$ and $m = 1$: $n = 59$. The smallest is $\boxed{17}$.

📌 Question: What is the greatest common divisor of $2^{12} - 1$ and $2^{18} - 1$?

📌 Solution: Use the identity $\gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a, b)} - 1$. Compute $\gcd(12, 18) = 6$. Thus, $\gcd(2^{12} - 1, 2^{18} - 1) = 2^6 - 1 = 64 - 1 = \boxed{63}$.

📌 Question: What is the smallest three-digit number divisible by 11 and 13?

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