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5 guard buzz cut - American Beagle Club

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Solution: Rewrite the equation as $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Possible integer values for $y$ are $-2, -1, 0, 1, 2$. For each $y$, solve for $x$:

📌 $y = 0$: $x^2 = 16 \Rightarrow x = \pm4$ (2 points).

📌 $y = \pm1$: $x^2 + 4(1) = 16 \Rightarrow x^2 = 12$ (no integer solutions).

📌 $y = \pm2$: $x^2 + 4(4) = 16 \Rightarrow x^2 = 0 \Rightarrow x = 0$ (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 $\boxed{17}$.

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