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Home » Pregunta: Si $a + b = 10$ y $a^2 + b^2 = 58$, encuentra $a^3 + b^3$. - Appfinity Technologies

Pregunta: Si $a + b = 10$ y $a^2 + b^2 = 58$, encuentra $a^3 + b^3$. - Appfinity Technologies

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Each divisor corresponds to a perfect square divisor $s^2$, so the number of such $s$ (i.e., integer side lengths of square grids dividing the area) is equal to the number of positive divisors of $2025$, which is $15$. However, each divisor $d$ of $2025$ corresponds to $s = \sqrt{d}$, but only those $d$ that are perfect squares yield integer $s$. Since $2025 = 45^2$, the number of square divisors equals the number of perfect square divisors.

📌 To count perfect square divisors: for a divisor to be a perfect square, all exponents in its prime factorization must be even.

📌 Exponent of $3$: choices $0, 2, 4$ → 3 options

📌 Exponent of $5$: choices $0, 2$ → 2 options

📌 Thus, number of perfect square divisors is $3 \cdot 2 = 6$.

📌 Therefore, there are $ \boxed{6} $ different square grid sizes with integer side lengths that evenly divide the area.

📌 Question: A quantum computing researcher is encrypting data using modular exponentiation in a protocol where the exponent is derived from quadratic residues. What is the remainder when $3^1 + 3^2 + 3^3 + \cdots + 3^{24}$ is divided by $7$?

📌 Solution: We compute the sum $ \sum_{k=1}^{24} 3^k \mod 7 $. This is a geometric series with first term $a = 3$, ratio $r = 3$, and $n = 24$ terms. The sum is:

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