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Ejercicio 1.

Enunciar el Teorema Fundamental de la Aritmética y utilizarlo para mostrar que 48 es el número natural más chico que admite exactamente 10 divisores positivos.
Hallar 
m inZ tal que el coeficiente que multiplica a 
x^(5) en 
(mx^(3)+x^(-2))^(10) es 8064 .

Ejercicio 11.\newline11) Enunciar el Teorema Fundamental de la Aritmética y utilizarlo para mostrar que 4848 es el número natural más chico que admite exactamente 1010 divisores positivos.\newline22) Hallar mZ m \in \mathbb{Z} tal que el coeficiente que multiplica a x5 x^{5} en (mx3+x2)10 \left(m x^{3}+x^{-2}\right)^{10} es 80648064 .

Full solution

Q. Ejercicio 11.\newline11) Enunciar el Teorema Fundamental de la Aritmética y utilizarlo para mostrar que 4848 es el número natural más chico que admite exactamente 1010 divisores positivos.\newline22) Hallar mZ m \in \mathbb{Z} tal que el coeficiente que multiplica a x5 x^{5} en (mx3+x2)10 \left(m x^{3}+x^{-2}\right)^{10} es 80648064 .
  1. State Fundamental Theorem of Arithmetic: State the Fundamental Theorem of Arithmetic and use it to show that 4848 is the smallest natural number with exactly 1010 positive divisors.\newlineThe Fundamental Theorem of Arithmetic states that every integer greater than 11 can be represented uniquely as a product of prime numbers, up to the order of the factors.\newline48=24×3148 = 2^4 \times 3^1\newlineThe number of divisors is given by (4+1)(1+1)=10(4+1)(1+1) = 10, which confirms that 4848 has exactly 1010 divisors.
  2. Expand Binomial Theorem: Expand (mx3+x2)10(mx^3 + x^{-2})^{10} using the binomial theorem to find the term containing x5x^5. We need the term where the powers of xx add up to 55, which is the term with x3x^3 from mx3mx^3 raised to the 77th power and x2x^{-2} raised to the 33rd power. The binomial coefficient for this term is C(10,3)C(10,3).
  3. Calculate Binomial Coefficient: Calculate the binomial coefficient C(10,3)C(10,3).C(10,3)=10!3!(103)!=10×9×83×2×1=120C(10,3) = \frac{10!}{3!(10-3)!} = \frac{10\times9\times8}{3\times2\times1} = 120
  4. Write Term with x5x^5: Write down the term containing x5x^5. The term is C(10,3)(mx3)7(x2)3C(10,3) \cdot (mx^3)^7 \cdot (x^{-2})^3. Substitute the binomial coefficient: 120m7x37x6120 \cdot m^7 \cdot x^{3\cdot7} \cdot x^{-6}. Simplify the powers of x: 120m7x216=120m7x15120 \cdot m^7 \cdot x^{21-6} = 120 \cdot m^7 \cdot x^{15}.
  5. Find Value of mm: Find the value of mm such that the coefficient of x5x^5 is 80648064. We need to equate the coefficient of x15x^{15} in the term to 80648064, but we made a mistake in the previous step; we need the coefficient of x5x^5, not x15x^{15}.

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