AMC12 2005 B

AMC12 2005 B · Q23

AMC12 2005 B · Q23. It mainly tests Linear equations, Logarithms (rare).

Let $S$ be the set of ordered triples $(x,y,z)$ of real numbers for which \[\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.\] There are real numbers $a$ and $b$ such that for all ordered triples $(x,y,z)$ in $S$ we have $x^{3}+y^{3}=a \cdot 10^{3z} + b \cdot 10^{2z}.$ What is the value of $a+b?$

令 $S$ 为满足以下条件的实数有序三元组 $(x,y,z)$ 的集合： \[\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.\] 存在实数 $a$ 和 $b$，使得对于 $S$ 中所有有序三元组 $(x,y,z)$，有 $x^{3}+y^{3}=a \cdot 10^{3z} + b \cdot 10^{2z}.$ 求 $a+b$ 的值。

(A) \dfrac{15}{2} \dfrac{15}{2}

(B) \dfrac{29}{2} \dfrac{29}{2}

(D) \dfrac{39}{2} \dfrac{39}{2}

(E) 24 24

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $x + y = s$ and $x^2 + y^2 = t$. Then, $\log(s)=z$ implies $\log(10s) = z+1= \log(t)$,so $t=10s$. Therefore, $x^3 + y^3 = s\cdot\dfrac{3t-s^2}{2} = s(15s-\dfrac{s^2}{2})$. Since $s = 10^z$, we find that $x^3 + y^3 = 15\cdot10^{2z} - (1/2)\cdot10^{3z}$. Thus, $a+b = \frac{29}{2}$ $\Rightarrow$ $\boxed{B}$

令 $x+y=s$ 且 $x^2+y^2=t$。则 $\log(s)=z$ 蕴含 $\log(10s)=z+1=\log(t)$，所以 $t=10s$。因此，$x^3+y^3=s\cdot\dfrac{3t-s^2}{2}=s(15s-\dfrac{s^2}{2})$。由于 $s=10^z$，得到 $x^3+y^3=15\cdot10^{2z}-(1/2)\cdot10^{3z}$。因此，$a+b=\frac{29}{2}$ $\Rightarrow$ $\boxed{B}$。

Topics