2016 AMC 10A 第 23 题

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23.

二元运算 \diamondsuit 满足 以及 aa=1a\,\diamondsuit \,a=1,对所有非零实数 a,ba, bcc 都成立。(这里 \cdot 表示乘法。)方程 的解可写成 pq\frac{p}{q},其中 ppqq 为互质正整数。求 p+qp+q 的值。 a(bc)=(ab)ca\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c 2016(6x)=1002016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100

A binary operation \diamondsuit has the properties that a(bc)=(ab)ca\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c and that aa=1a\,\diamondsuit \,a=1 for all nonzero real numbers a,b,a, b, and c.c. (Here \cdot represents multiplication). The solution to the equation 2016(6x)=1002016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100 can be written as pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. What is p+q?p+q?

109109

201201

301301

30493049

33,60133,601

答案:A
知识点:自定义运算函数方程
难度评级:1820
解答:

因为 aa=1a\diamondsuit a=1,在 中令 b=cb=c,得 a1=(ab)ba\diamondsuit1=(a\diamondsuit b)b。又由 a(aa)=(aa)aa\diamondsuit(a\diamondsuit a)=(a\diamondsuit a)a,可得 a1=aa\diamondsuit1=a,因此 ab=aba\diamondsuit b=\frac aba(bc)=(ab)ca\diamondsuit(b\diamondsuit c)=(a\diamondsuit b)c

因此代入方程得到 所以 x=2584x=\frac{25}{84}p+q=25+84=109p+q=25+84=1092016(6x)=20166x=20166/x=336x=100. \begin{aligned} 2016\diamondsuit(6\diamondsuit x) &= 2016\diamondsuit\frac6x \\ &= \frac{2016}{6/x} \\ &= 336x \\ &= 100. \end{aligned}

所以正确答案是 A

Since aa=1a\diamondsuit a=1, substituting b=cb=c in a(bc)=(ab)ca\diamondsuit(b\diamondsuit c)=(a\diamondsuit b)c gives a1=(ab)ba\diamondsuit1=(a\diamondsuit b)b. Also, using a(aa)=(aa)aa\diamondsuit(a\diamondsuit a)=(a\diamondsuit a)a gives a1=aa\diamondsuit1=a. Therefore ab=aba\diamondsuit b=\frac ab.

The equation becomes 2016(6x)=20166x=20166/x=336x=100. \begin{aligned} 2016\diamondsuit(6\diamondsuit x) &= 2016\diamondsuit\frac6x \\ &= \frac{2016}{6/x} \\ &= 336x \\ &= 100. \end{aligned} Thus x=2584x=\frac{25}{84}, so p+q=25+84=109p+q=25+84=109.

Thus, the correct answer is A.

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