2022年普通高等学校招生全国统一考试数学(北京卷)T21

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题目:

待补充(求发个$\LaTeX$的, 懒得码)

昨晚上刷到这题, 看了一眼就摸鱼去了反正不是京爷考不到我, 今早(12:00)起来做了一下

搜了一下没发现什么好的解答, 不是伪证就是太复杂

但还是感谢一下2022年北京高考数学压轴题赏析, 提到了富比尼原理, 指明了方向

(1)(2)自己做

(3):

实际上本题有个条件是无用的自己猜是哪个

只证$k=6$不成立, $k<6$的情况补$0$即为$k=6$的情况, 同时得知$k=6$时$\forall i\in{\{1,2,3,4,5,6\}}, a_{i}\neq 0$

$k=6$时

设$S$为所有连续和组成的可重集合

由

$|S|=21$

$\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}\subseteq S$

知

$\exists t\in{S}, \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}=S/\{t\}$

算两次

$\sum \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}=210=6\sum_{i\in{\{1,2,3,4,5,6\}}}a_{i}-t=\sum S/\{t\}$

$\sum_{i\in{\{1,2,3,4,5,6\}}}a_{i}=\frac{210+t}{6}$

若$\exists j\in{\{1,2,3,4,5,6\}}, a_{j}<0$

则$t=a_{j}<0$

$\sum_{i\in{\{1,2,3,4,5,6\}}, i\neq j}a_{i}=\frac{210-5t}{6}\geq\frac{210-5(-6)}{6}=40$

又$a_{j}$至多将$\{a_{n}\}$分为两部, 或有一部连续和大于$20$, 或有两部连续和均大于等于$20$

均矛盾

若$\forall i\in{\{1,2,3,4,5,6\}}, a_{i}>0$

则$t>0$

$\sum_{i\in{\{1,2,3,4,5,6\}}}a_{i}=\frac{210+t}{6}\geq\frac{210+6}{6}=36$

考察等于$20$的连续和

$l,r\in{\mathbb{Z}^{+}},l,r-1\in{\{1,2,3,4,5,6\}},\sum_{i\in{\mathbb{Z}^{+}},i\in{[l,r-1]}}a_{i}=20$

若$r-l<5$

则由$(r-l)+1<6$知$\exists L,R\in{\mathbb{Z}^{+}},L,R-1\in{\{1,2,3,4,5,6\}},[l,r-1]\subset[L,R-1],R-L=5$

$\sum_{i\in{\mathbb{Z}^{+}},i\in{[L,R-1]}}a_{i}>\sum_{i\in{\mathbb{Z}^{+}},i\in{[l,r-1]}}a_{i}=20$

又

$\sum_{i\in{\{1,2,3,4,5,6\}}}a_{i}\geq 36>20$

即

$\sum_{i\in{\mathbb{Z}^{+}},i\in{[L,R-1]}}a_{i}\notin \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$

$\sum_{i\in{\{1,2,3,4,5,6\}}}a_{i}\notin \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$

$\{i|i\in{\mathbb{Z}^{+}},i\in{[L,R-1]}\}\neq \{i|i\in{\{1,2,3,4,5,6\}}\}$

矛盾

且由

$\sum_{i\in{\{1,2,3,4,5,6\}}}a_{i}\notin \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$

知$r-l\neq 6$又$r-l\leq 6$故$r-l=5$

考察$j\notin{[l,r-1]}$

$a_{j}=\sum_{i\in{\{1,2,3,4,5,6\}}}a_{i}-\sum_{i\in{\mathbb{Z}^{+}},i\in{[l,r-1]}}a_{i}\geq 36-20=16$

$\exists L,R\in{\mathbb{Z}^{+}},L,R-1\in{\{1,2,3,4,5,6\}},j\in{[L,R-1]},R-L=5$

$\sum_{i\in{\mathbb{Z}^{+}},i\in{[L,R-1]}}a_{i}\geq 16+1+1+1+1=20$

$\{i|i\in{\mathbb{Z}^{+}},i\in{[L,R-1]}\}\neq \{i|i\in{\{1,2,3,4,5,6\}}\}$

$\{i|i\in{\mathbb{Z}^{+}},i\in{[L,R-1]}\}\neq \{i|i\in{\mathbb{Z}^{+}},i\in{[l,r-1]}\}$

或有两个连续和等于$20$, 或有两个连续和不属于$\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$

均矛盾

$Q.E.D.$