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# [Umi](https://umijs.org/) | ||
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[github]() |
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# 小学数学 | ||
> | ||
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## 模数 | ||
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### 2024-5-23 | ||
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$$ | ||
A = 1 \times 3 \times 5 \times 7 \times 9 \times 11 \times \cdots \cdots \times 2023 \text{的末两位是多少?} | ||
$$ | ||
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小学题应该找规律就可以了吧,十位乘十位就是整百,所以只要算1×3×5×7×9,然后用周期性规律得出,类似组合问题的本来就应该先简单枚举 去体会 再找到突破口。 | ||
这里我想利用同余的乘积性质来分析这个问题。 | ||
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求末两位数就是模100,就是求除以25和除以4的余数。A一定是25的倍数,这样末两位数只能是00,25,50,75. 由于默认是奇数2023,那么只能是25或75两个的一个。 | ||
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$\color{red} \text{如果这个数A模4余1就是25,A模4余3就是75.}$ | ||
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奇数模4的结果是1和-1的交叉出现 | ||
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$$ | ||
\text{余为正是} 1 \equiv 1 \pmod 4 \equiv 5 \pmod 4 \equiv 9 \pmod 4 \equiv \cdots \cdots \newline | ||
\text{余为负是} -1 \equiv 3 \pmod 4 \equiv 7 \pmod 4 \equiv 11 \pmod 4 \equiv \cdots \cdots \newline | ||
\text{而} \lceil {2023/4} \rceil = 2024 = 4 \times 506 \to 2023 = 4 \times 506 - 1 | ||
$$ | ||
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由此得出最后的余为负的共有506个,偶数个-1的结果是1,这样A模4的结果就是1,A的末两位数就是25. | ||
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把问题扩展一下,就是数A的末三位数是多少呢? | ||
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就是求模1000,分别除以125和除以8的余数。数A一定是125的倍数,且为奇数,那么它的末三位数一定是 | ||
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$125 \times 1 = 125, 125 \times 3 = 375, 125 \times 5 = 625, 125 \times 7 = 875 $ | ||
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$$ | ||
125 \equiv 5 \pmod 8, 375 \equiv 7 \pmod 8, 625 \equiv 1 \pmod 8, 875 \equiv 3 \pmod 8. | ||
$$ | ||
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现在的问题变成求数A莫8后的余数,由于连续8个整数是模8的一个完全剩余系,这里只有奇数,所以每四个数为一组,每组四个数除以8的余数分别是1,3,5,7,即1,3,-3,-1。它们的乘积除以8的余数是1. | ||
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1到2024是的自然数一共有2024个,其中一半的奇数,共有1012,1012恰好是4的倍数,刚好四个数一组分完。 | ||
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$$ | ||
1 \times 3 \times 5 \times 7 \times 9 \times 11 \times 13 \times 15 \times \cdots \cdots \times 2023 \newline | ||
= (1 \times 3 \times 5 \times 7) \times (9 \times 11 \times 13 \times 15) \times \cdots \cdots \times (2017 \times 2019 \times 2021 \times 2023) \newline | ||
\equiv = 1 \times 1 \times \cdots \cdots \times 1 \equiv 1 \pmod 8 | ||
$$ | ||
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所以答案是末三位数是625 | ||
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