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A set consists of all three-digit positive integers with the following properties.Each integer is of the form JKL,where J,K,and I are digits;all the digits are nonzero;and the two-digit integers JK and KL are each divisible by 9.How many integers are in the set?
当前版本由R妹更新于2024-04-15 16:16:31 感谢由R妹对此题目的解答所做出的贡献。
根据题目要求,要找到所有由三位数组成的正整数JKL,其中J、K和L都是非零数字,并且两位整数JK和KL都能被9整除。
首先来确定JK,即百位和十位上的数字。由于它们能被9整除,它们的和也必定能被9整除。只有当J和K的值分别为1到9时,才满足题目中所述的条件。百位J和十位K组成的数字JK被9整除,有如下几种可能(只列出其中一些):
18, 27, 36, 45, 54, 63, 72, 81
这些是符合条件的两位整数JK,可以看出每个十位数都可以与任意一个1-9之间的个位数组合形成一个符合条件的数(让该数能被9整除)。因此,存在9个符合条件的JK两位数。
接下来考虑KL。其实同理,所以也有9个符合条件的KL。
现在我们有了9个可能的JK和9个可能的KL,但这并不意味着我们总共有 9 * 9 = 81 个三位数,因为我们需要保证JK和KL中的K(JK中十位上的数和KL中百位上的数)是一样的。
看一下两组中每组有哪些选择:
- JK 的选择: 18, 27, 36, 45, 54, 63, 72, 81
- KL 的选择: 18, 27, 36, 45, 54, 63, 72, 81
我们知道K对于每对JK和KL来说是固定的,所以:
- 当K=1时,可选的J和L有 (1) [J=1,L=8]
- 当K=2时,可选的J和L有 (2) [J=2,L=7]
- ...
- 当K=8时,可选的J和L有 (8) [J=8,L=1]
总结一下,每个可能的K值实际上定义了一个唯一的三位数JKL。于是我们总共有与可能的K值数量相同的三位数个数,即9个可能个数。
所以答案是9个这样的三位正整数。
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