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CIE 9709 2026 June Paper 52 Q3

A Level / CIE / S1

CIE 9709 2026 June Paper 52 Paper · Question 3

题目

Problem

(a) Find the number of different 3-digit numbers greater than 200 that can be made from the digits {1,2,3,4,5,6,7}\{1,2,3,4,5,6,7\} if digits may be repeated and the number is divisible by 5.

(1)

(b) Find the number of different 3-digit numbers greater than 200 that can be made from the digits {1,2,3,4,5,6,7}\{1,2,3,4,5,6,7\} if no digit is repeated and the number is even.

(3)
题目中文翻译

(a) 用数字 {1,2,3,4,5,6,7}\{1,2,3,4,5,6,7\} 可以组成多少个大于 200 的不同三位数,要求数字可以重复使用,且该数能被 5 整除?

(1)

(b) 用数字 {1,2,3,4,5,6,7}\{1,2,3,4,5,6,7\} 可以组成多少个大于 200 的不同三位数,要求数字不重复使用,且该数为偶数?

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

能被5整除,个位只能为5;百位不能为1,十位可任取给出的7个数字,因为允许重复。

答题过程

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The units digit must be 5. The hundreds digit has 6 choices, from 2 to 7, and the tens digit has 7 choices. Thus the required number is

6×7×1=42.6\times7\times1=\boxed{42}.

(b)

解法一

思路

展开

先限制最强的个位:只有2、4、6可选。确定个位后,百位从2至7中排除该数字,十位再从剩余数字中选取。

答题过程

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The units digit has 3 choices: 2, 4 or 6.

For each choice, the hundreds digit has 5 choices, since it must be from 2 to 7 and must differ from the units digit. The tens digit then has 5 choices from the remaining digits, including 1.

Therefore the required number is

3×5×5=75.3\times5\times5=\boxed{75}.

解法二

思路

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也可以先按百位的奇偶性分类。百位为偶数会减少个位可选的偶数个数,百位为奇数则不会。

答题过程

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If the hundreds digit is even, it has 3 choices. There are then 2 choices for the even units digit and 5 for the tens digit, giving 3×2×5=303\times2\times5=30 numbers.

If the hundreds digit is odd, it must be 3, 5 or 7, giving 3 choices. There are 3 choices for the even units digit and 5 for the tens digit, giving 3×3×5=453\times3\times5=45 numbers.

Hence the required number is

30+45=75.30+45=\boxed{75}.