使用所有数字 9, 2, 0, 3 来组成
- (a)
最大的可能奇数
- (b)
最接近 3000 的数
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完整解法
1(a)
1. Determine the condition for the number to be odd
For the number to be odd, the last digit must be odd. The available odd digits are and .
2. Determine the arrangement for the greatest possible number
To make the number as large as possible, the largest available digit should be placed in the highest place value (thousands). Thus, the first digit must be . This leaves as the only available odd digit for the last position (ones).
3. Arrange the remaining digits to maximize the number
The remaining digits are and . To maximize the number, place the larger digit () in the hundreds place and the smaller digit () in the tens place. The number is .
1(b)
1. Find the smallest possible number starting with 3
Using the digits 9, 2, 0, 3, the smallest number starting with 3 is . The difference from is .
2. Find the largest possible number starting with 2
The largest number starting with 2 is . The difference from is .
3. Compare the differences to find the closest number
Since , is closer to .
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