AMC 8 2022 Problem 6
AMC8 2022 年第 6 题
数列与规律★★☆☆☆
Three positive integers are equally spaced on a number line. The middle number is 15, and the largest number is 4 times the smallest number. What is the smallest of these three numbers?
三个正整数在数轴线上等距排列。中间的数是 15，最大的数是最小的数的 4 倍。问这三个数中最小的数是多少？
- A.4
- B.5
- C.6
- D.7
- E.8
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
Let the smallest number be $x.$ It follows that the largest number is . Since , 15, and are equally spaced on a number line, we have $$$$$$$$
Let the common difference of the arithmetic sequence be . Consequently, the smallest number is 15-d and the largest number is 15+d. As the largest number is 4 times the smallest number, . Finally, we find that the smallest number is 15-9=6.
Let the smallest number be . Because and are equally spaced from 15, 15 must be the average. By adding and and dividing by 2, we get that the mean is also 2.5x. We get that 2.5x=15, and solving gets x=6.
Since the problem is giving us the options, we can try all of them. The differences MUST be equal because the numbers have to be evenly spaced on the number line. We see that 4 cannot work because 15 - 4 = 11, and 4 * 4 = 16. 16 - 15 is not 11, so this option does not work. Next, we can try 5. We see that this also does not work because 15 - 5 = 10, and 5 * 4 = 20. 20 - 15 = 5, and this is not 10. We can then try 6 to see that it does work. 15 - 6 = 9, and 6 * 4 = 24. 24 - 15 = 9. The differences are equivalent, so we can see that 6 indeed does work. Therefore, the answer is. ~~~
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