AMC 8 2023 Problem 18
AMC8 2023 年第 18 题
特殊数与不定方程★★★☆☆
Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump 5 pads to the right or 3 pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located 2023 pads to the right of her starting position?
蚱锰Greta坐在池塘里一排长长的睡莲叶上：从任何睡莲叶出发，Greta都可以向右跳5个叶片或者向左跳3个叶片. 问Greta最少跳多少次才能到达位于她起始位置右侧的第2023片睡莲叶上？
- A.405
- B.407
- C.409
- D.411
- E.413
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
We have 2 directions going 5 right or 3 left. We can assign a variable to each of these directions. We can call going right 1 direction and we can call going 1 left . We can build an equation of , where we have to limit the number of moves we do. We can do this by making more of our moves the 5 move turn than the 3 move turn. The first obvious step is to go some amount of moves in the right direction then subtract off in the left direction to land on 2023. The least amount of 3 's added to 2023 to make a multiple of 5 is 4 as 2023 + 4(3) = 2035. So now, we have solved the problem as we just go hops right, and just do 4 more hops left. Yielding 407 + 4 = 411 as our answer., SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
Notice that , and jumping to the left increases the value of Greta's position by 2. Therefore, the number of jumps to the left must be . As the number of jumps to the left increases, so does the number of jumps to the right, we must minimize both, which occurs when we jump 4 to the left and 407 to the right. The answer is 411.
5y - 2023 must be divisible by 3. The smallest value of that will achieve this is 407, which lands it at 2035. After that, it takes 4 jumps back, making a total of 411..1 Here is but worded differently: We can to go back steps from steps we took past 2023. The other way to say this is we go steps from 2023 to get to . What is the least value of ? We need 2023+3x to end in a 5 or 0. The least value for , which makes 2023+3x=2035, is 4. . Therefore, 407+4=411.
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