1997 AMC 12 第 22 题

先试着解答 1997 AMC 12 第 22 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1997 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

22.

阿什莉、贝蒂、卡洛斯、迪克和埃尔金一起去购物。每个人可花的钱数都是整数美元,他们共有 $56\$56。阿什莉和贝蒂可花金额之差的绝对值为 $19\$19。贝蒂与卡洛斯的金额之差的绝对值为 $7\$7,卡洛斯与迪克为 $5\$5,迪克与埃尔金为 $4\$4,埃尔金与阿什莉为 $11\$11。埃尔金有多少钱?
Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $56.\$56. The absolute difference between the amounts Ashley and Betty had to spend was $19.\$19. The absolute difference between the amounts Betty and Carlos had was $7,\$7, between Carlos and Dick was $5,\$5, between Dick and Elgin was $4,\$4, and between Elgin and Ashley was $11.\$11. How much did Elgin have?
$6\$6
$7\$7
$8\$8
$9\$9
$10\$10
答案:E
知识点:absolute differencessigned cases
小提示:
沿五人组成的环,为每个相邻金额之差选定正负号
Assign a sign to each successive difference around the five-person cycle
大提示:
在计算五人的金额总和之前,先利用这些带符号的差之和必须为零
The signed differences must total zero before the five amounts can be summed
解答:
环上各差的绝对值为 1919、77、55、44、1111,而带符号的差之和为 00。因此正负两组中的一组之和必须是 4646 的一半,即 2323。唯一的分组是 19+4=7+5+1119+4=7+5+11。按一种方向取符号时,各人的金额满足 A=E+11A=E+11、B=E−8B=E-8、C=E−1C=E-1、D=E+4D=E+4。总金额为 5E+65E+6,所以 5E+6=565E+6=56,从而 E=10E=10。若把所有符号反向,则有 5E−6=565E-6=56,得不到整数 EE。因此埃尔金有 $10\$10,正确答案是 E。
The signed differences around the cycle have magnitudes 19,19, 7,7, 5,5, 4,4, 1111 and sum 0.0. Thus one side of the sign split must total half of 46,46, namely 23.23. The only split is 19+4=7+5+11.19+4=7+5+11. In one orientation, the amounts are A=E+11,A=E+11, B=E−8,B=E-8, C=E−1,C=E-1, and D=E+4.D=E+4. Their total is 5E+6,5E+6, so 5E+6=565E+6=56 and E=10.E=10. Reversing every sign would give 5E−6=56,5E-6=56, not an integral E.E. Thus Elgin had $10\$10 and E is correct.
LIVE by Po-Shen Loh
LIVE项目

LIVE直播课程

Self-Paced录播课程

启明星Star计划

奖学金项目

关于我们

罗博深是谁?

邮件联系

订阅英文简报

加入我们

申请明星导师

For a more Thought Full world.

© 2026 Expii, Inc.