AMC 8 2024 Problem 15

AMC8 2024 年第 15 题

整除与因数★★★☆☆
Let the letters F, L, Y, B, U, and G represent distinct digits.
Suppose F‾ L‾ Y‾ F‾ L‾ Y‾\underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}}\,\underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}} is the greatest six-digit number that satisfies the following equation:
8⋅F‾ L‾ Y‾ F‾ L‾ Y‾=B‾ U‾ G‾ B‾ U‾ G‾8 \cdot \underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}}\,\underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}} = \underline{\mathrm{B}}\,\underline{\mathrm{U}}\,\underline{\mathrm{G}}\,\underline{\mathrm{B}}\,\underline{\mathrm{U}}\,\underline{\mathrm{G}}
What is the value of F‾ L‾ Y‾+B‾ U‾ G‾\underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}} + \underline{\mathrm{B}}\,\underline{\mathrm{U}}\,\underline{\mathrm{G}}?
用 F、L、Y、B、U、G 表示不同数字。
若 F‾ L‾ Y‾ F‾ L‾ Y‾\underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}}\,\underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}} 为满足下列等式的最大六位数：
8⋅F‾ L‾ Y‾ F‾ L‾ Y‾=B‾ U‾ G‾ B‾ U‾ G‾8 \cdot \underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}}\,\underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}} = \underline{\mathrm{B}}\,\underline{\mathrm{U}}\,\underline{\mathrm{G}}\,\underline{\mathrm{B}}\,\underline{\mathrm{U}}\,\underline{\mathrm{G}}
求 F‾ L‾ Y‾+B‾ U‾ G‾\underline{\mathrm{F}}\,\underline{\mathrm{L}}\,\underline{\mathrm{Y}} + \underline{\mathrm{B}}\,\underline{\mathrm{U}}\,\underline{\mathrm{G}} 的值。
  1. A.1089
  2. B.1098
  3. C.1107
  4. D.1116
  5. E.1125

正确答案 · Correct Answer

C

解析 · Solution

解法1

设 FLY‾\overline{FLY} 是三位数 100F+10L+Y。将一个三位数块重复两次得到 FLYFLY‾=1000⋅FLY‾+FLY‾=1001⋅FLY‾\overline{FLYFLY}=1000\cdot \overline{FLY}+\overline{FLY}=1001\cdot \overline{FLY}。类似地，BUGBUG‾=1001⋅BUG‾\overline{BUGBUG}=1001\cdot \overline{BUG}。已知 8⋅FLYFLY‾=BUGBUG‾8\cdot \overline{FLYFLY}=\overline{BUGBUG}，代入得：8(1001FLY‾)=1001BUG‾8(1001\overline{FLY})=1001\overline{BUG}。两边除以 1001：8FLY‾=BUG‾8\overline{FLY}=\overline{BUG}。由于 BUG‾\overline{BUG} 是三位数，所以 100≤8FLY‾≤999⇒13≤FLY‾≤124100\le 8\overline{FLY}\le 999 \quad\Rightarrow\quad 13\le \overline{FLY}\le 124。为了最大化 FLYFLY‾\overline{FLYFLY}，需要最大化 FLY‾\overline{FLY}。尝试最大可能的值：FLY‾=123⇒BUG‾=8⋅123=984\overline{FLY}=123 \quad\Rightarrow\quad \overline{BUG}=8\cdot 123=984。数字 {1,2,3,9,8,4}\{1,2,3,9,8,4\} 全部互异，因此该解成立且为最大。所以，FLY‾+BUG‾=123+984=1107\overline{FLY}+\overline{BUG}=123+984=1107。1107 温斯顿·丘吉尔中学

Solution 1

Let FLY‾\overline{FLY} be the 3 number 100F+10L+Y. Repeating a 3 block twice gives FLYFLY‾=1000⋅FLY‾+FLY‾=1001⋅FLY‾.\overline{FLYFLY}=1000\cdot \overline{FLY}+\overline{FLY}=1001\cdot \overline{FLY}. Similarly, BUGBUG‾=1001⋅BUG‾.\overline{BUGBUG}=1001\cdot \overline{BUG}. Given 8⋅FLYFLY‾=BUGBUG‾8\cdot \overline{FLYFLY}=\overline{BUGBUG}, substitute: 8(1001FLY‾)=1001BUG‾.8(1001\overline{FLY})=1001\overline{BUG}. Divide: 8FLY‾=BUG‾.8\overline{FLY}=\overline{BUG}. Since BUG‾\overline{BUG} is a 3 number, 100≤8FLY‾≤999⇒13≤FLY‾≤124.100\le 8\overline{FLY}\le 999 \quad\Rightarrow\quad 13\le \overline{FLY}\le 124. To maximize FLYFLY‾\overline{FLYFLY}, maximize FLY‾\overline{FLY}. Try the largest possible value: FLY‾=123⇒BUG‾=8⋅123=984.\overline{FLY}=123 \quad\Rightarrow\quad \overline{BUG}=8\cdot 123=984. The digits {1,2,3,9,8,4}\{1,2,3,9,8,4\} are all distinct, so this works and is maximal. Therefore, FLY‾+BUG‾=123+984=1107.\overline{FLY}+\overline{BUG}=123+984=1107. 1107 of Winston Churchill Middle School

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