HMMT 二月 2026 · 冲刺赛 · 第 12 题

HMMT February 2026 — Guts Round — Problem 12

[7] Let a , b , c , and d be positive real numbers such that ac = 100 and bd = 101 . Compute the largest possible value of log b log c log d log a 10 10 10 10 a · b · c · d . ©2026 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2026, February 14, 2026 — GUTS ROUND Organization Team Team ID#

解析

[7] Let a , b , c , and d be positive real numbers such that ac = 100 and bd = 101 . Compute the largest possible value of log b log c log d log a 10 10 10 10 a · b · c · d . Proposed by: Carlos Rodriguez 2 Answer: 101 = 10201 log b log c log d log a 10 10 10 10 Solution 1: Let f ( a, b, c, d ) = a b c d . Taking log base 10 of f ( a, b, c, d ) yields: log b log c log d log a 10 10 10 10 log ( f ( a, b, c, d )) = log ( a b c d ) 10 10 = log a log b + log b log c + log c log d + log d log a 10 10 10 10 10 10 10 10 = (log a + log c )(log b + log d ) 10 10 10 10 = log ( ac ) log ( bd ) 10 10 = log (100) log (101) 10 10 = 2 log (101) . 10 2 log (101) 2 10 Therefore, f ( a, b, c, d ) = 10 = 101 = 10201 for all positive real numbers a , b , c , and d with ac = 100 and bd = 101 . Solution 2: Define f as before. Substituting d = 101 /b and c = 100 /a directly into the expression log b log c log d log a 10 10 10 10 f ( a, b, c, d ) = a b c d yields: log b log (100 /a ) log (101 /b ) log a 10 10 10 10 f ( a, b, c, d ) = a b (100 /a ) (101 /b ) log b − log (101 /b ) log (100 /a ) − log a log (101 /b ) log a 10 10 10 10 10 10 = a b 100 101 ( ) ( ) ( ) 2 log b 2 log (1 /a ) log (1 / 101) log a log (100) log (1 /b ) log (101) 10 10 10 10 10 10 10 = a b · a 101 · b 100 · 100 log (101) 10 = 1 · 1 · 1 · 100 2 = 101 = 10201 .