HMMT 十一月 2024 · 冲刺赛 · 第 31 题

HMMT November 2024 — Guts Round — Problem 31

[17] Positive integers a , b , and c have the property that lcm ( a, b ) , lcm ( b, c ) , and lcm ( c, a ) end in 4 , 6 , and 7 , respectively, when written in base 10 . Compute the minimum possible value of a + b + c .

解析

[17] Positive integers a , b , and c have the property that lcm( a, b ), lcm( b, c ), and lcm( c, a ) end in 4, 6, and 7, respectively, when written in base 10. Compute the minimum possible value of a + b + c . Proposed by: Derek Liu Answer: 28 Solution: Note that a + b + c = 28 is achieved when ( a, b, c ) = (19 , 6 , 3). To show we cannot do better, first observe we would need a + c < 27 and lcm( a, c ) ≤ ac ≤ 13 · 13 = 169, which is only possible when lcm( a, c ) is 7, 17, 57, 77, or 117. We do casework on each value: • lcm( a, c ) = 7 . Then a and c are both 1 or 7, so lcm( a, b ) and lcm( b, c ) are both b or 7 b . It is impossible for one of b and 7 b to end in 4 and the other to end in 6. • lcm( a, c ) = 17 . The same argument as above proves this case is also impossible. • lcm( a, c ) = 77 . Since the divisors of 77 only end in 1 and 7, the same argument as above rules out this case. • lcm( a, c ) = 57 . As a + c < 27, we must have { a, c } = { 3 , 19 } . Then b is even and less than 6, so it’s easy to verify there are no solutions here. • lcm( a, c ) = 117 . As a + c < 27, we must have { a, c } = { 9 , 13 } . Then b is even and less than 6, so it’s easy to verify there are no solutions here. This rules out all cases, so 28 is optimal.