HMMT 十一月 2018 · THM 赛 · 第 4 题
HMMT November 2018 — THM Round — Problem 4
题目详情
- I have two cents and Bill has n cents. Bill wants to buy some pencils, which come in two different packages. One package of pencils costs 6 cents for 7 pencils, and the other package of pencils costs a dime for a dozen pencils (i.e. 10 cents for 12 pencils). Bill notes that he can spend all n of his cents on some combination of pencil packages to get P pencils. However, if I give my two cents to Bill, he then notes that he can instead spend all n + 2 of his cents on some combination of pencil packages to get fewer than P pencils. What is the smallest value of n for which this is possible? Note: Both times Bill must spend all of his cents on pencil packages, i.e. have zero cents after either purchase.
解析
- I have two cents and Bill has n cents. Bill wants to buy some pencils, which come in two different packages. One package of pencils costs 6 cents for 7 pencils, and the other package of pencils costs a dime for a dozen pencils (i.e. 10 cents for 12 pencils). Bill notes that he can spend all n of his cents on some combination of pencil packages to get P pencils. However, if I give my two cents to Bill, he then notes that he can instead spend all n + 2 of his cents on some combination of pencil packages to get fewer than P pencils. What is the smallest value of n for which this is possible? Note: Both times Bill must spend all of his cents on pencil packages, i.e. have zero cents after either purchase. Proposed by: James Lin Answer: 100 Suppose that Bill buys a packages of 7 and b packages of 12 in the first scenario and c packages of 7 and d packages of 12 in the second scenario. Then we have the following system: 6 a + 10 b = n 6 c + 10 d = n + 2 7 a + 12 b > 7 c + 12 d. Since the packages of 12 give more pencils per cent, we must have b > d . Subtract the first two equations and divide by 2 to get 3( c − a ) − 5( b − d ) = 1 . Note that the last inequality is 12( b − d ) > 7( c − a ). The minimal solutions to the equation with b − d > 0 are ( c − a, b − d ) = (2 , 1) , (7 , 4) , (12 , 7) , (17 , 10) . (17 , 10) is the first pair for which 12( b − d ) > 7( c − a ). Hence b ≥ 10 so n ≥ 100. We can easily verify that ( a, b, c, d, n ) = (0 , 10 , 17 , 0 , 100) satisfies the system of equations.