AMC 8 2015 Problem 20
AMC8 2015 年第 20 题
特殊数与不定方程★★★☆☆
Ralph went to the store and bought 12 pairs of socks for a total of $24 . Some of the socks he bought cost $1 a pair, some of the socks he bought cost $3 a pair, and some of the socks he bought cost $4 a pair. If he bought at least one pair of each type, how many pairs of $1 socks did Ralph buy?
Ralph 去商店买了 12 双袜子，总共花了 24 美元。在她买的这些袜子中，有的是 1 美元一双，有的是 3 美元一双，还有的是 4 美元一双。如果她每种袜子都至少买了 1 双，那么 1 美元一双的袜子她买了多少双？
- A.4
- B.5
- C.6
- D.7
- E.8
正确答案 · Correct Answer
D
解析 · Solution
Solution 1
So, let there be pairs of $1 socks, pairs of $3 socks, and pairs of $4 socks. We have x+y+z=12, x+3y+4z=24, and . Now, we subtract to find 2y+3z=12, and . It follows that is a multiple of 3 and is a multiple of 3. Since sum of 2 multiples of 3 = multiple of 3, so we must have 2y=6. Therefore, y=3, and it follows that z=2. Now, x=12-y-z=12-3-2=7, as desired.
Since the total cost of the socks was $24 and Ralph bought 12 pairs, the average cost of each pair of socks is . There are two ways to make packages of socks that average to $2. You can have: Two $1 pairs and one $4 pair (package adds up to $6 ) One $1 pair and one $3 pair (package adds up to $4 ) Now, we need to solve 6a+4b=24, where is the number of $6 packages and is the number of $4 packages. We see our only solution (that has at least one of each pair of sock) is a=2, b=3, which yields the answer of 2×2+3×1 = 7.
Since there are 12 pairs of socks, and Ralph bought at least one pair of each, there are 12-3=9 pairs of socks left. Also, the sum of the three pairs of socks is 1+3+4=8. This means that there are 24-8=16 dollars left. If there are only 1 dollar socks left, then we would have dollars wasted, which leaves 7 more dollars. If we replace one pair with a 3 dollar pair, then we would waste an additional 2 dollars. If we replace one pair with a 4 dollar pair, then we would waste an additional 3 dollars. The only way 7 can be represented as a sum of 2 s and 3 s is 2+2+3. If we change 3 pairs, we would have 6 pairs left. Adding the one pair from previously, we have pairs.
Let the amount of 1 dollar socks be , 3 dollar socks be , and 4 dollar socks be . We then know that a+b+c=12 and a+3b+4c=24. We can make a+b+c=12 into a=12-b-c and then plug that into the other equation, producing 12-b-c+3b+4c=24 which simplifies to 2b+3c=12. It's not hard to see b=3 and c=2. Now that we know and , we know that a=7, meaning the number of 1 dollar socks Ralph bought is 7. If Ralph bought one sock of each kind, he already used $8, so there are $16 left and 9 socks. If we split the $16 into four $4 sections, (as it is the smallest possible number that 1, 3, 4, can make in different ways that in all use at least each of the numbers once,) if Ralph bought a $3 pair, he would need to buy a $1 pair in order for it to add up to a multiple of four. Similarly, if Ralph bought a $1 pair, he would either need to buy three $1 pairs or a $3 pair. If Ralph bought a $4 pair, it would already make a group. Now, the problem is just how we can split 9 into 4 groups of 1, 2, or 4. We clearly see that 1 + 2 + 2 + 4 = 9, or a $4 pair, two $3 pairs, and six $1 pairs. Because we subtracted the necessary one of each kind, there are two $4 pairs, three $3 pairs, and seven $1 pairs. Therefore, the number of $1 pairs Ralph bought is.
To simplify the problem, we can reduce the price of each sock by $1, resulting in the prices being $0, $2 and $3. This means the total price Ralph spends becomes . Now the $0 socks aren't worth anything, but we still need to keep in mind there are at least one of them, so the question becomes: at most 11 of $2 and $3 socks add up to $12. It is obvious that , so the number of $0 socks (actually $1 socks) is 12-3-2=7.
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