1, buy a few at most if you have some money. Solution: first look at buying the whole book, how many books can be bought with this money, and how many books can be bought by buying the whole book. After buying the whole book, consider whether the remaining money is enough to pay the bill. If so, you can buy several copies. Finally, add up the amount bought by an integer and the amount bought by a single number, which is the maximum amount that can be bought.

2, calculate the total amount of money to buy a certain number of items. Solution: First buy the one that can make up the whole amount according to the preferential scheme, and then buy the rest according to the order, and then add up the money spent by the two schemes.

For example, look at the topic first and get key information. For example, a shopping mall is holding a "May Day" promotion. Every 3 yuan for a pair of socks, buy 5 pairs and get 1 pair. Aunt Xu has 50 yuan. How many pairs can she buy at most? The key message is that every time you buy 5 pairs in 3 yuan, you get 1 pair. That is, 5+ 1=6 pairs, and the cost is 3*5= 15 yuan.

So how many 15 yuan are there in 50 yuan? 50÷ 15=3+5, that is, * * * has 3 groups of 6 pairs and needs to go to 5 yuan, while 5 yuan can only buy 1 pair at most. You can buy 6×3+ 1= 19 pairs at most.

To solve the application problem of "buy a few and get a few free", the key is to find out how much you need to buy. Once this problem is discovered, it becomes the most basic shopping problem, which can be easily solved with "unit price × quantity = total price".