Monkeys are impatient and sometimes stand upright.
When the first monkey came to the beach, he wanted to take his share, so he divided the peaches into two piles. When he found another one, he threw the extra one into the sea and took his share.
When the second monkey came to the beach, he also wanted to take his share. A monkey is always a monkey. It can't know that its partner took a copy. So the second monkey divided the peaches into two piles and found one more, so he threw the extra one into the sea and took his share.
If the original number of peaches is not less than 100, how many peaches can the first monkey take away?
It may not be easy to solve it by arithmetic, but try the method of "listing algebraic expressions":
If the number of peaches taken by the second monkey is represented by a, then the number of peaches it faced before taking away should be 2a+1; Think about it, why? )
Since the number of peaches left by the first monkey is (2A+l), the number of peaches it took away should also be 2a+1;
Before the first monkey takes it away, the number of peaches it faces should be (2a+1)+(2a+1)+1,which is 4A+3.
This means that the original number of peaches on the beach is 4A+3, but this pile of peaches is not less than 100, so A is not less than 25. So the first monkey can get at least 5 1(=2×25+ 1) peaches.
Looking back on the whole process of solving problems, we always step by step "first express the words related to quantity in the problem with formulas containing numbers, letters and operational symbols", that is to say, "listing algebraic expressions" plays an important role in solving problems.
Thinking: If this pile of peaches is the property of three monkeys, how to solve the problem?
The answer. . Ideas:
If the number of peaches taken by the third monkey is represented by X, then the number of peaches it faced before taking away should be 2x+ 1.
Since the number of peaches left by the second monkey is 2x+ 1, the number of peaches it took away should also be 2x+1; Before the second monkey takes it away, the peach it faces should be 2x+1+2x+1+1,which is 4x+3. Since the number of peaches left by the first monkey is 4X+3, the number of peaches it takes away should also be 4X+3, so the peaches it faces should be 4x+3+4x+3. That is 8X+7.
This shows that the original number of peaches on the beach is 8X+7, but this pile of peaches is not less than 100, so x is not less than 12. So the first monkey can get at least 5 1=(4* 12+3) peaches.
Solution: suppose the third one takes X.
((2X+ 1)* 2+ 1)* 2+ 1 = 4(2X+ 1)+3 = 8X+7
48+7 >99
48 >； 92
X & gt 1 1.5
X= 12
(12 * 2+1) * 2+1= 51(only)