Divide the gold bar into three sections (that is, divide it twice or cut it twice), which are 1/7 and 2/7 4/7 of the whole gold bar respectively.
Day 1: For 1/7,
The next day: Give 2/7 and get back 1/7.
The third day is 1/7.
Day 4: Pay 4/7, take back 1/7 and 2/7.
Day 5: For 1/7
Day 6: Pay 2/7 and take back 1/7.
Day 7: For 1/7
2. Archimedes problem-the problem of dividing cattle.
Sun God has a herd of cows, white, black, flowered and brown.
Among the bulls, the number of white cattle is more than that of brown cattle, and the extra number is equivalent to1/2+1/3 of the number of black cattle; The number of black cattle is more than that of brown cattle, and the extra number is equivalent to1/4+1/5 of the number of flower cattle; The number of flower cattle is more than that of brown cattle, and the extra number is equivalent to 1/6+ 1/7 of the number of white cattle.
Among the cows, the number of white cows is1/3+1/4 of all black cows; The number of black cattle is1/4+1/5 of all flower cattle; The number of flower cattle is1/5+1/6 of all brown cattle; The number of brown cattle is 1/6+ 1/7 of the total number of white cattle.
How is this herd made up?
3. If three line segments with lengths A, B and C can form a triangle, can the root numbers A, B and C of the line segments form a triangle?
Assuming that A is the smallest and C is the largest, the necessary and sufficient condition for abc to form a triangle is A+B >; c；
At this time, the comparison between √a+√b and √c is actually the comparison between √ a+b+2√ab and C (both sides are squares). a+b is already greater than C, and obviously a triangle can be formed.
4. Someone has a triangular meadow. He divided the grassland into four parts: east, south, west and north. After a period of time, he found that the grassland in the west can graze five sheep, the grassland in the south can graze 10 sheep, and the grassland in the east can graze eight sheep. How many sheep can the grassland in the north graze?
Let the length of the bottom edge of this grassland be a, assume that the bottom edge runs east-west, and the opposite vertex is in the north, and then assume that the number of grazing grassland is proportional to the area of grassland (this assumption is reasonable).
Suppose this person divides the eastern and western grasslands into right triangles, the southern grassland is rectangular, the bottom of the rectangle is B, and the height is C, then the northern grassland is a triangle similar to the whole grassland. Set the northern grassland to graze X sheep.
Because the grassland in the west can graze 5 livestock and the grassland in the south can graze 10 livestock, one right-angled side of the grassland in the west is the height c of the rectangle, and the other is equal to the bottom b of the rectangle. Eight animals can graze from the east, knowing that one of the two right-angled sides of the east lawn is the height c of a rectangle and the other is equal to (8/5)b,
So a=b+b+ 1.6b=3.6b, thus assuming that there are:
(3.6/ 1)^2=(x+5+ 10+8)/x，
The solution x is approximately equal to 2.
A: You can graze 2 sheep.
Note: If the grassland apex is in the south, the equation [(13/x)+1] 2 = (x+23)/10, and the solution of x is approximately equal to 14. Can graze 14 sheep.