Solution to the problem of chicken and rabbit in the same cage in ancient times

Chicken and rabbit in the same cage is one of the famous mathematical problems in ancient China. About 1500 years ago, this interesting question was recorded in Sun Tzu's calculation. The book describes it like this:

Today, there are pheasant rabbits in the same cage, with 35 heads above and 94 feet below. Pheasant rabbit geometry?

These four sentences mean:

There are several chickens and rabbits in a cage, counting from the top, 35 heads, counting from the bottom, 94 feet. How many chickens and rabbits are there in each cage?

There is a simplest algorithm to calculate this.

(total number of feet-total number of heads × number of chicken feet) ÷ (number of rabbit feet-number of chicken feet) = number of rabbits.

(94-35× 2) ÷ 2 = 12 (number of rabbits) Total number (35)- number of rabbits (12)= number of chickens (23)

Description: Let rabbits and chickens lift their feet at the same time, so that the total number of feet in the cage is reduced by two. Because the chicken has only two feet, only the rabbit's feet are left in the cage, and then the number of rabbits is two.

Expanding the information of keeping chickens and rabbits in the same cage is one of the famous interesting topics in ancient China, which was recorded in Sun Tzu's calculation. The problem of chickens and rabbits in the same cage is a common problem in primary school Olympic mathematics. Many elementary school arithmetic application problems can be transformed into such problems, or solved by a typical solution-"hypothesis method". So we should learn its solutions and ideas.

The first question of chickens and rabbits in the same cage:

(1) suppose they are all chickens, and there are

Number of rabbits = (actual number of feet -2 × total number of chickens and rabbits) ÷ (4-2)

(2) Assume that all rabbits have

Number of chickens = (4 × total number of chickens and rabbits-actual number of feet) ÷ (4-2)

The second problem of chickens and rabbits in the same cage:

(1) suppose they are all chickens, and there are

Number of rabbits = (2 × total number of chickens and rabbits-foot difference between chickens and rabbits) ÷ (4+2)

(2) Assume that all rabbits have

Number of chickens = (4 × total number of chickens and rabbits+foot difference between chickens and rabbits) ÷ (4+2)