A math thinking question, a math thinking question

The following is a personal opinion.

There are 32 books in total, (17+7+6+2=), and they have to be divided into 4 equal parts at the end, so in the end, each pile must have (32 4=) 8 books.

Therefore, we observe the number of each stack, 17 7 6 2, two odd numbers and two even numbers.

And the stack of only 2 books accounted for two moves. The stack of six books is to be moved twice, the first time two books are removed, and the second time four books are brought in from other piles.

Two even-numbered piles, whether moved out or in, are moving even-numbered books.

Therefore, four shifts cannot satisfy the requirement of turning two odd number piles into even numbers, and at least five shifts are required to meet the requirements of the problem.

There are four copies. 17, 7, 6, 2 are added to get 32, divided by four to get 8 8, which is the final number of each pile: 17 move 7 to 7 7 becomes 14 17 becomes 10 14 shift 6 to 6 14 becomes 8 6 becomes 12 10 moves 2 to 2 10 becomes 8 2 becomes 4 12 moves 4 4 12 moves 4 to 4 12 becomes 8 4 becomes 8

The first floor is right, and the following is a random list of several moves 5 times.

I call 17 the first pile, 7 the second pile, 6 the third pile, and 2 the fourth pile.

The following list: 1. The first pile - 7 books - the second pile (the first pile takes 7 copies to the fourth pile, at this time the second pile is 14 books, and the first pile is 10 books);

2. The first pile--- 2 copies--- the fourth pile (8 copies of the first pile, 4 copies of the fourth pile) 3, 6 copies ---of the second pile--- 3rd pile (8 copies of the second pile, 12 copies of the third pile) 4, 4 copies of the third pile--- 4 copies--- fourth pile (8 copies of the third pile, 8 copies of the fourth pile) and finally eight copies.

Ideas: A total = 17 + 7 + 6 + 2 = 32 books, 32 4 = 8 books, so the main thing is how to make up 8.

2 is the smallest, and basically needs other piles to move into 2 piles twice.

Then try a few more times and you're good to go.

It is also possible to push it backwards backwards.

Initial 17 7 6 2

1st move 17 - > 7 : 10 14 6 2 2nd move 10 - > 2 : 8 14 6 4 3rd move 14 - > 6 : 8 8 12 4 4 4 4 12 - > 4 : 8 8 8 8

Let the cube root of s s=1995 * x = the cube root of 1996 * y = the cube root of 1997*z, then 1995x +1996y + 1997z = (the cube root of 1995 + the cube root of 1996 + the cube root of 1997) * s, so s = (the cube of 1995 contains the noisy root + the cube root of 1996 + the cube root of 1997), s=the cube root of 1995 + the cube root of 1996 + the cube root of 1997, 1 x =The cube root of 1995 s, the cube root of 1 y = 1996 s, the cube root of 1 z = 1997 s, 1 x + 1 y + 1 z = 1

Let 1995x = 1996y = 1997z = k, i.e., 1995=k x, 1996=k y, 1997=k z (1995x +1996y +1997z) = k x + k y + k z1995 cube root + 1996 cube root + 1997 cube root.

The cube root of k*1 x+k*1 y+k's cube root leaks*1 zk's cube root (1 x+1 y+1 z) (1995x +1996y +1997z) = k's cube root * (1 x+1 y+1 z).

The cube root of (1995x +1996y +1997z) = the cube root of 1995 + the cube root of 1996 + the cube root of 1997.

The cube root of k * (1 x + 1 y + 1 z) = the cube root of k * (1 x + 1 y + 1 z).

1 x+1 y+1 z=(1 x+1 y+1 z).

And xyz 0, 1995x = 1996y = 1997z x bright file 0, y 0, z 0

And the positive number of the cube root equal to itself is only 1

i.e. 1 x+1 y+1 z=1

The volume of the rising water surface is equal to the volume of the body speed, so 240*80*4=40*40*h

The solution is h=48 centimeters=decimeters.

923970, hello:

Iron block volume: 240 80 4 76800 (cubic centimeters).

The height of the iron block is: 76800 (40 40) 48 (cm).

The volume of the iron block: 240 80 4 = 76800 (cubic centimeters).

The height of the iron block: 76800 40 40 = 48 (cm).

The first 1+1=1 should be 1=1, so you can observe the following relationship.

1+3+5+……2n-1=n squared.