What are the mathematical problems in the nine chapters of arithmetic

There are 246 mathematical problems in jiuzhangsuanshu, which are divided into nine chapters. Their main contents are as follows: the first chapter is "square field", which studies the calculation of acreage area; the second chapter is "corn", which studies the conversion of grain according to proportion; the third chapter is "declining Division", which studies the problem of proportional distribution; the fourth chapter is "Shaoguang", which studies the problem of "small and wide", In chapter five, we study the earth rock engineering and volume calculation; in chapter six, we study the reasonable apportionment of taxes; in chapter seven, we study the problem of surplus and deficiency, that is, the problem of double solutions; in chapter eight, we study the problem of first-order equations; in chapter nine, we use the Pythagorean theorem to solve the problem

There is a pool. The water surface is a square with a side length of 10 feet. In the center of the pool, there is a new reed, which is 1 foot higher than the water surface. If the reed is pulled vertically to the bank, its top just reaches the water surface of the bank. What is the depth of the pool and the length of the reed?

Let the depth of the pool be x feet, then the length of the reed is (x + 1) feet. According to the meaning of the title, we get: (102) & nbsp; 2 + x2 = (x + 1) 2, and the solution is: x = 12, so the length of the reed is: 12 + 1 = 13 (feet). A: the depth of the pool is 12 feet, and the length of the reed is 13 feet

There is a pool. The water surface is a square with a side length of 10 feet. In the center of the pool, there is a new reed, which is 1 foot higher than the water surface. If the reed is pulled vertically to the bank, its top just reaches the water surface of the bank. What is the depth of the pool and the length of the reed?

A natural number, using it to remove 63, 90130, has remainder, what is the smallest of the three remainder?

Suppose that the natural number is m, and the remainder of m after removing 63, 90130 is a, B, C respectively, then 63-a, 90-b, 130-c are all multiples of M
(63-a) + (90-b) + (130-c) = 283 - (a + B + C) = 283-25 = 258 is also a multiple of M. if 258 = 2 × 3 × 43, it may be 2 or 3 or 6 or 43;
A + B + C = 25, so at least one of a, B, C is greater than 8;
According to the fact that the divisor must be greater than the remainder, we can determine = 43. Thus a = 20, B = 4, C = 1. Obviously, 1 is the smallest of the three remainder
So the answer is: 1

There is a natural number, which can be used to remove 63, 90130 and have remainder respectively. The sum of the three remainder is 25. What is the largest of the three remainder?

Let this natural number be m, and the remainders obtained by removing 63, 90130 from m are a, B, and C respectively. Then 63-a, 90-b, and 130-c are all multiples of M. we can get that: (63-a) + (90-b) + (130-c) = 283 - (a + B + C) = 283-25 = 258 is also a multiple of M. and 258 = 2 × 3 × 43. Then it may be 2 or 3 or 6 or 43; a +

If 1059, 1417 and 2312 are divided by natural number x respectively, the remainder is y, then the value of X-Y is equal to () (the following four options)
A.15
B.1
C.164
D.174

Let the quotient of three known numbers divided by X be natural numbers a, B and C respectively
ax+y=1059,①
bx+y=1417,②
cx+y=2312.③
② (B-A) x = 358 = 2 × 179, 4
③ (C-B) x = 895 = 5 × 179, ⑤
⑤ (C-A) x = 1253 = 7 × 179. 6
From formula 4, 5 and 6, we can know that x = 179, and then we can easily get y = 164,
So X-Y = 179-164 = 15

A two digit number divided by 10 is 9, divided by 9 is 8, what is the two digit number

This number plus 1 divides 10 and 9, so this number is 10x9-1 = 89

What is the quotient of 124 divided by 8 and the remainder is 4?

A:
fifteen
Methods: (124-4) / 8 = 15

What number divided by 15, the quotient and the remainder are equal, and what number can the divisor be

16 32 48 .
The divisor can be 16N (n is an integer)

The sum of quotient and remainder is 100, the divisor is (), the quotient is ()

If an integer is divided by 15 and there are 2, then the integer can be set as X
(X-2) / 15 = k, K is any integer
(x-2)=15k
If the sum of the quotient and the remainder is 100, there is:
k+2=100
k=98
x=15k+2=15*98+2=1472
That is, the divisor is 1472 and the quotient is 98

What are the mathematical problems in the nine chapters of arithmetic