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Vertical calculation problem with remainder division
Vertical calculation problems with remainder division, such as:

I. 2 1 ÷ 5 = 4... 1

Second, 32 ÷ 6 = 5...2

In the division of integers, there are only two situations: divisible and non-divisible. When it is not divisible, it will produce a remainder, so the remainder problem is very important in primary school mathematics.

When it is not divisible, a remainder is generated. The remainder operation: a mod b = c(b is not 0) means that the remainder obtained by dividing the integer a by the integer b is c, for example, 7 ÷ 3 = 2 1.

The specific problem solving process is shown in the figure of this article, as follows:

Extended data:

The remainder has the following important properties (A, B and C are all natural numbers):

1, the absolute value of the difference between the remainder and the divisor is less than the absolute value of the divisor (applicable to real number fields);

Dividend = Divider× Quotient+Remainder;

Divider = (dividend-remainder) ÷ quotient;

Quotient = (dividend-remainder) divider;

Remainder = dividend-divisor × quotient.

2. If the remainder of a and b divided by c is the same, then the difference between a and b can be divisible by c. For example, the remainder of 17 and1divided by 3 is 2, then17-1can be divisible by 3.

3. The sum of A and B divided by the remainder of C (except when A and B divided by C have no remainder) is equal to the sum of the remainder of A and B divided by C respectively (or the remainder of this sum divided by C).

For example, 23, the remainder of 16 divided by 5 is 3 and 1 respectively, so the remainder of (23+ 16) divided by 5 is equal to 3+ 1=4. Note: When the sum of remainders is greater than the divisor, the remainders are equal to the sum of remainders and divided by the remainder of c ... For example, the remainders of 19 divided by 5 are 3 and 4 respectively, so the remainders of (23+ 19) divided by 5 are equal to the remainders of (3+4) divided by 5.