Lesson OVERVIEW of Divisibility rules by 2, 3, 4, 5, 6, 9, 10 and 11

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OVERVIEW of Divisibility rules by 2, 3, 4, 5, 6, 9, 10 and 11

The rules for divisibility by 2, 3, 4, 5, 6, 9, 10, and 11 are

    An integer number is divisible by 2 if and only if its last digit is divisible by 2.

    An integer number is divisible by 3 if and only if the sum of its digits is divisible by 3.

    An integer number is divisible by 4 if and only if the number formed by its two last digits is divisible by 4.

    An integer number is divisible by 5 if and only if its last digit is 5 or 0.

    An integer number is divisible by 6 if and only if its last digit is even and the sum of the digits is divisible by 3.

    An integer number is divisible by 9 if and only if the sum of its digits is divisible by 9.

    An integer number is divisible by 10 if and only if its last digit is 0.

    An integer number is divisible by 11 if and only if the alternate sum of its digits is divisible by 11.

(The alternate sum is the algebraic sum of digits where the sign is changed to the opposite from each digit to the next one).

My lessons in this site on divisibility numbers are

    - Divisibility by 2 rule
    - Divisibility by 3 rule
    - Divisibility by 4 rule
    - Divisibility by 5 rule
    - Divisibility by 6 rule
    - Divisibility by 9 rule
    - Divisibility by 10 rule
    - Divisibility by 11 rule
    - Restore the omitted digit in a number in a way that the number is divisible by 9
    - Restore the omitted digit in a number in a way that the number is divisible by 11
    - Can there be a perfect square ?
    - Math circle level problems on divisibility numbers
    - Math circle level problem on restoring digit in the product of two 16-digit numbers
    - Math circle level problem on finding remainders

Below are short annotations to the lessons with problems.

Restore the omitted digit in a number in a way that the number is divisible by 9

        Problem 1 In the number 12345_67 one digit was omitted, and you see the blank placeholder in the corresponding position.
                           Restore the digit in the number in a way that the number is divisible by 9.

        Problem 2. In the number 12345_678 one digit was omitted, and you see the blank placeholder in the corresponding position.
                           Restore the digit in the number in a way that the number is divisible by 9.
                           Is the solution unique? How many solutions are there?

Restore the omitted digit in a number in a way that the number is divisible by 11

        Problem 1 In the number 123456_87 one digit was omitted, and you see the blank placeholder in the corresponding position.
                           Restore the digit in the number in a way that the number is divisible by 11.

        Problem 2 In the number 12349_92 one digit was omitted, and you see the blank placeholder in the corresponding position.
                           Restore the digit in the number in a way that the number is divisible by 11.

Can there be a perfect square ?

        Problem 1 Can there be a perfect square whose digits consist of exactly 4 ones, 4 twos and 4 zeros in any order?

Math circle level problems on divisibility numbers

        Problem 1 Find the number of 7-digit positive integers, where the sum of the digits is divisible by 3.

        Problem 2 Find the number of 7-digit numbers, where the sum of the digits is divisible by 9.

        Problem 3 A five digit number is the fourth power of an integer. The sum of the first, third, and fifth digits
                           equals the sum of the second and fourth digits. What digit is in the thousands place ?

Math circle level problem on restoring digit in the product of two 16-digit numbers

        Problem 1 Here is a multiplication involving two 16-digit numbers:
                                   3851902343886132 * 5221791683705111
                           The 32-digit product is   20 113 831 625 748 8_8 690 240 050 420 652.
                           What is the missed digit in the product?

Math circle level problem on finding remainders

        Problem 1 Find the remainder when 2 raised to 512 degree is divided by 14.

Lessons that are closely adjacent to these are

    - Product of two consecutive integers is divisible by 2
    - Product of three consecutive integers is divisible by 6
    - Problems dealing with the product of two consecutive integers
    - Problems dealing with the product of three consecutive integers

When you learn the listed lessons on divisibility rules, you will be able to formulate and to establish many other simple and useful divisibility rules like these:

    An integer number is divisible by 8 if and only if the number formed by its three last digits is divisible by 8.

    An integer number is divisible by 15 if and only its last digit is 5 or 0 and the sum of its digits is divisible by 3.

    An integer number is divisible by 20 if and only if its two last digits are 20, 40, 60, 80 or 00.

    An integer number is divisible by 25 if and only if its two last digits are 25, 50, 75 or 00.

    An integer number is divisible by 50 if and only if its two last digits are 50 or 00.

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