Divisibility Rules Practice Quiz

The number 236 is divisible by 2 because it ends in an even digit, which means it is an even number. In contrast, the other numbers listed (123, 237, and 345) do not end in even digits and are therefore not divisible by 2.

A number is divisible by another if it can be divided evenly, without leaving a remainder. 36 is divisible by 2 because 36 ÷ 2 = 18, which is a whole number. However, 36 is not divisible by 10 (36 ÷ 10 = 3.6), 5 (36 ÷ 5 = 7.2), or 7 (36 ÷ 7 ≈ 5.14), as none of these result in a whole number. Therefore, 2 is the correct answer.

1000 can be divided evenly by both 2 and 5. When a number is divisible by 2, it means that it can be divided into two equal parts without leaving any remainder. Similarly, when a number is divisible by 5, it means that it can be divided into five equal parts without leaving any remainder. In the case of 1000, it can be divided by 2 to give 500 and by 5 to give 200. Therefore, the correct answer is that 1000 is divisible by both 2 and 5.

For a number to be divisible by both 2 and 3, it must meet the following conditions:

Divisible by 2: The number must be even (i.e., its last digit must be 0, 2, 4, 6, or 8).

Divisible by 3: The sum of its digits must be divisible by 3.

Let’s check each number:

126:

Last digit is 6 (even), so it is divisible by 2.

Sum of digits = 1 + 2 + 6 = 9, which is divisible by 3.

Therefore, 126 is divisible by both 2 and 3.

100:

Last digit is 0 (even), so it is divisible by 2.

Sum of digits = 1 + 0 + 0 = 1, which is not divisible by 3.

Therefore, 100 is not divisible by both 2 and 3.

123:

Last digit is 3 (odd), so it is not divisible by 2.

Therefore, 123 is not divisible by both 2 and 3.

133:

Last digit is 3 (odd), so it is not divisible by 2.

Therefore, 133 is not divisible by both 2 and 3.

Thus, the correct answer is 126.

The number 203 is not divisible by 3 because the sum of its digits (2+0+3) is not divisible by 3. Additionally, 203 is divisible only by 7. 

The number 15000 can be divided evenly by 2 because it is an even number. It can also be divided evenly by 3 because the sum of its digits (1+5+0+0+0) is divisible by 3. Additionally, it can be divided evenly by 5 because it ends in 0 or 5. Therefore, all of the given numbers (2, 3, and 5) can divide 15000 evenly.

A number is divisible by 9 if the sum of its digits is divisible by 9.

7 + 8 + 5 + 3 + 1 = 24. 24 is not divisible by 9.

7 + 8 + 5 + 3 + 2 = 25. 25 is not divisible by 9.

7 + 8 + 5 + 3 + 4 = 27. 27 is divisible by 9.

Therefore, 78534 is divisible by 9.

For a number to be divisible by both 3 and 9, the sum of its digits must be divisible by 9. Let's check each number:

78531:

Sum of digits = 7 + 8 + 5 + 3 + 1 = 24

24 is divisible by 3, but not by 9.

Therefore, 78531 is divisible by 3 but not by 9.

78534:

Sum of digits = 7 + 8 + 5 + 3 + 4 = 27

27 is divisible by both 3 and 9.

Therefore, 78534 is divisible by both 3 and 9.

3333:

Sum of digits = 3 + 3 + 3 + 3 = 12

12 is divisible by 3, but not by 9.

Therefore, 3333 is divisible by 3, but not by 9.

1234:

Sum of digits = 1 + 2 + 3 + 4 = 10

10 is not divisible by either 3 or 9.

Thus, the correct answer is 78534.

If the sum of the digits of a number is divisible by 3, then the number will be divisible by 3 as well. This is because when a number is divisible by 3, the sum of its digits is also divisible by 3. Therefore, if the sum of the digits of a number is divisible by 3, it implies that the number itself is divisible by 3.

The numbers that are divisible by 2, 3, and 10 at the same time are called common multiples. In this case, 30 and 120 are the common multiples. Both 30 and 120 can be divided evenly by 2, 3, and 10, making them the correct answers.