Problem 6 - Intermediate Algebra

Question

In the following exercises, use the divisibility tests to determine whether each number is divisible by 2 , by \(3,\) by \(5,\) by \(6,\) and by 10 . 39,075

Step-by-Step Solution

Answer

39,075 is divisible by 3 and 5 only.

Step 1: Divisibility by 2

A number is divisible by 2 if its last digit is even. The last digit of 39,075 is 5. Since 5 is not an even number, 39,075 is not divisible by 2.

Step 2: Divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3. Sum the digits of 39,075: 3 + 9 + 0 + 7 + 5 = 24. Since 24 is divisible by 3, 39,075 is also divisible by 3.

Step 3: Divisibility by 5

A number is divisible by 5 if its last digit is 0 or 5. The last digit of 39,075 is 5, so 39,075 is divisible by 5.

Step 4: Divisibility by 6

A number is divisible by 6 if it is divisible by both 2 and 3. From Step 1, 39,075 is not divisible by 2. Therefore, 39,075 is not divisible by 6.

Step 5: Divisibility by 10

A number is divisible by 10 if its last digit is 0. The last digit of 39,075 is 5. Hence, 39,075 is not divisible by 10.

Key Concepts

Divisibility by 2

To check if a number is divisible by 2, you need to look at its last digit. A number is divisible by 2 if its last digit is an even number. Even numbers are 0, 2, 4, 6, and 8. If the last digit is any of these, the number is divisible by 2. For example, if the number ends in 4, like 1234, it is divisible by 2.
In the exercise, we checked the last digit of 39,075 and found it to be 5. Since 5 is not an even number, 39,075 is not divisible by 2.

Divisibility by 3

To determine if a number is divisible by 3, sum all its digits. If the sum of the digits is divisible by 3, then the original number is also divisible by 3.
For instance, consider the number 123. First, add the digits: 1 + 2 + 3 = 6. Since 6 is divisible by 3, we say that 123 is also divisible by 3.
In the given exercise, we summed the digits of 39,075: 3 + 9 + 0 + 7 + 5 = 24. Since 24 is divisible by 3, 39,075 is also divisible by 3.

Divisibility by 5

A number is divisible by 5 if its last digit is either 0 or 5. This is perhaps one of the easiest divisibility rules to remember.
For example, numbers like 40, 75, and 100 are all divisible by 5 because their last digits are either 0 or 5.
In the exercise example, the number 39,075 ends in 5. Therefore, it is divisible by 5.

Divisibility by 6

To find out if a number is divisible by 6, it must meet two conditions: it has to be divisible by both 2 and 3.

• First, check for divisibility by 2 by looking at the last digit. If it is even, proceed to the next step.
• Second, check for divisibility by 3 by adding up the digits and ensuring the sum is divisible by 3.

For example, take the number 42. Its last digit, 2, is even, fulfilling the rule for divisibility by 2. Summing its digits gives us 4 + 2 = 6, and since 6 is divisible by 3, the number 42 is divisible by 6.
In the exercise case, we found that 39,075 was not divisible by 2 (from Step 1). Therefore, 39,075 is not divisible by 6.

Divisibility by 10

You can determine if a number is divisible by 10 by checking if its last digit is 0. This rule is very straightforward and easy to apply.
For instance, numbers like 20, 50, and 100 all end in 0 and are therefore divisible by 10.
In the provided exercise, the last digit of 39,075 is 5. Since it does not end in 0, 39,075 is not divisible by 10.