Question

let the natural number n have the decimal numeral 123,454,26d, where d is the units digit. use divisibility tests to parts (a) through (h) below.

0,2,4,6,8 (use a comma to separate answers as needed.)

(b) for what value(s) of d is n divisible by 3?

0,3,6,9 (use a comma to separate answers as needed.)

(c) for what value(s) of d is n divisible by 4?

0,4,8 (use a comma to separate answers as needed.)

(d) for what value(s) of d is n divisible by 5?

0,5 (use a comma to separate answers as needed.)

(e) for what value(s) of d is n divisible by 6?

0,6 (use a comma to separate answers as needed.)

(f) for what value(s) of d is n divisible by 8?

0, (use a comma to separate answers as needed.)

Explanation:

Analyze the number and divisibility criteria

The natural number is \(n = 123,454,26d\), where \(d\) is a single digit: \(d \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}\).
We need to solve parts (b), (c), (d), (e), and (f) based on the visible questions.

Solve part (b): Divisibility by 3

A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
The sum of the digits of \(n\) is:

$$ 1 + 2 + 3 + 4 + 5 + 4 + 2 + 6 + d = 27 + d $$

For \(27 + d\) to be divisible by 3, since 27 is already divisible by 3, \(d\) must be a multiple of 3.
Thus, \(d \in \{0, 3, 6, 9\}\).

Solve part (c): Divisibility by 4

A number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4.
The last two digits of \(n\) form the number \(2d\), which represents \(20 + d\).
We test values of \(d \in \{0, 1, \dots, 9\}\) such that \(20 + d\) is divisible by 4:

• For \(d = 0\): \(20\) is divisible by 4.
• For \(d = 4\): \(24\) is divisible by 4.
• For \(d = 8\): \(28\) is divisible by 4.

Thus, \(d \in \{0, 4, 8\}\).

Solve part (d): Divisibility by 5

A number is divisible by 5 if and only if its units digit is 0 or 5.
Since \(d\) is the units digit, we must have:

$$ d \in \{0, 5\} $$

Solve part (e): Divisibility by 6

A number is divisible by 6 if and only if it is divisible by both 2 and 3.

• For divisibility by 2, the units digit \(d\) must be even: \(d \in \{0, 2, 4, 6, 8\}\).
• For divisibility by 3, from part (b), we have \(d \in \{0, 3, 6, 9\}\).

Taking the intersection of these two sets:

$$ d \in \{0, 6\} $$

Solve part (f): Divisibility by 8

A number is divisible by 8 if and only if the number formed by its last three digits is divisible by 8.
The last three digits of \(n\) form the number \(26d\), which represents \(260 + d\).
We divide 260 by 8 to find the remainder:

$$ 260 = 8 \times 32 + 4 $$

So, \(260 + d = 8 \times 32 + (4 + d)\).
For this to be divisible by 8, the term \(4 + d\) must be a multiple of 8.
Since \(d\) is a single digit (\(0 \le d \le 9\)), the only value that satisfies this is:

$$ 4 + d = 8 \implies d = 4 $$

Thus, \(d \in \{4\}\).

Answer:

Question b

0,3,6,9

Question c

0,4,8

Question d

0,5

Question e

0,6

Question f

4