28.1k views
5 votes
What is the value of the digit k that will make the number 481, 5k6 divisible by 2, 3, 4, and 9?

User DurkoMatko
by
8.3k points

1 Answer

3 votes

Final answer:

The value of k that makes the number 481, 5k6 divisible by 2, 3, 4, and 9 is 3. This is based on divisibility rules for these numbers, requiring the sum of the digits to be divisible by 3 and 9, and the last two digits to be divisible by 4.

Step-by-step explanation:

The value of the digit k that makes the number 481, 5k6 divisible by 2, 3, 4, and 9 must satisfy certain divisibility rules for each of these numbers:

Divisibility by 2: The last digit of the number must be even. Here, it is 6, so the number is divisible by 2.

Divisibility by 3: The sum of the digits must be divisible by 3. For the number 481,5k6, the sum of the digits is 4+8+1+5+k+6.

Divisibility by 4: The last two digits of the number must form a number divisible by 4. Since the last digit is 6, k must be 2 or 6 to make the last two digits either 26 or 66, both divisible by 4.

Divisibility by 9: The sum of the digits must be divisible by 9. This is a more restrictive rule compared to divisibility by 3, and we must consider it to find the correct value of k.

According to the sum for divisibility by 3 and 9, we have 4+8+1+5+k+6 = 24+k. In order for this sum to be divisible by 9, the only possible values for k from the earlier step (2 or 6) that would work are 3 or 6 because 24+3=27 and 24+6=30, both of which are divisible by 9. However, since the number must also be divisible by 4 and 4826 is not divisible by 4, the correct value of k must be 3. Therefore, the value of k that makes 481,536 divisible by 2, 3, 4, and 9 is 3.

User NeilG
by
7.4k points

Related questions

asked Feb 16, 2020 46.7k views
Brad Colthurst asked Feb 16, 2020
by Brad Colthurst
8.4k points
2 answers
5 votes
46.7k views