Divisibility Rules: Finding Numbers Divisible By 3, 5, 4, & 9

Nov 2, 2025 by Admin 62 views

Hey guys! Let's break down how to find numbers that fit specific divisibility rules. We're going to tackle two main problems here: finding numbers in the form 4x3y that are divisible by both 3 and 5, and finding numbers in the form 1x2y3x that are divisible by 4 and 9. Don't worry, it's not as scary as it sounds! We'll go through each step nice and slow.

Part 1: Finding Numbers of the Form 4x3y Divisible by 3 and 5

Okay, so we need to find numbers that look like 4x3y, where x and y are digits, and these numbers have to be divisible by both 3 and 5. This means they need to play by the rules of both the '3 club' and the '5 club'. Let’s dig into what that means. When dealing with divisibility, understanding the rules for each number is crucial. Divisibility rules act like shortcuts, allowing us to quickly determine if a number can be divided evenly by another number without actually performing the division. This is super handy for problems like this one!

First, let's focus on the divisibility rule for 5. A number is divisible by 5 if its last digit is either 0 or 5. This gives us a big clue about what 'y' can be. Then, we'll consider the divisibility rule for 3. This rule states that a number is divisible by 3 if the sum of its digits is divisible by 3. Combining these two rules will help us narrow down the possibilities for 'x' and 'y'. Remember, we are looking for natural numbers, which are positive whole numbers (1, 2, 3, and so on).

Applying the Divisibility Rule of 5

The divisibility rule for 5 is pretty straightforward: a number is divisible by 5 if its last digit is either 0 or 5. In our case, the number is 4x3y, so 'y' is the last digit. This means 'y' can only be 0 or 5. So, we have two possibilities to explore: 4x30 and 4x35. This simplifies the problem a lot because now we only have to figure out the 'x' that makes these numbers work with the divisibility rule for 3.

Applying the Divisibility Rule of 3

Now, let's bring in the divisibility rule for 3. A number is divisible by 3 if the sum of its digits is divisible by 3. This means we need to add up the digits of our numbers (4, x, 3, and either 0 or 5) and see what values of 'x' make that sum a multiple of 3. For example, multiples of 3 are 3, 6, 9, 12, 15, and so on. Let's consider our two possibilities for 'y' separately.

Case 1: y = 0

If y is 0, our number is 4x30. The sum of the digits is 4 + x + 3 + 0 = 7 + x. Now we need to find values for 'x' that make 7 + x divisible by 3. Remember, 'x' is a single digit, so it can be any number from 0 to 9. Let's go through the possibilities:

If x = 0, 7 + 0 = 7 (not divisible by 3)
If x = 1, 7 + 1 = 8 (not divisible by 3)
If x = 2, 7 + 2 = 9 (divisible by 3!)
If x = 3, 7 + 3 = 10 (not divisible by 3)
If x = 4, 7 + 4 = 11 (not divisible by 3)
If x = 5, 7 + 5 = 12 (divisible by 3!)
If x = 6, 7 + 6 = 13 (not divisible by 3)
If x = 7, 7 + 7 = 14 (not divisible by 3)
If x = 8, 7 + 8 = 15 (divisible by 3!)
If x = 9, 7 + 9 = 16 (not divisible by 3)

So, when y = 0, the possible values for x are 2, 5, and 8. This gives us the numbers 4230, 4530, and 4830.

Case 2: y = 5

If y is 5, our number is 4x35. The sum of the digits is 4 + x + 3 + 5 = 12 + x. Now we need to find values for 'x' that make 12 + x divisible by 3. Let's go through the possibilities again:

If x = 0, 12 + 0 = 12 (divisible by 3!)
If x = 1, 12 + 1 = 13 (not divisible by 3)
If x = 2, 12 + 2 = 14 (not divisible by 3)
If x = 3, 12 + 3 = 15 (divisible by 3!)
If x = 4, 12 + 4 = 16 (not divisible by 3)
If x = 5, 12 + 5 = 17 (not divisible by 3)
If x = 6, 12 + 6 = 18 (divisible by 3!)
If x = 7, 12 + 7 = 19 (not divisible by 3)
If x = 8, 12 + 8 = 20 (not divisible by 3)
If x = 9, 12 + 9 = 21 (divisible by 3!)

So, when y = 5, the possible values for x are 0, 3, 6, and 9. This gives us the numbers 4035, 4335, 4635, and 4935.

Solution for Part 1

Combining both cases, the numbers of the form 4x3y that are divisible by both 3 and 5 are: 4230, 4530, 4830, 4035, 4335, 4635, and 4935.

Part 2: Finding Numbers of the Form 1x2y3x Divisible by 4 and 9

Now, let's move on to the second part of the problem: finding numbers in the form 1x2y3x that are divisible by both 4 and 9. This one is a bit trickier, but we can handle it! Again, we'll use the divisibility rules for 4 and 9 to figure out what digits 'x' and 'y' can be.

First up, the divisibility rule for 4 focuses on the last two digits of the number. The last two digits must be divisible by 4. Next, we'll use the divisibility rule for 9, which is similar to the rule for 3: the sum of all the digits must be divisible by 9. By applying these rules systematically, we can narrow down the possible values of 'x' and 'y'. Let's get to it!

Applying the Divisibility Rule of 4

The divisibility rule for 4 states that a number is divisible by 4 if its last two digits are divisible by 4. In our case, the number is 1x2y3x, so the last two digits are 3x. This means we need to figure out what values of 'x' make 3x divisible by 4. Remember, 'x' is a single digit.

Let's list the possibilities for 3x (keeping in mind that 'x' is a digit): 30, 31, 32, 33, 34, 35, 36, 37, 38, 39. Out of these, the numbers divisible by 4 are 32 and 36. So, 'x' can be either 2 or 6. This gives us two possible forms for our number: 122y32 and 162y36.

Applying the Divisibility Rule of 9

Now let's use the divisibility rule for 9. A number is divisible by 9 if the sum of its digits is divisible by 9. We have two cases to consider, based on the possible values of 'x' we found earlier.

Case 1: x = 2

If x = 2, our number is 122y32. The sum of the digits is 1 + 2 + 2 + y + 3 + 2 = 10 + y. Now we need to find values for 'y' that make 10 + y divisible by 9. Remember, 'y' is also a single digit (0 to 9).

Let's go through the possibilities:

If y = 0, 10 + 0 = 10 (not divisible by 9)
If y = 1, 10 + 1 = 11 (not divisible by 9)
If y = 2, 10 + 2 = 12 (not divisible by 9)
If y = 3, 10 + 3 = 13 (not divisible by 9)
If y = 4, 10 + 4 = 14 (not divisible by 9)
If y = 5, 10 + 5 = 15 (not divisible by 9)
If y = 6, 10 + 6 = 16 (not divisible by 9)
If y = 7, 10 + 7 = 17 (not divisible by 9)
If y = 8, 10 + 8 = 18 (divisible by 9!)
If y = 9, 10 + 9 = 19 (not divisible by 9)

So, when x = 2, the only possible value for y is 8. This gives us the number 122832.

Case 2: x = 6

If x = 6, our number is 162y36. The sum of the digits is 1 + 6 + 2 + y + 3 + 6 = 18 + y. Now we need to find values for 'y' that make 18 + y divisible by 9. Again, 'y' is a single digit (0 to 9).

Let's go through the possibilities:

If y = 0, 18 + 0 = 18 (divisible by 9!)
If y = 1, 18 + 1 = 19 (not divisible by 9)
If y = 2, 18 + 2 = 20 (not divisible by 9)
If y = 3, 18 + 3 = 21 (not divisible by 9)
If y = 4, 18 + 4 = 22 (not divisible by 9)
If y = 5, 18 + 5 = 23 (not divisible by 9)
If y = 6, 18 + 6 = 24 (not divisible by 9)
If y = 7, 18 + 7 = 25 (not divisible by 9)
If y = 8, 18 + 8 = 26 (not divisible by 9)
If y = 9, 18 + 9 = 27 (divisible by 9!)

So, when x = 6, the possible values for y are 0 and 9. This gives us the numbers 162036 and 162936.

Solution for Part 2

Combining both cases, the numbers of the form 1x2y3x that are divisible by both 4 and 9 are: 122832, 162036, and 162936.

Final Answer

Alright, we've cracked both parts of the problem!

a) The numbers of the form 4x3y divisible by 3 and 5 are: 4230, 4530, 4830, 4035, 4335, 4635, and 4935.

b) The numbers of the form 1x2y3x divisible by 4 and 9 are: 122832, 162036, and 162936.

I hope this breakdown helps you understand divisibility rules better! Remember, practice makes perfect, so try out some similar problems and you'll be a pro in no time!