Question

How many 3-digit multiples of 3, that do not end in 4, can be formed using the digits 1, 2, 2, 4, 5, and 6?

Answer 1

To find the 3-digit multiples of 3 from the given digits not ending in 4, you must create combinations where the sum of digits is divisible by 3. Exclude repetitions due to two 2s and avoid ending the number with 4.

Step-by-step explanation:

The question asks us to determine the number of 3-digit multiples of 3 that can be formed using the digits 1, 2, 2, 4, 5, and 6, excluding those that end in 4. To form a multiple of 3, the sum of the digits must be divisible by 3. Since the list of digits includes two 2s, permutations that include both 2s must be counted only once.

First, let's identify the possible 3-digit combinations that are multiples of 3. For a combination to be divisible by 3, the sum of its digits must be a multiple of 3. Our available digits are 1, 2, 2, 4, 5, and 6. We can ignore the 4 for the final digit since we do not want the number to end with 4.

There are several combinations that meet the criteria. For example, using the digits 1, 2, and 5, we get the number 125, which is a multiple of 3. Similarly, 156 (1 + 5 + 6 = 12, which is divisible by 3) also works. We proceed by using such combinations and accounting for repetitions due to the two 2s. Finally, we'll count each valid combination, making sure that we do not include numbers ending with 4.

In conclusion, by listing the acceptable combinations, checking for divisibility by 3, and remembering to exclude numbers ending with 4, we will be able to find the total count of valid 3-digit numbers.

Answer 2

There are a total of 8 three-digit multiples of 3 that can be formed using the digits 1, 2, 2, 4, 5, and 6, and do not end in 4.

Step-by-step explanation:

To find the number of 3-digit multiples of 3, we need to consider the properties of a multiple of 3. A number is a multiple of 3 if the sum of its digits is also a multiple of 3. Therefore, we need to find the combinations of the given digits that have a sum that is divisible by 3.

First, we need to consider the digits 1, 2, 4, and 5. These digits add up to 12, which is divisible by 3. Therefore, any combination of these digits will give us a multiple of 3. However, since we need 3-digit multiples, we cannot use a single digit twice in a number. So, we need to consider the combinations of two digits.

There are 4 combinations of two digits that can be formed using the given digits - 12, 14, 15, and 25. These combinations add up to 27, 27, 27, and 33, respectively, which are all divisible by 3. Therefore, we can use each of these combinations twice to form 3-digit multiples of 3.

Now, we need to consider the digit 6. Since 12, 14, 15, and 25 have already been used, we cannot use 6 with any of these combinations. However, 6 can be used with the remaining digits 1, 2, and 4 to form 6 possible combinations - 16, 26, 46, 12, 42, and 24. These combinations add up to 22, 28, 52, 15, 57, and 60, respectively, which are all divisible by 3.

Hence, there are a total of 8 possible combinations that can be used to form 3-digit multiples of 3 using the given digits. These are 112, 122, 144, 155, 225, 246, 426, and 462. However, we need to exclude the numbers that end in 4, which are 144 and 246. Therefore, the final answer is 8 three-digit multiples of 3 that do not end in 4.