Key Concepts and Formulas

• Permutations: The number of ways to arrange r objects from a set of n distinct objects, where order matters and repetition is not allowed, is given by $P(n, r) = \frac{n!}{(n-r)!}$.
• Divisibility Rule for 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

Step-by-Step Solution

Step 1: Calculate the total number of 3-digit numbers without repetition.

We need to find the total number of 3-digit numbers that can be formed using the digits 2, 3, 4, 5, and 7, without repetition. This is a permutation problem because the order of the digits matters. We are choosing 3 digits out of 5 and arranging them. The number of such permutations is given by: $P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60$ Therefore, there are 60 possible 3-digit numbers that can be formed from the digits 2, 3, 4, 5, and 7 without repetition.

Step 2: Identify combinations of digits whose sum is divisible by 3.

A number is divisible by 3 if the sum of its digits is divisible by 3. We need to find all combinations of 3 digits from the set {2, 3, 4, 5, 7} whose sum is divisible by 3. Let's list all possible combinations and their sums: \begin{itemize} \item 2 + 3 + 4 = 9 (divisible by 3) \item 2 + 3 + 5 = 10 \item 2 + 3 + 7 = 12 (divisible by 3) \item 2 + 4 + 5 = 11 \item 2 + 4 + 7 = 13 \item 2 + 5 + 7 = 14 \item 3 + 4 + 5 = 12 (divisible by 3) \item 3 + 4 + 7 = 14 \item 3 + 5 + 7 = 15 (divisible by 3) \item 4 + 5 + 7 = 16 \end{itemize} The combinations whose sums are divisible by 3 are: {2, 3, 4}, {2, 3, 7}, {3, 4, 5}, and {3, 5, 7}.

Step 3: Calculate the number of 3-digit numbers divisible by 3.

For each combination of 3 digits whose sum is divisible by 3, we can form $3! = 3 \times 2 \times 1 = 6$ different 3-digit numbers. Since we have 4 such combinations, the total number of 3-digit numbers divisible by 3 is: $4 \times 6 = 24$

Step 4: Calculate the number of 3-digit numbers not divisible by 3.

To find the number of 3-digit numbers that are not divisible by 3, we subtract the number of 3-digit numbers that are divisible by 3 from the total number of 3-digit numbers that can be formed. $60 - 24 = 36$

Step 5: Re-evaluate the problem statement and correct the initial answer The original answer given was 3, but our calculations clearly show the answer should be 36. Let's revisit the divisibility by 3 sums to see if there's an error. The combinations whose sums are divisible by 3 are: {2, 3, 4}, {2, 3, 7}, {3, 4, 5}, and {3, 5, 7}. The number of 3-digit numbers that can be formed from each set of 3 digits is: 3! = 6. So, the total number of 3-digit numbers divisible by 3 is: 4 sets * 6 permutations per set = 24. To find the 3-digit numbers not divisible by 3, we subtract the number of those divisible by 3 from the total number of 3-digit numbers: 60 - 24 = 36.

It appears the original answer provided ("Correct Answer: 3") is incorrect.

Common Mistakes & Tips

• Carefully list all possible combinations to avoid missing any.
• Remember that permutations are used when the order of the digits matters.
• Double-check the divisibility rule for 3.

Summary

We first calculated the total number of 3-digit numbers that can be formed using the given digits without repetition. Then, we identified the combinations of digits whose sum is divisible by 3 and calculated the number of 3-digit numbers that can be formed from these combinations. Finally, we subtracted the number of 3-digit numbers divisible by 3 from the total number of 3-digit numbers to find the number of 3-digit numbers not divisible by 3. The number of 3-digit numbers formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3, is equal to 36.

Final Answer

The final answer is \boxed{36}.