How many five digit numbers that are divisible by 125 can be formed using all the digits 2, 3, 8, 7 and 5 exactly once each?

Introduction / Context:
This question combines ideas from permutations and divisibility rules. We must count how many five digit numbers divisible by 125 can be formed when each digit from a given set is used exactly once. Understanding the divisibility rule for 125 and constructing valid endings is the key step, followed by counting permutations of remaining digits.

Given Data / Assumptions:
- The digits available are 2, 3, 8, 7 and 5. - Each five digit number must use all five digits exactly once. - The number must be divisible by 125.

Concept / Approach:
- A number is divisible by 125 if and only if its last three digits form a number that is divisible by 125. - The multiples of 125 between 100 and 999 that use only the digits from the given set are limited. - First identify valid three digit endings that are multiples of 125 and use only the available digits. - For each valid ending, arrange the remaining two digits in the first two places in all possible orders.

Step-by-Step Solution:
Step 1: Recall common three digit multiples of 125: 125, 250, 375, 500, 625, 750, 875. Step 2: From these, pick those that use only digits from the set {2, 3, 8, 7, 5}. The candidates are 375 and 875 because 3, 7, 5 and 8, 7, 5 are all available digits. Step 3: The ending 125 uses digit 1 which is not in the set. Endings 250, 500, 625 and 750 involve digit 0 or 6 which are not allowed by the digit set. Step 4: Therefore, the only possible last three digit blocks for our five digit numbers are 375 and 875. Step 5: Consider the ending 375. The digits used at the end are 3, 7 and 5, leaving digits 2 and 8 for the first two positions. Step 6: The first two positions can be arranged as 28 or 82. So with ending 375, the valid numbers are 28375 and 82375. Step 7: Now consider the ending 875. The digits used are 8, 7 and 5, leaving digits 2 and 3 for the first two positions. Step 8: The first two positions can be arranged as 23 or 32. So with ending 875, the valid numbers are 23875 and 32875. Step 9: In total, we have 2 numbers from ending 375 and 2 numbers from ending 875, giving 2 + 2 = 4 valid five digit numbers.

Verification / Alternative check:
Check one example numerically. 28375 / 125 = 227 exactly, so it is divisible by 125. Similarly, 82375 / 125 = 659, 23875 / 125 = 191 and 32875 / 125 = 263, each an integer. All four constructed numbers are divisible by 125, and no other multiples of 125 using only these digits exist.

Why Other Options Are Wrong:
Option B (5): This would require one more valid arrangement from some other three digit ending, but no other eligible ending uses only the digits from the given set. Option C (6): Overcounts by assuming more patterns than actually exist for the three digit endings. Option D (7): Significantly overestimates the count and ignores the strict divisibility condition. Option E (3): Undercounts, since we have explicitly constructed four distinct valid numbers.

Common Pitfalls:
- Forgetting that divisibility by 125 is determined solely by the last three digits, not the whole number at once. - Trying to permute all five digits freely and then test divisibility one by one instead of narrowing down valid endings first. - Overlooking that digits cannot be repeated and must be used exactly once.

Final Answer:
Exactly 4 five digit numbers can be formed using the digits 2, 3, 8, 7 and 5 once each that are divisible by 125.