2003 AIME II Problem 2 | Summary and Q&A

TL;DR

The problem asks for the remainder when the greatest integer multiple of eight, with no repeated digits, is divided by 1000.

Key Insights

• 🔁 The problem involves finding the greatest integer multiple of eight with no repeated digits.
• 🪈 By arranging the digits in descending order (9876543210), the maximum value is obtained.
• ✅ Divisibility by eight can be checked by considering the last three digits of the number.
• #️⃣ Rearranging the last three digits (210 to 120) yields a number divisible by eight.

Transcript

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Questions & Answers

Q: How do we approach finding the greatest integer multiple of eight with no repeated digits?

To find the greatest integer multiple, we select the largest possible digits and arrange them in descending order (9876543210) to maximize the number's value.

Q: How do we check for divisibility by eight?

Instead of assessing the entire number, we focus on the last three digits. In this case, by rearranging 210 to 120, it becomes divisible by eight.

Q: Why is it important to avoid borrowing numbers from the larger digits when rearranging?

Borrowing from the larger digits would reduce their values and ultimately diminish the magnitude of the resulting number. To achieve the greatest integer multiple, it's necessary to keep the larger digits as large as possible.

Q: What is the remainder when the derived number (9876543210) is divided by 1000?

The remainder is simply the value of the number in the hundreds, tens, and ones place, which in this case is 120.

Summary & Key Takeaways

• The problem requires finding the greatest integer multiple of eight with no repeated digits.

• By selecting the largest possible digits and arranging them, a number (9876543210) is derived.

• To determine divisibility, the last three digits are considered, and rearranging them results in 120, which is divisible by eight.

• The remainder when the derived number (9876543210) is divided by 1000 is 120.