Find the least number with mixed remainder conditions The smallest number leaves remainder 3 when divided by 5, 6, 7, and 8, but leaves no remainder when divided by 9. What is that number?

Introduction / Context:
This is a Chinese Remainder Theorem style problem in an elementary form. We use LCM to handle the common remainder part, then apply the additional divisibility condition with 9 to select the correct instance.

Given Data / Assumptions:

• N ≡ 3 (mod 5, 6, 7, 8).
• N ≡ 0 (mod 9).

Concept / Approach:
Numbers satisfying the first set can be written as N = LCM(5, 6, 7, 8) * k + 3. Then we search for k such that N is divisible by 9.

Step-by-Step Solution:
LCM(5, 6, 7, 8) = 840.Hence N = 840k + 3.Impose divisibility by 9: 840k + 3 ≡ 0 (mod 9). Since 840 ≡ 3 (mod 9), we have 3k + 3 ≡ 0 ⇒ k + 1 ≡ 0 (mod 3) ⇒ k ≡ 2 (mod 3).Smallest k = 2 ⇒ N = 840 * 2 + 3 = 1683.

Verification / Alternative check:
1683 ÷ 5 leaves 3; ÷ 6 leaves 3; ÷ 7 leaves 3; ÷ 8 leaves 3; and 1683 is divisible by 9 because 1 + 6 + 8 + 3 = 18, a multiple of 9.

Why Other Options Are Wrong:
1677, 2523, 3363, and 8403 violate at least one of the remainder conditions or the divisibility by 9 condition.

Common Pitfalls:
Using the sum 5 + 6 + 7 + 8 instead of LCM, or forgetting to test the added divisibility requirement by 9.

Final Answer:
1683