AMC 8 Daily Practice Round 6

Complete problem set with solutions and individual problem pages

Problem 20 Easy

How many integers between 401 and 1000, inclusive, leave a remainder of 1 when divided by 8?

  • A.

    72

  • B.

    73

  • C.

    74

  • D.

    75

  • E.

    76

Answer:D

These integers form an arithmetic sequence with first term 1, common difference 8, and general term 8n-7.

Within the range 401–1000:

The smallest term is 401 (when n=51: 8 \times 51 - 7 = 401).

The largest term is 993 (when n=125: 8 \times 125 - 7 = 993).

The number of terms is calculated as: \frac{993 - 401}{8} + 1 = \frac{592}{8} + 1 = 74 + 1 = 75

Final result: \boxed{75}

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