AMC 8 Daily Practice Round 6

Complete problem set with solutions and individual problem pages

Problem 23 Medium

37 students line up to count consecutively starting at 1, with each subsequent number increasing by 3. One student erroneously subtracts 3 instead, resulting in a total sum of 2011. Which student made the counting error?

  • A.

    32

  • B.

    33

  • C.

    34

  • D.

    35

  • E.

    36

Answer:C

If all students counted correctly, the sequence forms an arithmetic progression with: First term a_1 = 1, Common difference d = 3 ，37^{\text{th}} term:   a_{37} = a_1 + d(n-1) = 1 + 3 \times (37-1) = 109.

The correct sequence would be 1, 4, 7, 10, \dots, 109. The total sum is: S_{37} = \frac{(a_1 + a_{37}) \times 37}{2} = \frac{(1 + 109) \times 37}{2} = 2035

The discrepancy between the correct sum and actual sum is: 2035 - 2011 = 24

Starting from the erroneous student, each subsequent student's number is 6 less than expected.

The number of affected terms is: \frac{24}{6} = 4

Thus, the error occurs at position: 34

Final result: \boxed{34}

Related Problems
Problem 21 AMC 8 Daily Practice Round 6
Problem 22 AMC 8 Daily Practice Round 6
Problem 24 AMC 8 Daily Practice Round 6
Problem 25 AMC 8 Daily Practice Round 6
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