AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 13 Medium

Jane plans to enter the numbers 1, 2, \cdots, N into the computer to calculate their average. When she thought she had finished entering the numbers, the computer showed that only (N - 1) numbers had been entered, and the average was 35\frac{5}{7}. Assuming the (N - 1) numbers were entered correctly, what is the missing number?

  • A.

    10

  • B.

    53

  • C.

    56

  • D.

    65

  • E.

    67

Answer:C

First, estimate the value of N:

 

If the missing number is N (the largest possible value), then the average would be

\frac{1 + 2 + \cdots + (N - 1)}{N - 1} = \frac{N}{2};

 

If the missing number is 1 (the smallest possible value), then the average would be

\frac{2 + 3 + \cdots + N}{N - 1} = \frac{N}{2} + 1.

 

This shows that the actual average, 35\frac{5}{7}, must lie between \frac{N}{2} and \frac{N}{2} + 1.

That means N can only be 70 or 71.

 

Moreover, since the fraction 35\frac{5}{7} must be derived from a total divided by (N - 1),

and the denominator (N - 1) must be divisible by 7, we conclude that N = 71.

 

Now calculate:

35\frac{5}{7} \times (N - 1) = 35\frac{5}{7} \times 70 = 2500.

 

This should equal the sum from 1 to 71 minus the missing number.

The sum from 1 to 71 is \frac{71 \times 72}{2} = 2556,

so the missing number is:

2556 - 2500 = 56.

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