AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 14 Hard

Mr. Wang wrote a sequence of consecutive natural numbers starting from 1 on the blackboard: 1, 2, 3, 4, \cdots. Then he erased three numbers, two of which were prime numbers. If the average of the remaining numbers is 19\frac{8}{9}, what is the maximum possible sum of the two prime numbers that were erased?

  • A.

    40

  • B.

    50

  • C.

    60

  • D.

    70

  • E.

    80

Answer:C

1 + 2 + 3 + \ldots + n = \frac{(1 + n) \cdot n}{2}

The average of these numbers is: \frac{\frac{(1 + n) \cdot n}{2}}{n} = \frac{1 + n}{2}

So the average is around 20, which suggests that n is around 40.

Since 3 numbers were erased, the number of remaining numbers should be a multiple of 9.

Therefore, n = 39.

The sum of numbers from 1 to 39 is: 1 + 2 + 3 + \ldots + 39 = 780

After erasing 3 numbers, 36 remain. The total sum of the remaining numbers is: 19\frac{8}{9} \times 36 = 716

So the sum of the 3 erased numbers is: 780 - 716 = 64

Among the prime numbers less than 39, the maximum possible sum of two primes that add up to no more than 64 is: 37 + 23 = 60 \quad \text{or} \quad 31 + 29 = 60.

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