AMC 10 Daily Practice Round 3

Complete problem set with solutions and individual problem pages

Problem 9 Easy

Sally arranges the numbers from 1 to 8 in a circle and then calculates the sum of each set of 3 adjacent numbers. She would like an arrangement in which the smallest of these sums is as large as possible. If she chooses such an arrangement, what will this smallest sum be?

  • A.

    10

  • B.

    11

  • C.

    12

  • D.

    13

  • E.

    14

Answer:C

Let the smallest of the sums of three adjacent numbers be x.

Then adding from the number 1 around the circle and back to 1 adds 9 numbers in a row, so this value is at least 3x.

However these 9 numbers add to 1+1+2+3+\cdots +8 = 37.

Hence 37 ⩾ 3x so that 12\dfrac{1}{3}\geqslant x  and since x is a whole number, x ⩽ 12.

Once it is known that 12 has a chance of being the largest minimum, it is straightforward to find one of the many arrangements where all sums of 3 adjacent numbers (shown in blue) are 12 or more, such as the arrangement shown. This confirms that the smallest possible sum of 3 adjacent numbers is 12.