2022 AMC 8

Complete problem set with solutions and individual problem pages

Problem 6 Easy

Three positive integers are equally spaced on a number line. The middle number is 15, and the largest number is 4 times the smallest number. What is the smallest of these three numbers?

  • A.

    4

  • B.

    5

  • C.

    6

  • D.

    7

  • E.

    8

Answer:C

Solution 1

Let the smallest number be x. It follows that the largest number is 4x.

Since x,15, and 4x are equally spaced on a number line, we have

\begin{align*} 4x-15 &= 15-x \\ 5x &= 30 \\ x &= \boxed{\textbf{(C) } 6}. \end{align*}

 

Solution 2

Let the common difference of the arithmetic sequence be d. Consequently, the smallest number is 15-d and the largest number is 15+d. As the largest number is 4 times the smallest number, 15+d=60-4d\implies d=9. Finally, we find that the smallest number is 15-9=\boxed{\textbf{(C) } 6}.

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