2020 AMC 8

Complete problem set with solutions and individual problem pages

Problem 16 Hard

Each of the points A,B,C,D,E, and F in the figure below represents a different digit from 1 to 6. Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is 47. What is the digit represented by B?

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

  • E.

    5

Answer:E

Solution 1

We can form the following expressions for the sum along each line:

\begin{cases} A + B + C \\ A + E + F \\ C + D + E \\ B + D \\ B + F \end{cases}

Adding these together, we must have 2A+3B+2C+2D+2E+2F=47, i.e. 2(A+B+C+D+E+F)+B=47. Since A,B,C,D,E,F are unique integers between 1 and 6, we obtain A+B+C+D+E+F=1+2+3+4+5+6=21 (where the order doesn't matter as addition is commutative), so our equation simplifies to 42 + B = 47. This means B = \boxed{\textbf{(E) }5}.

 

Solution 2

Following the first few steps of Solution 1, we have 2(A+C+D+E+F)+3B=47. Because an even number 2(A+C+D+E+F) subtracted from an odd number (47) is always odd, we know that 3B is odd, showing that B is odd. Now we know that B is 1, 3, or 5. If we try B=1, we get 43=47, which is false. Testing B=3, we get 45=47, which is also false. Therefore, we have B = \boxed{\textbf{(E) }5}.