2014 AMC 8

Complete problem set with solutions and individual problem pages

Problem 21 Hard

The 7-digit numbers \underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1} and \underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C} are each multiples of 3. Which of the following could be the value of C?

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    5

  • E.

    8

Answer:A

Solution 1

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. To be a multiple of 3, A + B has to be either 2 or 5 or 8 \cdots and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10 \cdots and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7 \cdots and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1, so \boxed{\textbf{(A) }1} is our answer.

 

Solution 2

If n is divisible by 3, the sum of it's digits should also be divisible by 3. This means that 19+A+B (\mod 3) \equiv 0 or A+B+1 (\mod 3) \equiv 0. For equation 2, 15+A+B+C (\mod 3) \equiv 0 or A+B+C (\mod 3) \equiv 0. Logically, you can see the correlation between our first and second equations so we can make the assumption, C =\boxed{\textbf{(A) }1}