2021 AMC 10 A Fall

Complete problem set with solutions and individual problem pages

Problem 12 Medium

The base-nine representation of the number N is 27,006,000,052 nine. What is the remainder when N is divided by 5 ?(2021 AMC Fall 10A, Question #12)

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:D

Solution 1:

Recall that 9 \equiv-1(\bmod 5). We expand N by the definition of bases: \begin{array}{rlr} N & =27,006,000,052_{9} & \\ & =2 \cdot 9^{10}+7 \cdot 9^{9}+6 \cdot 9^{6}+5 \cdot 9+2 & \\ & \equiv 2 \cdot(-1)^{10}+7 \cdot(-1)^{9}+6 \cdot(-1)^{6}+5 \cdot(-1)+2 & (\bmod 5) \\ & \equiv 2-7+6-5+2 & (\bmod 5) \\ & \equiv-2 & (\bmod 5) \\ & \equiv(\text{D}) 3 & (\bmod 5) . \end{array}

Solution 2:

We need to first convert N into a regular base- 10 number: N=27,006,000,052_{9}=2 \cdot 9^{10}+7 \cdot 9^{9}+6 \cdot 9^{6}+5 \cdot 9+2 . Now, consider how the last digit of 9 changes with changes of the power of 9 : \begin{aligned} &9^{0}=1 \\ &9^{1}=9 \\ &9^{2}=81 \\ &9^{3}=729 \\ &9^{4}=6561 \end{aligned} Note that if x is odd, then 9^{x} \equiv 4(\bmod 5) . On the other hand, if x is even, then 9^{x} \equiv 1(\bmod 5) Therefore, we have \begin{array}{rlr} N & \equiv 2 \cdot(1)+7 \cdot(4)+6 \cdot(1)+5 \cdot(4)+2 \cdot(1) & (\bmod 5) \\ & \equiv 2+28+6+20+2 & (\bmod 5) \\ & \equiv 58 & (\bmod 5) \\ & \equiv(\text { D) } 3 & (\bmod 5) \end{array} Note that for the odd case, 9^{x} \equiv-1(\bmod 5) may simplify the process further, as given by Solution 1 .

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