2019 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 18 Medium

For some positive integer k, the repeating base-k representation of the (base-ten) fraction \frac{7}{51} is 0.\overline{23}_{k}=0.232323\cdots_{k}. What is k? (2019 AMC 10A Problem, Question#18)

  • A.

    13

  • B.

    14

  • C.

    15

  • D.

    16

  • E.

    17

Answer:D

We can expand the fraction 0.\overline{23}_{k} as follows: 0.\overline{23}_{k}=2\cdot k^{-1}+3\cdot k^{-2}+2\cdot k^{-3}+3\cdot k^{-4}+\cdots Notice that this is equivalent to 2(k^{-1}+k^{-3}+k^{-5}+\cdots)+3(k^{-2}+k^{-4}+k^{-6}+\cdots)

By summing the geometric series and simplifying, we have \frac{2k+3}{k^2-1}=\frac{7}{51}. Solving this quadratic equation (or simply testing the answer choices) yields the answer k=16.

Let a=0.2323\cdots_k. Therefore, k^2a=23.2323\cdots_k.

From this, we see that k^{2}a-a=23_k, so a=\frac{23_k}{k^2-1}=\frac{2k+3}{k^2-1}=\frac{7}{51}.

Now, similar to in Solution 1, we can either test if 2k+3 is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is 16.

We can simply plug in all the answer choices as values of k, and see which one works. After lengthy calculations, this eventually gives us 16 as the answer.

Just as in Solution 1, we arrive at the equation \frac{2k+3}{k^2-1}=\frac{7}{15}.

We can now rewrite this as \frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot17}. Notice that 2k+3=2(k+1)+1=2(k-1)+5. As 17 is a prime, we therefore must have that one of k-1 and k+1 is divisible by 17. Now, checking each of the answer choices, this gives 16.

Assuming you are familiar with the rules for basic repeating decimals, 0.232323\cdots=\frac{23}{99}. Now we want our base, k, to conform to 23=7(mod\ k) and 99=51(mod\ k), the reason being that we wish to convert the number from base 10 to base k. Given the first equation, we know that k must equal 9, 16, 23, or generally, 7n+2. The only number in this set that is one of the multiple choices is 16. When we test this on the second equation, 99=51(mod\ k) , it comes to be true. Therefore,our answer is 16.

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