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Number theory Difficulty 2.3 Junior Find the answer Canada

The product $8 \times 48 \times 81$ is divisible by $6^k$ . The largest possible integer value of $k$ is

Pick one

Solution

We note that $8 \times 48 \times 81 = 2^3 \times (2^4 \times 3) \times 3^4 = 2^7 \times 3^5 = 2^2 \times 2^5 \times 3^5 = 2^2 \times (2\times 3)^5 = 2^2 \times 6^5$ .

After $6^5$ is divided out from $8 \times 48 \times 81$ , the quotient has no factors of 3 and so no further factors of 6 can be divided out.

Therefore, the largest integer $k$ for which $6^k$ is a divisor of $8 \times 48 \times 81$ is $k=5$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.