Maths Olympiad Prep

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Problem 500

Number theory Difficulty 2.5 Find the answer CEMC Cayley

What is the tens digit of the smallest positive integer that is divisible by each of 20, 16, and 2016?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

We note that 20=22⋅520=2^{2} \cdot 5 and 16=2416=2^{4} and 2016=16⋅126=25⋅32⋅72016=16 \cdot 126=2^{5} \cdot 3^{2} \cdot 7. For an integer to be divisible by each of 22⋅52^{2} \cdot 5, 242^{4}, and 25⋅32⋅72^{5} \cdot 3^{2} \cdot 7, it must include at least 5 factors of 2, at least 2 factors of 3, at least 1 factor of 5, and at least 1 factor of 7. The smallest such positive integer is 25⋅32⋅51⋅71=100802^{5} \cdot 3^{2} \cdot 5^{1} \cdot 7^{1}=10080. The tens digit of this integer is 8.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.