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Number Theory Level 2

Find the smallest number k k such that the given statements give remainder 0 0

k − 1 1 \frac{k-1}{1}

k − 2 2 \frac{k-2}{2}

k − 3 3 \frac{k-3}{3}

k − 4 4 \frac{k-4}{4}

k − 5 5 \frac{k-5}{5}

k − 6 6 \frac{k-6}{6}

k − 7 7 \frac{k-7}{7}

k − 8 8 \frac{k-8}{8}

k − 9 9 \frac{k-9}{9}

k − 10 10 \frac{k-10}{10}


The answer is 2520.

Ajit Singh by Ajit Singh , 20, India

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3 solutions

Curtis Clement
Sep 17, 2015

Note that the highest power of 2 is 3, the highest power of 3 is 2 and finally the highest power for 5 & 7 is 1. (Our aim is to find the prime factorisation of the smallest k) ∴ m i n ( k ) = 2 3 × 3 3 × 5 × 7 = 2520 \therefore\ min(k) = 2^3 \times\ 3^3 \times\ 5 \times\ 7 = 2520

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Chris Cooper
Sep 3, 2015

It may be worth saying "find the smallest number k" since at the moment there are an infinite number of solutions (e.g. 10!)

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Zhaochen Xie
Sep 2, 2015

The only thing you have to do is find the smallest common multiple which is 2520

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