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    OlympiadMediumNumber Theory7–9

    What is gcd⁡(210−1,215−1)\gcd(2^{10} - 1, 2^{15} - 1)?

    Qual é o mdc(210−1,215−1)\text{mdc}(2^{10} - 1, 2^{15} - 1)?

    Answer options

    Solution

    For numbers of this shape, gcd⁡(2m−1,2n−1)=2gcd⁡(m,n)−1\gcd(2^{m} - 1, 2^{n} - 1) = 2^{\gcd(m,n)} - 1. Here gcd⁡(10,15)=5\gcd(10, 15) = 5, so the answer is 25−1=312^{5} - 1 = 31.