Compute gcd(215−1,210−1)\gcd(2^{15} - 1, 2^{10} - 1)gcd(215−1,210−1) using the property that gcd(2a−1,2b−1)=2gcd(a,b)−1\gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a,b)} - 1gcd(2a−1,2b−1)=2gcd(a,b)−1.
25−1=312^5 - 1 = 3125−1=31
210−1=10232^{10} - 1 = 1023210−1=1023
23−1=72^3 - 1 = 723−1=7
22−1=32^2 - 1 = 322−1=3