20. The number $N=3^{16}-1$ has a divisor of 193. It also has some divisors between 75 and 85 inclusive. What is the sum of these divisors?
20. 247 First factorise $N$ twice using the difference of two squares i.e. $N=3^{16}-1=\left(3^{8}-1\right)\left(3^{8}+1\right)=\left(3^{4}-1\right)\left(3^{4}+1\right)\left(3^{8}+1\right)=80 \times 82 \times\left(3^{8}+1\right)$. This shows that both 80 and 82 are divisors of $N$ in the required range. The question tells us that 193 is a divisor of $\mathrm{N}$ and, since 193 is prime, it must be a divisor of $3^{8}+1=81 \times 81+1=6562$. Now observe that $6562=2 \times 3281$ and that $3281 \div 193=17$. Therefore $N=80 \times 82 \times 2 \times 17 \times 193$ or $N=\left(2^{4} \times 5\right) \times(2 \times 41) \times 2 \times 17 \times 193$.
Next consider the integers from 75 to 85 inclusive to see which could be divisors of $N$. Because $N$ has no prime factors of 3 or 7 , we know that $75,77,78,81$ and 84 are not divisors of $N$ while the initial argument established that 80 and 82 are divisors of $N$. Both 79 and 83 are prime and $76=4 \times 19$ so, since $N$ does not have a prime factor of 79,83 or 19 , these must also be excluded. This only leaves 85 to be considered. Note that $85=5 \times 17$ and both 5 and 17 are prime factors of $N$ so 85 is a divisor of $N$. Hence the divisors of $N$ in the required range are 80,82 and 85 with sum 247 .