The smallest prime number that is a divisor of 17408 is 2 (17408 = 17 × $$2^{10}$$​). What is the sum of digits of the greatest prime number, which is a divisor of 17391?

Correct option is A

Given:

17408 = $$17 \times 2^{10}$$

Solution:

17391 = 17408 - 17

= $$17 \times 2^{10} - 17$$​​

=  $$17(2^{10} - 1)$$​​

= 17(1024 - 1)

= $$17 \times 1023$$​​

= $$17 \times 3 \times 11 \times 31$$​​

Thus, the greatest prime number = 31.

Sum of digits = (3 + 1) = 4

Thus, the correct option is (a) 4