Let $$a_{1}<a_{2}<.... <a_{n}$$ be the list of all prime numbers less than 25. Define $$X_{i}=\frac{b_{i}}{a_{i}}$$, where $$b_{i}$$ is the sum of all $$a_{k}$$ where k ranges from 1 to n, $$k \neq i$$. Let B be the set of all integer-valued $$X_{i}$$. What is the Smallest element of B?

List of all prime numbers less than 25 = 2,3,5,7,11,13,17,19,23 = 9 numbers

$$a_1$$ = 2, $$a_2$$ = 3, $$a_3$$ = 5, $$a_4$$ = 7, $$a_5$$ = 11, $$a_6$$ = 13, $$a_7$$ = 17, $$a_8$$ = 19, $$a_9$$ = 23

Sum of all the above prime numbers ($$a_1$$ + $$a_2$$ + ...... + $$a_9$$) = 100

$$X_i=\ \frac{\ b_i}{a_i}$$ ,where, $$b_i$$ = Sum of all prime numbers $$a_1\ to\ a_{n\ }$$ except $$a_i$$

Example : $$b_3$$ is the sum of all the prime numbers $$a_1$$ to $$a_9$$ except $$a_3$$. 

$$X_1$$ = $$\ \frac{\ b_1}{a_1}$$ = $$\ \frac{\ 100-a_1}{a_1}$$ = $$\ \frac{\ 100-2}{2}$$ = 49

$$X_2$$ = $$\ \frac{\ b_2}{a_2}$$ = $$\ \frac{\ 100-a_2}{a_2}$$ = $$\ \frac{\ 100-3}{3}$$ = $$\ \frac{\ 97}{3}$$

$$X_3$$ = $$\ \frac{\ b_3}{a_3}$$ = $$\ \frac{\ 100-a_3}{a_3}$$ = $$\ \frac{\ 100-5}{5}$$ = 19

$$X_4$$ = $$\ \frac{\ b_4}{a_4}$$ = $$\ \frac{\ 100-a_4}{a_4}$$ = $$\ \frac{\ 100-7}{7}$$ = $$\ \frac{\ 93}{7}$$

$$X_5$$ = $$\ \frac{\ b_5}{a_5}$$ = $$\ \frac{\ 100-a_5}{a_5}$$ = $$\ \frac{\ 100-11}{11}$$ = $$\ \frac{\ 89}{11}$$

$$X_6$$ = $$\ \frac{\ b_6}{a_6}$$ = $$\ \frac{\ 100-a_6}{a_6}$$ = $$\ \frac{\ 100-13}{13}$$ = $$\ \frac{\ 87}{13}$$

$$X_7$$ = $$\ \frac{\ b_7}{a_7}$$ = $$\ \frac{\ 100-a_7}{a_7}$$ = $$\ \frac{\ 100-17}{17}$$ = $$\ \frac{\ 83}{17}$$

$$X_8$$ = $$\ \frac{\ b_8}{a_8}$$ = $$\ \frac{\ 100-a_8}{a_8}$$ = $$\ \frac{\ 100-19}{19}$$ = $$\ \frac{\ 81}{19}$$

$$X_9$$ = $$\ \frac{\ b_9}{a_9}$$ = $$\ \frac{\ 100-a_9}{a_9}$$ = $$\ \frac{\ 100-23}{23}$$ = $$\ \frac{\ 77}{23}$$

B is the set of all integer-valued $$X_i$$ = {$$X_1$$, $$X_3$$} = {49, 19}

The smallest element of B = 19

$$\therefore\ $$ The required answer is B.