Two numbers, $$297_{B}$$ and $$792_{B}$$ , belong to base B number system. If the first number is a factor of the second number then the value of B is:

In Base B, $$297_B = 2B^2 + 9B + 7$$

and $$792_B = 7B^2 + 9B + 2$$

It is given that $$297_{B}$$ is a factor of $$792_{B}$$

=> $$\frac{7B^2 + 9B + 2}{2B^2 + 9B + 7}$$ must be an integer    

=> $$\frac{(2B^2 + 9B + 7) + (5B^2 - 5)}{2B^2 + 9B + 7}$$

=> $$\frac{5B^2 - 5}{2B^2 + 9B + 7} + 1 = k$$

=> $$5B^2 - 5 = (2B^2 + 9B + 7) k$$      (where $$k$$ is factor)

Put $$k = 1$$

=> $$5B^2 - 5 = 2B^2 + 9B + 7$$

=> $$B^2 - 3B - 4 = 0$$

=> $$(B - 4) (B + 1) = 0$$

=> $$B = 4 , -1$$

Since, B is a base,so B must be greater than 9. Hence, it is not possible

Put $$k = 2$$

=> $$5B^2 - 5 = 4B^2 + 18B + 14$$

=> $$B^2 - 18B - 19 = 0$$

=> $$(B - 19) (B + 1) = 0$$

=> $$B = 19 , -1$$

$$\therefore B = 19$$