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  Â
=> $$\frac{(2B^2 + 9B + 7) + (5B^2 - 5)}{2B^2 + 9B + 7}$$
=> $$\frac{5B^2 - 5}{2B^2 + 9B + 7} + 1 = k$$
=> $$5B^2 - 5 = (2B^2 + 9B + 7) k$$ Â Â Â (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 $$k = 2$$
=> $$5B^2 - 5 = 4B^2 + 18B + 14$$
=> $$B^2 - 18B - 19 = 0$$
=> $$(B - 19) (B + 1) = 0$$
=> $$B = 19 , -1$$
$$\therefore B = 19$$
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