Find the greatest number which divides 2706, 7041 and 8250 leaving remainder 6, 21 and 42 respectively.

When 2706 is divided by the required number, 6 is left as remainder. So 2706 − 6 = 2700 is exactly divisible by that number.

Similarly, 7041 − 21 = 7020 and 8250 − 42 = 8208 are exactly divisible by that number.

Therefore, the required number is the HCF of 2700, 7020 and 8208.

First, find HCF of 2700 and 7020:

$\begin{array}{l} 2700\overline{\smash{\big)}7020\smash{\big(}}\phantom{}2 \ \phantom{x}\phantom{())}\underline{-5400} \ \phantom{{x^2 } 1+)}1620\overline{\smash{\big)}2700\smash{\big(}}\phantom{}1 \ \phantom{{x} +1xa()}\underline{-1620} \ \phantom{{x^2 a} + 2x+} 1080\overline{\smash{\big)}1620\smash{\big(}}\phantom{}1 \ \phantom{{x} +1xa+++}\underline{-1080} \ \phantom{{x^2 a} + 2x++++} 540\overline{\smash{\big)}1080\smash{\big(}}\phantom{}2 \ \phantom{{x} +1xa++++(())}\underline{-1080} \ \phantom{{x^2 a} + 2x+++++++} 0 \ \end{array}$

So, HCF of 2700 and 7020 = 540.

Now, find HCF of 540 and 8208:

$\begin{array}{l} 540\overline{\smash{\big)}8208\smash{\big(}}\phantom{}15 \ \phantom{x}\phantom{))}\underline{-8100} \ \phantom{{x^2 } 1+)}108\overline{\smash{\big)}540\smash{\big(}}\phantom{}5 \ \phantom{{x} +1xa(}\underline{-540} \ \phantom{{x^2 a} + 2x+)} 0 \ \end{array}$

So, HCF of 540 and 8208 = 108.

Hence, the required greatest number is 108.