Find the largest number that will divide 623, 729 and 841 leaving remainders 3, 9 and 1 respectively.

When 623 is divided by the required number, 3 is left as a remainder. So 623 − 3 = 620 is exactly divisible by that number.

Similarly, 729 − 9 = 720 and 841 − 1 = 840 are exactly divisible by that number.

Therefore, the required number is the HCF of 620, 720 and 840.

First, find HCF of 620 and 720:

$\begin{array}{l} 620\overline{\smash{\big)}720\smash{\big(}}\phantom{}1 \ \phantom{x}\phantom{))}\underline{-620} \ \phantom{{x^2 } 1()}100\overline{\smash{\big)}620\smash{\big(}}\phantom{}6 \ \phantom{{x} +1xa}\underline{-600} \ \phantom{{x^2 a} + 2x()} 20\overline{\smash{\big)}100\smash{\big(}}\phantom{}5 \ \phantom{{x} +1xa++}\underline{-100} \ \phantom{{x^2 a} + 2x+++} 0 \ \end{array}$

So, HCF of 620 and 720 = 20.

Now, find HCF of 20 and 840:

$\begin{array}{l} 20\overline{\smash{\big)}840\smash{\big(}}\phantom{}42 \ \phantom{x}\phantom{)}\underline{-840} \ \phantom{{x^2 } 1+)}0 \ \end{array}$

So, HCF of 20 and 840 = 20.

Hence, the largest required number is 20.