Find the greatest number which can divide 257 and 329 so as to leave a remainder 5 in each case.

When 257 is divided by the required number, 5 is left as a remainder. So 257 − 5 = 252 is exactly divisible by that number.

Similarly, 329 − 5 = 324 is exactly divisible by that number.

Therefore, the required number is the HCF of 252 and 324.

$\begin{array}{l} 252\overline{\smash{\big)}324\smash{\big(}}\phantom{}1 \ \phantom{x}\phantom{))}\underline{-252} \ \phantom{{x^2 }() 1)}72\overline{\smash{\big)}252\smash{\big(}}\phantom{}3 \ \phantom{{x} +1xa}\underline{-216} \ \phantom{{x^2 a} + 2x()} 36\overline{\smash{\big)}72\smash{\big(}}\phantom{)}2 \ \phantom{{x} +1xa+()}\underline{-72} \ \phantom{{x^2 a} + 2x++()} 0 \ \end{array}$

Hence, the required greatest number is 36.