Answer :
Let the remainder in each case be x.
And the greatest number which divides (1935-x) and (59-x) be "a".
Then
(1935-x) = n × a
(59 - x) = m × a
Subtracting
we get (1935-59) = (n-m) × a where (n-m) is an integer.
1935-59 = 1876
prime factorization of 1876 = 2² × 7 × 67
"a" can't be greater than 59.
So maximum possible value of "a" = 2² × 7 = 28 (and (n-m) would be 67 for this case).
The greatest number is 28 and the remainder is 3.
And the greatest number which divides (1935-x) and (59-x) be "a".
Then
(1935-x) = n × a
(59 - x) = m × a
Subtracting
we get (1935-59) = (n-m) × a where (n-m) is an integer.
1935-59 = 1876
prime factorization of 1876 = 2² × 7 × 67
"a" can't be greater than 59.
So maximum possible value of "a" = 2² × 7 = 28 (and (n-m) would be 67 for this case).
The greatest number is 28 and the remainder is 3.