The last two digit of the multiplication of 1*3*5*7.........97*99 will be

1*3*5*...*99 = 25k 25k mod 100 = 25 if k mod 4 = 1 25k mod 100 = 75 if k mod 4 = 3 In this case, 1*3*...*23*27*...*99 mod 4 = 1*-1*1*-1*...*1 (25 pairs) = -1 = 3 So, remainder = 75