Cryptology - Unofficial St. Mary's College of California Web Site

3.1. THE REMAINDER OPERATOR 35
Performing the %26 operation is not particularly difficult, but what if the
number we are trying to reduce is 48924 Then repeatedly subtracting 26 is
rather unappealing. Fortunately, we are not interested in the number of times
we subtract 26, but rather in the number that is remaining when we are done.
Just as multiplication is addition done quickly (4 × 3 just means 4 sets of 3, or
3 added to itself 4 times), subtracting 26 many times is very closely connected
to dividing by 26.
To illustrate, we divide 85 by 26 using the form we learned as children:
3 r = 7
26 √ 85
78
7
In multiplicative form this says 85 = 3 × 26 + 7. The 3 represents the quotient,
how many times 26 goes into 85, and 7 is the remainder. Compare this with
the subtraction method. 85%26 = Subtracting once: 85 − 26 = 59, and 59
is larger than 26. Subtracting again: 59 − 26 = 33, still too large. Once more
33 − 26 = 7. Small enough. So 85%26 = 7. Notice that the quotient 3 is the
same as the number of subtractions, and that the remainder 7 is the same as
the modulus answer.
Let us try again with 109%26.
4 r = 5
26 √ 109
104
5
so 109 = 4 × 26 + 5. On the other hand,
109 − 26 = 83
83 − 26 = 57
57 − 26 = 31
31 − 26 = 5.
So 109%26 = 5. Again the quotient indicates how many 26’s need to be removed,
and the remainder in the division gives the same result as the remainder
operator.
For numbers larger than about fifty the division method is generally quicker