The following material can be reworded and incorporated into this article:--MCrawford 18:54, 30 June 2006 (EDT)
A General Algorithm
In the example above, we were fortunate to find a power of -- namely, -- that is congruent to mod . What if we aren't this lucky? Suppose we want to solve the following problem:
What are the tens and units digits of ?
Again, we will solve this problem by computing modulo . The first few powers of mod are
This time, no pattern jumps out at us. Is there a way we can find the power of mod without taking this list all the way out to the term -- or even without patiently waiting for the list to yield a pattern?
Suppose we condense the list we started above; and instead of writing down all powers of mod , we write only the powers , where is a power of . We have the following (all congruences are modulo ):
(Observe that this process yields a pattern of its own, if we carry it out far enough!)
Now, observe that, like any positive integer, can be written as a sum of powers of two:
We can now use this powers-of-two expansion to compute :
So the tens and units digits of are and , respectively.
We can use this method to compute modulo , for any integers and , with . The beauty of this algorithm is that the process takes, at most, approximately steps -- at most steps to compute the values modulo for a power of two less than , and at most steps to multiply the appropriate powers of according to the binary representation of .
This method can be further refined using Euler's Totient Theorem.
Why is assumed here, it's never needed. It's in general better to neglect this restriction after one has mastered the first steps (and as this is Intermediate, I would assume this to be true).
And I'm still strongly against using instead of , the first one is already used for -adic integers and rarely seen in non-olympiad books.
But as both is some kind of disputed, I didnÄt edit it untill now.
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