# Junior – Backpedaling: reverse cycling

Hey what’s up?  So far this year we’ve talked a lot about cycling, which in my opinion is one of the strongest strategies in high school Equations because it’s repeatable and can create problems that are very difficult to solve using arithmetic.  This is an idea to take it up a notch if you are playing against someone really good who you anticipate will be able to solve a normal quickie K problem.

The general idea is you take a cycle with a large cycle length like (5^n)/7 K7, and you make n equal to a number a bit smaller than the cycle length.  Since the cycle length for this problem is the lcm of 6 and 7, a good n to choose would be 41.  Since you know (5^42)/7 k’s down to 1/7, that means when you multiply the answer for (5^41)/7 by 5, it will k down to 1/7; in other words 1/7+7n=(x*5)/7, where n is any whole number.  You can multiple both sides by 7 to get 1+49n=x*5.  The answer here is 10 because it satisfies that equation, thus your answer is 10/7.

A more complicated version involves using this method with the methods to determine a half cycle used in the previous posts.  An example would be (7^46)/root with K11 and base 12.  Since lambda equals the lcm of 10 and 11, your cycle length could be 110 or any factor of that.  If you calculate the cycle length of 7^n with K11, you can narrow down the possibilities to 10 and 110, but we will discount 10 because only rarely will that be a factor of 11^2. (the only example I’ve found so far is (3^n)/root with K11, where 3^10=121n+1).  If your cycle length is 110 then 7^55=-1.  Then you have to calculate a number that times 7 K’s down to -1 to find the answer (once you base 46 you get 54).

Problems:

(5^40)/7, K7

(5^46)/root, K11, B12

(7^48)/root, k11, B12