Last practice we were discussing a problem that was pretty tricky, and pretty much impossible to solve without knowing this method. The problem was (1/7)*54, 5 cubes with POB and K11 called. Applying a formula you can solve this problem pretty quickly, but let me get into the reasons why you can use shortcuts cycling.

Say you have a problem b^x with k. For the last number in your cycle, λ**, **you are saying that b^λ= nk+1, or that when you divide by a certain multiple of k, you get a remainder of 1. Say our actual problem is (5^33)/7 with K 11. If we do 5^33 K7 and 5^33 K11, we get these cycles:

K7

1 | 5 |

2 | 4 |

3 | 6 |

4 | 2 |

5 | 3 |

6 | 1 |

K11

1 | 5 |

2 | 3 |

3 | 4 |

4 | 9 |

5 | 1 |

So with k7, not only is 5^6=1, but 5^12, 5^18, etc. So for each of these, 5^6n= k+1. With k11, 5^5n= k+1. The power of 5 where these both equal 1 is 30, so if 5^30= nk+1 for both k7 and k11, that means nk is divisible by 77 and we can subtract it out. Thus λ=30, and we can solve for (5^3)/7.

If you do a problem like (5^44)/7 K7, the cycle length for k7 is 6 but 5^6 is only 7n+1, not 49n+1. So what we have to do with this one is raise both sides to the 7th power, getting (5^6)^7=(7n+1)^7. The left is just 5^42, and the right you can expand where you get 1+7n*7(the second 7 is from Pascal’s triangle) + 7n^2*21 +…. + 7n^7. All of the terms except for the 1 are multiples of 7^2, so you can reduce it to 1 and you find λ=42, and we reduce the goal to 5^2/7. If you do k=7^3, you would do it to the 49th power instead of the 7th, and your cycle length would be 6*7^2.

This is the basis of the formula for calculating λ**, **for which you take the prime factorization of your effective k A^a * B^b *… and you take the lcm of (A^(a-1), B^(b-1), A-1, B-1,…)

It is important to note that this only works if your base and k are relatively prime. For example if you take (8*40)/5 K10, you will never be able to cycle down to 1 because both the base and the multiple you subtract of 50 will both be even, thus you can only get even numbers. In this case the cycle length will be lcm(5,4) and you ignore the 2s, but the last number in your cycle will not be 1. In this case it’s 26.

I hope this explains a little bit better how K works. Here are some problems for you to practice:

(1/7)^5, k7, POB 5 cubes

(5^56)/9 K9 5 cubes