Category Archives: Equations

Cycling Practice

Cycling Practice

These problems are listed with ascending difficulty.

  1. 5^99 K7
  2. 3^87 K11
  3. 7^73 K9
  4. 5^(2^87) K7
  5. 3^(3^87) K11
  6. (3^99)/√ K11 B12
  7. (8^99)/7 K7
  8. (5^46)/7 K7
  9. (5^93)/7 K11
  10. (5^55)/9 K9
  11. (5^65)/7 K7
  12. (9^65)/7 K7
  13. (5^81)/√ K11 B12
  14. (7^46)/√ K11 B12
  15. (7^68)/√ K11 B12

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

Junior – Explaining cycling and quickie K

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, λyou 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 λfor 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

Junior – Quickie K revealed

Alright so we are playing Senior Eq and I call base 12 and k11, and set the goal (6^93)/root.  What if I told you that you could solve this in about 30 seconds?

To figure out the cycle length what you have to do is take the effective K, which is 121 in this instances, and prime factorize it, so you get 11^2.  To find the cycle length, you take the the lcm of the factors minus one and the power with the exponent reduced by one, so in this case (11-1) and (11^1), which is 110.  Making sure the k and base are relatively prime, you know your cycle length is 110.  Since the goal adjusted for base is 6^111 / 11, your answer should equal 6/11.

Example problem: 5^88/7 K7