If you have any good information/tips or ideas to share, please add them to the page!

WHAT GOALS/SOLUTIONS DID YOU SEE AT NATIONALS THAT WERE PRETTY COOL? EVEN IF YOU AREN'T SURE OF THE ANSWERS, POST THEM AT THE BOTTOM AND MAKE SURE YOU LIST THE VARIATIONS THAT YOU REMEMBER WERE CALLED.


Take a look at how Michigan studies for Equations.

• Factorials
• Cycling
• Double Divide
• Double Signs
  

Check this page for tips on

• Factorials
• Cycling (below)

Abby Lusk!*

Fun with Factorials! (All Divisions) One of the most popular variations in all levels of Equations is factorial. Students enjoy it because it is easy to understand, but there are a lot of different things you can do with it. Remember that factorial is represented by a ! that may be placed anywhere in the solution or goal (elementary and middle only). Factorial is defined as the product of all the numbers between 1 and the number being acted upon. That is 5! = 5 x 4 x 3 x 2 x 1 = 120, and 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040. Note that in the junior and senior division, factorials can not be used in the Goal, just the Solution.
Here are a few interesting goal - solution combinations that you can use with factorial:
1) 17 / 16 = 17 Explanation: Goal is interpreted as 17! divided by 16!. 17! = 17x16x15x14x13... When divided by 16x15x14x13..., all the numbers cancel out and the only the 17 is left. This goal can be used with any two consecutive numbers. The higher the numbers, the more effective the goal. This strategy could not be used in the junior or senior divisions. 2) 24 x 23 = 24! Explanation: Goal is interpreted as 24 x 23!, which equals 24x23x22x21x20..., or 24! Like the previous goal, it can be used with any two consecutive numbers. With both of these soultions, be careful not to use too many factorials in the solution. This strategy could not be used in the junior or senior divisions. 3) 54 x 56 = 9! / 5! Explanation: Goal equals (9 x 6) x (7 x 8), or 9 x 8 x 7 x 6, which equals 9! (9x8x7x6x5x4x3x2) divided by 5!(5x4x3x2). There are a number of different number combinations that this factor rearrangement works with. Factorial also works very well with a number of different variations, including LCM, GCF, # of factors, Powers of the Base, Multiple of K. Experiment a little in your practices, and you will probably find a few new tricks of your own.


Cycling (Jun/Sen EQ) What is cycling? Cycling is a way to have your solution equal a large goal when multiple of K is called; or, have a large solution equal a small goal when K is called. It’s a reliable and fast shortcut that keeps you from having to calculate 81*97, for example.
Why should I care? Cycling is a strategy used by many teams. Although there are more sophisticated strategies out there, cycling will win its fair share of games.
Why haven’t I learned this in middle? A few people have learned this, but cycling isn’t generally used in middle because the goal must be between -1000 and 1000. It is possible to use cycling in your solution, for example using 3*7 ÷ 4 = 3 ÷ 4 with K7 (solution = goal). Most middle players don’t learn this because it is a lot of work compared to some other strategies. However, things change in junior because lots of other people know it.

How do I cycle? Cycling is based on finding patterns. Typically, you start with some goal like 3*97, 57*63, or 97*98 (something * something) with multiple of K in effect. The bigger the goal the better, since you don’t want people writing some easy solution after the trouble you took to learn cycling. Then you multiply the entire goal out and just use K on each number, for example: Tyler Dunigon!!
Goal: 3*97 with K7
3*1 = 3
3*2 = 9 – 7 = 2 (the remainder after K)
3*3 = 6 (you only have to multiply the remainder by the original number, since the rest is a multiple of K and does not matter)
3*4 = 18 – 14 = 4
3*5 = 12 – 7 = 5
3*6 = 1
3*7 = 3
3*8 = 9 – 7 = 2
This could get boring pretty quickly, but there is a shortcut. Notice that the numbers are repeating starting with 3*7. In fact, see if you can guess what 3*10 equals with K7 without calculating the rest… If you said 4, you were right. The numbers repeat, so 3*10 = 3*4 (notice how 3*7 corresponds to 3*1) = 4. If you calculated the rest, you would find that 3*70 = 3*4 = 4. The key here is that the numbers repeat after 6 powers, which we call cycle length 6. The easiest way to find the cycle length is to just take the power when it equals 1, since the next power will equal the original number. Once you have the cycle length, divide the original power (97) by the cycle length, and take the remainder (1). Then just calculate 3*1 to get your answer. As it turns out, 3*97 = 3 with K7. Here’s another example:
Goal: 83*97 with K8 (you can use K on the 87 to get 3*97, since the rest is just a multiple of K and doesn’t matter)
Adjusted goal: 3*97
3*1 = 3
3*2 = 9 -6 = 1
Cycle length = 2
97 / 2 = Remainder 1
3*1 = 3 = 3*97

What if the cycle never goes to one? Goal: 2*95 K6
2*1 = 2
2*2 = 4
2*3 = 8 – 6 = 2
2*5 = 4
2*6 = 8 – 6 = 2
Well, this isn’t going to 1 any time soon. However, cycling is about finding out when the numbers repeat, not just searching for the 1. Since the numbers repeat after 2 exponents, the cycle length is 2. Then, just do the rest of the problem accordingly.
95 / 2 = remainder 1
2*1 = 2 = 2*95
How can I use this in a game? The easiest way is to just set a big goal using an exponent or a root after you call K. Cycle it out before you put cubes on the goal line, then just set it if it works out to a reasonable answer. You should avoid setting smaller goals like 3*5, since this can be solved pretty easily by regular calculating. Once you set the goal, you should try to guess what your opponent’s solution is. Generally, they will try to recreate the goal if they don’t know cycling, which is a reasonable strategy. Move the cubes that they need and you don’t need to forbidden, typically the exponent or root. If you think someone knows cycling (as in they aren’t scribbling furiously on their paper), try to solve for something a little more complicated. If the goal simplifies to 5 K7, solve for 40 or even 68. The requirements for this strategy are small, just an exponent and a few number cubes. Two x or ÷ cubes are nice but not necessary. It’s very hard for your opponents to solve if they don’t know cycling, but easy to solve if they do know cycling. Cycling has many steps and is very easy to mess up at first. However once you've practiced it enough it is straightforward and can be a very effective strategy.
borrowed from the Michigan Academic Games League