Chapter 6 - Riddles and answers

Create a riddle

Say that I want to create a riddle (about numbers, just because...). I could start with the stupidest riddle ever and say: "I have some sheep. The number of sheep is 5. How many sheep do I have?".
Not an impressive challenge... but wait. I can do better, by transforming a bit my riddle, for example I could multiply everything by ten. That would give me:

Moral of the story so far: the simplest possible riddle IS its own solution here; and every difficult riddle is just a simple riddle made to LOOK more complex, by transforming it multiple times.

Solve the riddle

However, this idea of going backwards from an answer to a riddle, using a transformation after another is very general in math, and it is even more interesting when you imagine the process inverted. Could I take a riddle, transform it and transform it, until I eventually get an answer?
As an example I could try to solve: "10 times the number of sheep and 4 is 54"... but where to start?
I can look at the two main types of transformations I used to build this riddle:

• *k , multiplication by a number (any number, positive or negative, or fraction, just not zero)
• +n , addition of a number

Note: we covered inverse and opposite numbers in chapter 3.

Did you notice anything about the two kinds of transformations, playing with the playground above?
What? Perhaps something about the order in which you decide to apply the transformations and the precision of the calculations?...

Navigation tricks

When you play with the playground above, it seems that it is possible to find the answer in two steps, every time. Cool.
But can we be sure that it is the case EVERY time? For us, every riddle is a sentence of the form: "a times the number of sheep and b sheep is c". Now, consider the special cases: "0 times the number of sheep is 5", and "0 times the number of sheep is 0".
The first riddle has no answer: it is IMPOSSIBLE for a number MULTIPLIED by zero to become five.
The second riddle is different. What number of sheep would make the sentence true? I can try with 1 and I will get "0 times 1 is 0", which is true. And what about 2? Well, "0 times 2 is 0" is also true... and so with ANY number. So this riddle has INFINITE answers, because any number of sheep will satisfy it.
Fine, I will have to look out for those two special cases... but for all other cases it looks like I can solve the riddle in two steps. So here we have a procedure that starts from a riddle sentence, and following two steps, always gives me the answer to the riddle (eventually treating the two special cases in a different way). Now: a procedure that starts from some data about a problem, and follows a specific number of steps to always produce the answer to that problem, is called an algorithm. The word "algorithm" comes from the name of al-Khwarizmi a Persian mathematician, astronomer, geographer; according to Wikipedia: Al-Khwarizmi's popularizing treatise on algebra (The Compendious Book on Calculation by Completion and Balancing, c. 813–833 CE) presented the first systematic solution of linear and quadratic equations. Spoiler: our riddles are what math calls "linear equations" or "first degree equations"!
So playing with the playground above we have actually found an algorithm for solving any of our riddles! Terrific. Let's write down clearly and step-by-step how the algorithm works:

1. take a riddle written like: "a times the number of sheep and b sheep is c";
   for example: "10 times the number of sheep and 4 sheep is 54".
2. use a transformation to subtract whatever number b is, getting a simpler riddle of this form: "a times the number of sheep is (c-b)";
   in the example the transformation is add -4, and the result is:
   "10 times the number of sheep is 50".
3. then use another transformation to divide by a, getting:
   "the number of sheep is (c-b):a";
   in the example the transformation is multiply by ¹_a: "the number of sheep is 50:10", or "the number of sheep is 5"!

Those two kinds of cases are a problem for my algorithm because a procedure to calculate something is NOT an algorithm if it can get STUCK even in a single case. And here I have two "families" of riddles that would make my algorithm stuck... not good.
So to avoid my algorithm to get stuck I have to write it better, so that the steps are more solid, and I can avoid the trap of these two special cases. Here is the correct algorithm to solve linear equations (AKA any of our riddles):

1. take a riddle written like: "a times the number of sheep and b sheep is c"
   1. In case a, b and c are all 0, then stop and answer "any number of sheep would do for this riddle".
   2. In case a and b are 0, but c is not, then stop and answer "there is no solution to this riddle".
2. transform the riddle to "a times the number of sheep is (c-b)";
3. transform again to get "the number of sheep is (c-b):a", the answer to the riddle.

Two-pan balance

First of all, let me tell you that in math the variable x is traditionally used to stand for thing we don't know (for example the amount of sheep, in our riddles) mainly for historical reasons. Adopting this convention, I could change the way I write my riddles, from: "2 times the number of sheep and 1 is 3" to something like this:

And now let's play with the playground below.

Before moving away from the two-pan balance, I would like to see how the two special cases ("no solution" and "infinite solutions") look in this representation.
Starting from the case with no solutions, which is a riddle of the form "0 times x and 0 is c", I get this drawing:

Moral of the story so far: riddles (AKA linear equations) can be written as sentences about sheep or drawn as two-pan balances, with specific weights on the pans; but a riddle is also an equivalence between two linear combinations, both potentially with the variable x. We have an algorithm to solve these equations, and there are two special cases we have to watch out for. We also discovered that there are only three possible types of answers for a riddle: zero, one or infinite solutions.

But what about the real world?

In your school math lectures you might have encountered equations in a very different context: not as riddles, but as a way to express and solve "real life" problems. And now that we have looked at the two-pan balance, we can too. Take a situation where you have a lot of bricks (identical bricks, same size and weight) and would like to know how much a brick weights. But the only thing you know is that when you put three bricks on one pan, you have to put one brick and eight kilograms on the other to balance them. What is the riddle (AKA equation) that represent that situation?
It should be:

Write this stuff shorter

Time to write this stuff shorter, drop the sentences and the balances, and focus on the linear combinations (AKA polynomials), and invent a good normal form for an equation. And since we are at it, let's also define properly what the meaning-preserving transformations are in this case, and how they can change an equation.
What can I use for as normal form for an equation? I would like to go back to the interesting fact that a solution to a riddle is a (simpler) riddle. Therefore, a good normal form for an equation should be its solution: x=5, for example!
But, wait... what a surprise! This is a linear combination of x and am I saying that it must be the same as (AKA equal to) an ordinary number: 1*x=5. We can use the same notation we invented in chapter 5 to represent math expressions with variables, the labeled boxes, and write the equation 1*x=5 as: ^x₁│ ₅ and I will have to remember that these two boxes contain each a linear combination of x and some numbers, and that the left and right box must be kept EQUAL for the equation to make sense.
OK, and if that is the normal form, it means there are alternative ways to write the SAME thing (as we discussed already in chapter 1, about whole numbers and abacus-numbers). We have already used some transformation on riddles, like: multiply by a number, add a number, add x. But since we are talking about linear combinations now, I can do those things and better (and more general) stuff too:

• multiply both linear combinations (left and right of the "=" sign) by a non-zero number k.
  For example: (1*x=5) *10 → (1*x) *10=(5) *10 → 10*x=50, where k is 10,
  or if you prefer the boxes: (^x₁│ ₅) *10 → (^x_1*10│ _5*10) → (^x₁₀│ ₅₀)
• add the same number to n to both sides of the equation,
  for example: (1*x=5) +10 → (1*x) +10=(5) +10 → 1*x+10=15,
  or with the boxes: (^x₁│ ₅) +10 → (^x₁│ ₁₅)
• add k-time x to both sides of the equation, where k is any number.
  Example: (1*x=5) +2*x → (1*x) +2*x=(5) +2*x → 3*x=5+2*x
  Boxes: (^x₁│ ₅) +^x₂ → (^x₃│^x_2 5)
• add a linear combination in x that looks like m*x+n to both sides of the equation.
  Example: (1*x=5) +(2*x+3) → (1*x) +(2*x+3)=(5) +(2*x+3) →
  (1+2)*x+3=2*x+3+5 → 3*x+3=2*x+8
  Boxes: (^x₁│ ₅) +(^x_2 3) → (^x₁^x_2 3│ ₅^x_2 3) → ^x_3 3│^x_2 8
• swapping the two sides is also OK.
  Example: 1*x+5 = 2*x+3 → 2*x+3 = 1*x+5
  Boxes: ^x₁│ ₅ →  ₅│^x₁
• move something from one side of the "=" to the other is also possible, and it changes the look but not the meaning of the equation.
  Example: 1*x+5 = 2*x → (1*x+5) -1*(2*x)=(2*x) -1*(2*x) → (1*x+5)-2*x=0 → ect.
  Boxes: ^x_1 5│^x₂ → (^x_1 5│^x₂) -^x₂ → ^x_1-2 5│ ₀ → ^x_-1 5│ ₀
  So the short answer is: to move something from side to side, just flip its sign.

Finishing the algorithm

Now we can look back at our algorithm to solve linear equations (AKA riddles). It works fine, but it requires the equation to be in the form "a times the number of sheep and b sheep is c"; however, we have extended the way we write equations, to include equations that are made of two linear combinations, one on each side of the "=" sign. Our algorithm cannot cope with those yet. Time to finish the algorithm, so it can solve any kind of linear equations in x!
The idea here is to add a step in the very beginning of the old algorithm, to rewrite the equation into the nice form "a times the number of sheep and b sheep is c", then simply continue with the rest of the old algorithm (which worked just fine).
So if I have an equation like this: 1*x+5 = 2*x+3 the smartest thing to do is to move all the x-related stuff to the left and all the ordinary numbers to the right. In this way my equation will FIT a form that my algorithm can work with! Here I get: 1*x-2*x = 3-5, thanks to the rule that says than when things move across the "=" sign, their sign flips.
And you might have realized by now that moving terms across the "=" is not really a NEW transformation, but rather a clever use of an add n transformation, only that n here is chosen to be the opposite of the term that I want to move; for instance: 1*x + 5 = 2*x + 3
⇓ -2*x
1*x + 5 -(2*x) = 2*x + 3 -(2*x) →
1*x + 5 - 2*x = 3 Good. Now I can calculate a bit on each side: 1*x-2*x = 3-5 →
-1*x = -2 And finally I can divide both sides of the equation by -1 (because it is the number that multiplies x in the left part of the equation), and get: (-1*x):(-1) = (-2):(-1) →
1*x = 2 And that's the solution of the equation, or if your prefer its normal form.
Generalizing from this example (but also keeping in mind the two special cases), our revised algorithm to solve linear equations is:

1. take a linear equation written like: i*x + j = m*x + n and move stuff around until you get something that looks like a*x=b
   1. In case a and b are both 0, stop and answer "any number is a solution to this equation", or "this equation has infinite solutions".
   2. In case a is 0, but b is not, stop and answer "there is no solution to this equation".
2. transform the equation in this form: x = (b:a), and ^b_a is the solution.

1. start by: 1*x+5 = 2*x+3. It needs cleaning up, so move stuff around: 1*x+5 -2*x = 2*x+3 -2*x →
   1*x+5 -2*x -5 = 2*x+3 -2*x -5 →
   1*x-2*x = 3-5 →
   -1*x = -2 Now looks like a*x=b, with a=-1 and b=-2.
   1. Are a and b both 0? Nope... then go on.
   2. Is a equal to 0? No... go no.
2. The soltuion is ^b_a, and here that is ^-2_-1 → x=2.

Tables and drawings

OK, so by now it is clear that an equation is the equivalence of two linear combinations (in the variable x). For example I know that the solution of 1*x+5 = 2*x+3 is x=2, and that must mean that if I were to replace every COPY of the variable x in both linear combinations with the number 2, I would get the same result on both sides. (In math they call this replacement "a substitution".) Let's see, when x=2 the equation becomes: 1*(2)+5 = 2*(2)+3 →
2+5 = 2*2+3 →
7 = 7 which is true; so I can say "the value 2 SATISFIES the equation", or "the value 2 makes my equation true". You can try with any other number, but ONLY 2 will satisfy this equation. After all we discovered before that a linear equation can have zero, one or infinite solutions; this one has one solution, and only one. Let me show you, for example with x=0 (also because zero makes the calculations simpler!): 1*(0)+5 = 2*(0)+3 →
0+5 = 0+3 →
5 = 3 which unfortunately is false; hence, x=0 cannot be the solution... and I knew that.
However, this idea of TESTING is interesting. I could try substituting x by a few numbers, and hopefully get an some hints of the solution of the equation.

The same is true if we use a nicer form for the equation: a*x=b. In our example that would be: -1*x=-2. The table in this case would look something like this:

      |  x   | -1*x |  -2  | Equal? |
      |  0   |   0  |  -2  | false  |
      |  1   |  -1  |  -2  | false  |
 ---> |  2   |  -2  |  -2  |  true  | <---
      |  3   |  -3  |  -3  | false  |

      |  x   | 1000*x+7 |  130  | Equal? |
      |  -1  |   -993   |  130  | false  |
      |   0  |      7   |  130  | false  |
      |   1  |   1007   |  130  |  true  |

      |  x   | 1000*x+7 |  130  | Equal? |
      |   0  |      7   |  130  | false  |
      | 0.1  |    107   |  130  | false  |
      | 0.2  |    207   |  130  | false  |

      |  x   | 1000*x+7 |  130  | Equal? |
      | 0.1  |    107   |  130  | false  |
      | 0.11 |    117   |  130  | false  |
      | 0.12 |    127   |  130  | false  |
      | 0.13 |    137   |  130  | false  |

      |   x   | 1000*x+7 |  130  | Equal? |
      | 0.12  |    127   |  130  | false  |
      | 0.121 |    128   |  130  | false  |
      | 0.122 |    129   |  130  | false  |
      | 0.123 |    130   |  130  |  true  |

Drawing the solution

We can do better (and faster) than a table: we could turn a table into a drawing.
Here is the idea: take an equation that looks like an equivalence between two linear combinations, use a coordinate system (AKA a sheet of squared paper with a couple of arrows crossing on it) to DRAW each linear combination, and look for where the two drawing cross (the math term is INTERSECT).
When I say that I want to "turn" a table into a drawing, I meant just that: let's turn the table on its side and create a rectangle for each value, to get a bar chart. Let me take the (by now classic) example: 1*x+5 = 2*x+3 (spoiler!... the solutions is still x=2). I want to take the table for 1*x+5, with values for x that go from x=-3 to x=5, with step 1, and turn it on its side. And since we are doing this, let's calculate also the table for 2*x+3, in the same interval and turn it on its side as well.

Moral of the story so far: riddles (AKA equations) can be written as sentences, represented by a two-pan balance, or written as an equivalence between two linear combinations. The solution to an equation can be found using an algorithm, and there are two special cases that we have to take into account. As an alternative to using the algorithm, we can look for the solution to an equation also by making a table (eventually a very detailed table, very annoying to calculate). It turns out that the table is easier to calculate if we draw its contents on squared paper (AKA a coordinate system). And when you draw a linear combination on a coordinate system you get a straight line; so an equation can be drawn as two lines. The solution to the equation lays where the lines meet (AKA where they intersect). Finally, the two special cases are: no solutions and infinite solutions. When the lines are parallel (they meet nowhere) we have zero solutions; and when the two lines are really the same line, any point on one of the lines is a good solution, hence, infinite solutions.

Puzzles

(1) Solve riddles

Solve the following linear equations, using the final algorithm presented in this chapter.

• 5*x+10 = 2*x+46 → ?
• 7*x+7 = -8*x-2 → ?
• When I walk, I have to eat two candies every kilometer, for energy. However, before I start my walk I always eat 3 candies, just to convince myself to get out of my house. If I ate 11 candies in my last tour, how many kilometers did I walk?
  The equation to solve here is: 3+2*x = 11, and x gives the kilometers walked.
• 2*(4x+.3)-3x=45.6
• (3-^x₂)*2 = (10+x)-2*(x+2)
• 3*(x+x) = 1*2*3*4 + 6*x
• Alice can eat a whole cake in 4 minutes; Bob can also eat a whole cake, but it takes him 6 minutes. How long will it take to Alice and Bob together to eat one cake?
  Here x is the time it takes to eat the whole cake.
  ¹₄ + ¹₆ = ¹_x
  To help you I can rewriting the equation in this way:
  ¹₄ + ¹₆ = ¹_x → (¹₄ + ¹₆)*x = 1 assuming that x is not zero.

(2) Plot the solution

Using the plotting playground (the last one of the chapter) solve these equations:

• -14*x-18 = 16*x
• Can there be a number that divided by three times itself, becomes a third of itself? x:(3*x) = ^x₃
• 9 = 1+2*x

(2.2) Plot the solution