This youtube video https://www.youtube.com/watch?v=Mx0x7MoN4qw shows a mathematical puzzle.
You have 101 coins, 51 of which are genuine, 50 are counterfeit.
They all look exactly the same, they only differ in weight: a genuine coin weighs either 1 gram more or 1 gram less than a counterfeit one.
One random coin is chosen. You need to determine if it's genuine or counterfeit.
The only tool you have at your disposal is a weighing balance with a left and right plate, that displays the difference in grams between its left and right plate. You can use it only once.
The solution is:
Leave out the chosen coin. Put the remaining coins on the two plates, 50 left and 50 right. If the displayed difference is an even number, the chosen coin is genuine, otherwise it's counterfeit.
The guy gives the solution and then proves why it works.
It's all very clever, but I have a fundamental issue with this approach: it looks as though to solve the problem one has to know its solution in advance, which sounds quite circular to me.
In this particular case, the problem in essence is to know which and how many coins to put on each plate, and how the displayed difference is linked to the desired answer.
I do not see, in the video, anything that explains how one decides in the first place what the correct coins to put on the two plates are. It's clear that once one has that figured out, the a-posteriori explanation of why it works sort of follows naturally.
So here is my question: is there a general, formal/systematic approach to solving any mathematical problem, or is the 'try one that feels right' approach shown by the video guy sometimes the only possible one?.
To be clear, I am not asking whether one should be able to solve anything from first principles: I am aware that some mathematical knowledge is often necessary to solve some problems (e.g. I imagine it would be quite hard to solve a differential equation without knowing anything about calculus).
What I am trying to understand is whether there is a general process by which one can find the correct mathematical theory to apply to each problem, and proceed step-by-step, rationally, to its solution, without making any random guesses or leaps of faith.
In this case I think I figured out a process, although I am not sure about it.
I am reporting it below, for reference.
After selecting the random coin, 100 coins are left over.
Let $n_L, n_R$ be the numbers of left over coins one places on the left and right plate, respectively.
Let $w_g$ be the weight of a genuine coin.
Let $w_c$ be the weight of a counterfeit coin. Then $w_c = w_g + d, |d| = 1$.
Let $n_u$ be a binary variable indicating whether the randomly chosen coin, of unknown weight $w_u$ that can be either $w_g$ or $w_g + d$, is used or not; without loss of generality, assume that it is placed on the left plate, when used.
The overall numbers of coins on the two plates can be split into genuine and counterfeit coins:
$n_L = n_{L,g} + n_{L,c}$
$n_R = n_{R,g} + n_{R,c}$
The general formula for difference $D$ displayed by the balance is then: $D = n_u w_u + n_{L,g} w_g + n_{L,c} (w_g + d) - n_{R,g} w_g - n_{R,c} (w_g + d)$
The overall numbers of left over genuine and counterfeit coins on the balance, $n_g, n_c$, must fit the following:
$n_g = n_{L,g} + n_{R,g}$
$n_c = n_{L,c} + n_{R,c}$
If the unknown coin is genuine:
$w_u = w_g$
$D = n_u w_g + n_{L,g} w_g + n_{L,c} (w_g + d) - n_{R,g} w_g - n_{R,c} (w_g + d) =$
$= (n_u - n_{R,g} - n_{R,c} + n_{L,g} + n_{L,c}) w_g - d (n_{R,c} - n_{L,c}) =$
$= (n_u - n_R + n_L) w_g - d (n_c - 2 n_{L,c})$
If the unknown coin is counterfeit:
$w_u = w_g + d$
$D = n_u (w_g + d) + n_{L,g} w_g + n_{L,c} (w_g + d) - n_{R,g} w_g - n_{R,c} (w_g + d) =$
$= (n_u - n_{R,g} - n_{R,c} + n_{L,g} + n_{L,c}) w_g - d (- n_u + n_{R,c} - n_{L,c}) =$
$= (n_u - n_R + n_L) w_g - d (- n_u + n_c - 2 n_{L,c})$
After this point I am less sure, but that's how I would reason.
As we do not know $w_g$, to be able to use $D$ we need to make it independent from $w_g$, and the only way I see is to make its coefficient (same in both cases) equal to 0.
$n_u - n_R + n_L = 0$
Solving for $n_u$:
$n_u = n_R - n_L$
Knowing that $n_u$ is a binary variable, $\{0,1\}$, either $n_R = n_L$ or $n_R = n_L + 1$
Substituting $n_u$ in $D$:
If the unknown coin is genuine:
$D = - d (n_c - 2 n_{L,c})$
If the unknown coin is counterfeit:
$D = - d (- n_R + n_L + n_c - 2 n_{L,c})$
Regardless of how $n_R, n_L$ are chosen, in any randomly selected set of left over coins there would still be an unknown $n_c$, except in the single case where all left over coins are put on the balance: then $n_c$ is a constant.
If the unknown coin is genuine:
$n_c = 50$
$D = - d (50 - 2 n_{L,c})$
If the unknown coin is counterfeit:
$n_c = 49$
$D = - d (- n_R + n_L + 49 - 2 n_{L,c})$
Finally, examining the two possibilities for how to divide the left over coins between left and right:
If $n_R = n_L$
If the unknown coin is genuine:
$D = - d (50 - 2 n_{L,c}) = - 2 d (25 - n_{L,c}) \to \text{even integer}$
If the unknown coin is counterfeit:
$D = - d (49 - 2 n_{L,c}) = - d [1 + 2 (24 - n_{L,c})]\to \text{odd integer}$
If $n_R = n_L + 1$
If the unknown coin is genuine:
$D = - d (50 - 2 n_{L,c}) = - 2 d (25 - n_{L,c}) \to \text{even integer}$
If the unknown coin is counterfeit:
$D = - d (- 1 + 49 - 2 n_{L,c}) = - d (48 - 2 n_{L,c}) = - 2 d (24 - n_{L,c}) \to \text{even integer}$
So the only possibility that allows to identify the randomly chosen coin is:
$n_R = n_L = \frac {100} 2 = 50$
$n_u = 0$
The simplest answer is no. No, because if such an approach would exist, then mathematitians would not exist because they would not be needed.
That said, more technically, the answer is sorta yes (but not really), sorta no.
Sorta yes (but not really):
The set of all provable statements following from any set of axioms is countable, therefore it is technically possible to write a computer program that lists all provable statements, and wait for it to spit out the right one.
Note of course, that you have no guarantee when it will produce the statement you need, nor that it will happen before the heat death of the universe.
Sorta no:
Notice that I was talking about provable statements above, not true statements. That's because in every sufficiently advanced system of axioms (and that "sufficiently advanced" is a very low bar, this small set already suffices), there will always exist statements that are true, but cannot be proven.