Crossing Numbers

Logic Level 1

$\large{\begin{array}{ccccccc} & & \boxed{\phantom0} & & & & \\ & & - & & & & \\ \boxed{\phantom0} &+& \boxed{\phantom0} & = &\boxed{\phantom0} & & \\ & & = & & & & \\ & & \boxed {\phantom0} & & & & \\ \end{array}}$

Can we fill in the boxes above with distinct integers from 1 to 5 such that the array of equations above holds true?

Yes No

5 solutions

Angela Fajardo
Jul 16, 2016

Relevant wiki: Arithmetic Puzzles - Fill in the Blanks

The boxes may be filled as follows:

$\Large{\begin{matrix} \begin{array}{ccccccc} & & \boxed { 5 } & & & & \\ & & - & & & & \\ \boxed { 1 } & + & \boxed { 3 } & = & \boxed { 4 } & & \\ & & = & & & & \\ & & \boxed { 2 } & & & & \\ & & & & & & \end{array} & \begin{matrix} & & \end{matrix} & \begin{array}{ccccccc} & & \boxed { 5 } & & & & \\ & & - & & & & \\ \boxed { 2 } & + & \boxed { 1 } & = & \boxed { 3 } & & \\ & & = & & & & \\ & & \boxed { 4 } & & & & \\ & & & & & & \end{array} \\ \begin{matrix} \\ \\ \end{matrix} & & \\ \begin{array}{ccccccc} & & \boxed { 4 } & & & & \\ & & - & & & & \\ \boxed { 2 } & + & \boxed { 3 } & = & \boxed { 5 } & & \\ & & = & & & & \\ & & \boxed { 1 } & & & & \\ & & & & & & \end{array} & & \begin{array}{ccccccc} & & \boxed { 3 } & & & & \\ & & - & & & & \\ \boxed { 4 } & + & \boxed { 1 } & = & \boxed { 5 } & & \\ & & = & & & & \\ & & \boxed { 2 } & & & & \\ & & & & & & \end{array} \end{matrix} }$

And if written separately:

$\Large{{ \begin{array}{ccccccc} & & \boxed { 5 } & & & & \\ & & - & & & & \\ \boxed { 1 } & + & \boxed { 3 } & = & \boxed { 4 } & & \\ & & = & & & & \\ & & \boxed { 2 } & & & & \\ & & & & & & \end{array} }}$

$\Large{{ \begin{array}{ccccccc} & & \boxed { 5 } & & & & \\ & & - & & & & \\ \boxed { 2 } & + & \boxed { 1 } & = & \boxed { 3 } & & \\ & & = & & & & \\ & & \boxed { 4 } & & & & \\ & & & & & & \end{array} }}$

$\Large{{ \begin{array}{ccccccc} & & \boxed { 4 } & & & & \\ & & - & & & & \\ \boxed { 2 } & + & \boxed { 3 } & = & \boxed { 5 } & & \\ & & = & & & & \\ & & \boxed { 1 } & & & & \\ & & & & & & \end{array} }}$

$\Large{{ \begin{array}{ccccccc} & & \boxed { 3 } & & & & \\ & & - & & & & \\ \boxed { 4 } & + & \boxed { 1 } & = & \boxed 5 & & \\ & & = & & & & \\ & & \boxed { 2 } & & & & \\ & & & & & & \end{array} }}$

Angela Fajardo - 4 years, 11 months ago

Great! But I meant for you to present also the principle before finding them , meaning what is the reasoning by which you find all the solutions ?

A A - 4 years, 11 months ago

Only used common sense hahaha and quickly thought a solution and pressed Yes. But here's some idea:

Knowing the number $1, 2, 3, 4, 5$ are used only once:

Starting with the Subtraction Part:

$\Large{\begin{matrix} \boxed { } \\ - \\ \boxed { } \\ = \\ \boxed { } \end{matrix}}$

The minimum value of the minuend would be 3 because:

If 1 is the minimum value, then 1 - ? = negative number (and all the numbers are positive and to be used only once)
If 2 is the minimum value, then 2 - ? = negative number (and can't be 1 since $2-1=1$ and same reason as above)

So the possible values are: 3, 4, and 5.

Then with the Addition Part:

$\Large{\begin{matrix} \boxed { } & + & \boxed { } & = & \boxed { } \end{matrix}}$

The maximum value of the addend would be 4 because:

If 5 is the maximum value, then 5 + ? = number $>5$ (and we are only limited to numbers 1 to 5)

So the possible values are: 1, 2, 3, and 4

Finding One of the Solutions:

Let's try 5 first:

$\Large{{ { \begin{array}{ccccccc} & & \boxed { 5 } & & & & \\ & & - & & & & \\ \boxed { } & + & \boxed { } & = & \boxed { } & & \\ & & = & & & & \\ & & \boxed { } & & & & \\ & & & & & & \end{array} } }}$

$5 - 1 = 4$
$5 - 2 = 3$
$5 - 3 = 2$
$5 - 4 = 1$

The possible values would be $5 - 3 = 2 and 5 - 1 = 4$ because:

Taking into consideration the addition part of the problem:

$5 - 2 = 3$ would leave $1 + 2 = 4$ or $4 + 2 = 1$ which are both not true
$5 - 4 = 1$ would leave $2 + 4 = 3$ or $3 + 4 = 2$ which are both not true

So, using the possible values above, we are able to have at least one solution and prove the problem true

Angela Fajardo - 4 years, 11 months ago

@Angela Fajardo – @A A though I don't know how to figure out all the solutions of it

Angela Fajardo - 4 years, 11 months ago

@Angela Fajardo – Well then , I suppose I shouldn't let you do all the work alone anyway. Actually the problem is a good illustration of combinatorial thinking and by the reflexive notes which it implies also for understanding determinism where "determinism" means understanding which covers both , the abstract and concrete ways things are determined so because of this suppose the problem asks you to find all the possible solutions and also show all the working configurations (constructions) of it.

Anyway maybe this problem is better as it requires and asks for a complete understanding because in order to solve it you would have to think at what makes all the possible solutions and therefore arrive at a complete understanding of it thing which is more interesting that finding some solutions I suppose anyway. So then it may therefore seem interesting to ask how to think in order to solve such a problem that meaning therefore in what terms which almost immediately implies that anyway it would be needed some sort of organization of reasoning based on rules should the problem be seen such that it can be easily understood and solved.

Anyway , this is more of an illustration of combinatorial thinking. The solution of finding all the possible solutions isn't to hard and can be done in a few lines. Anyway , in other words is an illustration of how you think to solve such problems. It is therefore done more in the reflexive scope of understanding than in solving.

I'm pretty sure that there are pretty many ways of solving the problem. One way of understanding things by principle would be to think of the two equations of the problem independently at first finding all their possible solutions and then seeing when and what also the 2 equations can be synthetically considered together. Leading this reasoning you would find that the only possible configurations which works for addition (and because subtraction is the same as addition) for subtraction also anyway are the triplets (1,4,5) (1,3,2) (1,4,3) , (2,3,5) from which taking them one after the other when you construct the solutions by some restrictive rules and conditions you obtain all the solutions.

Now let's use some other reasoning to udnerstand when 2 solutions for the 2 independently conceived equations at first doesn't work together . Another way of thinking in a combinatorial fashion at the problem and therefore to find some kind of organization centered on rules for this problem which works and fit together is to try to understand it as a deterministic function of one certain element , by this I mean that you take some element and observe how starting from making statements about it anyway meaning to say it more formally making predications about it , the configuration is influenced by those changes to the point it gives you some constructions or some others.

As a note which has nothing to do with the solution to this problem anyway note that such a technique or orientation of understanding doesn't always helps you in dealing with combinatorial problems and especially isn't very helpful when there are more complicated cases when you deal with many element which have many relations cases in which you either form more abstract statements which usually are seen as emergent characteristics or you make very abstract predications.

Start from the banal , general observation that it is known that any possible working configuration uses all the available numbers from 1 to 5. Is there some special number or place in those anyway which behaves differently to the point by which the predication can be made such that it organizes nicely the understanding of the entire problem ? Of course there is anyway. Meaning by that the central square which admits just certain numbers , as you already observed , such that the 2 equations to be valid and because 5 and 4 can't be used in some of the equations in the place corresponding to the central square it can be cocnluded that the only nubmers that can be placed in that central square are just anyway the numbers 1 , 2 , 3.

Then because you know that (1,4,5) (1,3,2) (1,4,3) , (2,3,5) are the possible solutions and for any available solution one number is common which belongs in the central squar you can easily obtain all possible solutions along with an understanding of their construction. The same reasoning can be done in various way I suppose and considering many other stuff but it's pretty much determine din the same way considering this characteristics of the system so finding all possible solutions isn't hard at all anyway.

A A - 4 years, 11 months ago

This is a cute problem if you want to find all possible solutions.

Can you show which are all the possible solutions ?

A A - 4 years, 11 months ago