Hello, new to this forum, I really need some help!

[Beatleschick1996]

Hello!! I am new here, and I am not good at math at all. My strenghths are in other areas but I have these "Math Tiles" That are almost like Sudoku (if thats appropriate spelling) and they are really hard for me to understand. We also can only use numbers 1-9, and each number can only be used once. Here's an example...

[___ x ___]+___= 14
__ x __+___ = 51
___ x [___ + ___] = 27

And on most there's numbers also going down, so they want it to work vertically as well. If someone could help me with this that would be amazing. I am struggling with these a lot. Thank you so much!

-Emmi

 

[Denis]

Please post CLEARLY an actual problem.

Your example has this solution:
2 * 5 + 4 = 14
6 * 7 + 9 = 51
3 *(1 + 8) = 27
...but nothing can be done about the verticals.

 

[Beatleschick1996]

Oh ok, sorry if it wasn't clear. I know how frustrating unclear posts can be for Mods. :? But I was wondering if anyone knew of some type of formula to solve these problems, if nothing is known I'll just try to do them myself

 

[chrisr]

I know what you mean,
games like this are good practice.
You have 9 single-digit "variables" though.

The best way is probably to start with the last line and write the factors of 27,
which are 9 and 3, so you'd begin with writing those possibilities down.
3(5+4), 3(6+3), 3(7+2), 3(8+1)
(1+2)(9).
You cannot use 3(6+3).
Your last line must be one of the remaining 4.
You could pick one and then see if you can write the first 2 lines using the remaining 6 digits.
You find, the remaining 6 digits only combine correctly if you choose 3(8+1).

It's quite easy to start out by choosing digits in "add" and "multiply" combinations
and writing down the answers, but working backwards to discover the numbers chosen requires
some detective work.

In the solution as Denis gave, imagine combining the verticals in different combinations
of add and multiply.
You have horizontal answers, you could be given vertical ones instead.

 

[soroban]

Hello, Beatleschick1996!

Here is a solution:

. . \(\displaystyle \begin{array}{ccccccc} \bigg(\boxed{2} & \times & \boxed{5}\bigg) & + & \boxed{4} & = & 14 \\ \\[-3mm] \bigg(\boxed{6} & \times & \boxed{7}\bigg) & + & \boxed{9} &=& 51 \\ \\[-3mm] \boxed{3} & \times & \bigg(\boxed{1} &+& \boxed{8}\bigg) &=& 27 \end{array}\)