# Thread: Formula to always get 28

1. ## Formula to always get 28

I need to create a formula where the user would start with x and do at least 4 steps (multiplication, division or subtraction... no *0 allowed) to that number and always get 28. For example
x = any number
x+5 = n
2n+10 = y
2y+ 6 = z
final step
z+3 = 28

obviously that is nor correct.

Thank you!

2. You want the constant function $\displaystyle f(x)=28$ that doesn't depend on x. Now you can't multiply by 0, but subtracting x from itself x-x=0 always.

3. ## Create Algebra Puzzles - Mathemagic

The trick to creating your own algebra puzzles is just keep track of what you are doing to the original number, then undo it in a different order.

For example:
Pick a number (n)
Multply by 4 so that b=4a (and $\displaystyle b=4(n+3)=4n+12$)
Add 120 so that c=b+120 (and $\displaystyle c=4n+12+120$... how many 4's is that.?. $\displaystyle c=4(n+3+30)=4(n+27)$)
Divide by 2 so that d=c/2 (and $\displaystyle d=2(n+27)$)
Subtract 14 so that e=d-14 (and $\displaystyle e=2(n+27)-2(7)=2(n+20)=2n+40$)
Subtract twice the original number so that f=e-2n (and $\displaystyle f=40$)
Subtract 12 so that g=f-12 (and $\displaystyle g=28$)