Missing Operators: May 2021 Challenge Question

This is the one we knew of.

Good job to the other folks on this!

19+90 = 99999999/999999+9

Not quite but close enough for a good answer for introducing the idea of manipulating before the equals as well as after.

It is correct iff the calculator used is set to display no more than three digits after the decimal point.

1990=((9999-9)x((9+9)/9)-(9x9-(9/9)))/(9+(9/9))

But StandardsGuy has opened up a whole load of answer

First notice that if you can manage with an odd number of 9s less than 15 on the right hand side then you can easily waste the remaining ones.

0:-} 1 - (9/9) + 0 = ((9+9)/9) - (9/9) - (9/9)

1:-} 1 + 9 - 9 + 0 = ((9+9+9)/9) - ((9+9)/9)

2:-} 1 + (9/9) + 0 = (9+9)/9

10:-} 19 - 9 + 0 = (99+(9/9))/(9+(9/9)

Although already answered, here is my stream of thought:

1990 = 1000+900+90 = 1000 + 1000-100 + 100-10 =

999+1 + 999+1-(99+1) + 99+1-(9+1) = 999+999-99+99-9+1

So far so good, but this 1 at the end spoils the fun. Moreover, there are also another four 9's to take into account. Why not substituting 1 with 99:99 then? We can thus finally get:

1990 = 999+999-99+99-9+99:99

One could play with the cancelling +99 and -99 and add an extra 9 to each like this:

1990 = 999+999-999+999-9+9:9

ore remove a 9 from each and get:

1990 = 999+999-9+9-9+999:999

By using parentheses, one can change +'s to -'s and v.v. like e.g.:

1990 = 999+999-(9-9)-9+999:999

and so on. Of course one can also change position of the numbers, like e.g.:

1990 = 9:9-9+999+999+999-999

and so on, but I guess the latest examples do not provide any significant contribution.