# Thread: I am stucked by this question

1. ## I am stucked by this question

Is it possible to complete the following expression

2 **** 9 = 2009

and make it true by replacing the asterisks with a selection of the four operational symbols (+ - x / ) and the two other asterisks with digits 0,1,2....,9? Operational symbols and digits can be the same.

Thank you

2. What about 2+3*669

3. ## I am stuck with this question

Hi Drichie,

Sorry, there are 3 digits and 2 operator in your answer unfortunately. Please try again.

4. 2018-9

5. ## I am stucked with this question

Hi Driche,

Thank you for your reply again, but unfortunately, your answer has 3 digit and 1 operator and it still cannot meet the conditions of the question asked. i.e. 2 operator and 2 digits only.

6. how about 2 x 10 cubic+9 =2009

7. ## 2 cents

I'd like to throw in my 2cents and look at this in a case by case basis, if I correctly understand the question.

You want to complete the expression 2****9=2009, using two digits from 0-9 and two operational symbols from (+,-,x,/).
Two operational symbols cannot go together, so I'll break this into 3 cases.
($\displaystyle D=Digit, S=Symbol$)

Case 1
$\displaystyle 2DSDS9=2009$

If the final S is a x, the equation is not viable as 2DSD can only produce an integer, while 2009/9=223.22....
If it is a +, that would require 2DSD=2000, not going to happen.
If it is a -, 2DSD=2018, ditto.
If it is a /, 2DSD=18081
Case 1 is will not work.

Case 2
$\displaystyle 2SDDS9=2009$
If final S is a x, the same result as Case 1
If it is a +, then 2SDD=2000, same as Case 1
If it's a -,same result as Case 1
If it's a /, same result as Case 1
Case 2 will not work.

Case 3
$\displaystyle 2SDSD9=2009$
If final S is a x, then the final D must be a 4, as 49 is a factor of 2009.
Then 2SD=41, not going to happen
If it is a +, then 2SD=2009-D9. The RHS can be at the least 1910, while the LHS can be at most 18, this won't work.
If it is a -, then, then it won't work again because of the same reason as the +.
If it's a /, it won't work just like the previous cases.
Case 3 will not work

As these are all the possible cases, I conclude that there is no solution to this question.

This is my line of reasoning and I could be wrong, so try to get a second opinion.
If exponents are allowed, then glen's solution would be correct, though it seems they aren't allowed.

Btw, who gave you get this question?