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How do you solve 28+5-10=200?

 Oct 25, 2016

Best Answer 

 #2
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+5

Are you sure you put it in right

 Oct 25, 2016
 #1
avatar+257 
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Huh i tried to solve that in the calc but i couldn't though it just gave me this {}

 Oct 25, 2016
 #2
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+5
Best Answer

Are you sure you put it in right

Guest Oct 25, 2016
 #3
avatar+257 
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Yep

 Oct 25, 2016
 #4
avatar+37152 
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THAT is not 'solveable'     it is incorrect !!!

28+5-10 = 23    NOT 200 !!

 Oct 25, 2016
 #5
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You'd have to figure out what base you're in, first.

 

After that, you'd have to assign new values to each of the symbols given- itd have to be in base five, because there are five terms given.

What I'm asking is  more like this:

 

AB + C - DE = AEE

 

Is this possible to be true, assigning any value to the given terms? Let's find out!

First things first- A has to be low. In fact, A could be zero! Let's try that.

If A is zero, we're left with the following:
B + C - DE = EE

Much easier.

Moving forward from here, we'd have to figure that B+C is EE greater than DE. Let's say that E is one- giving us the following:

B + C - D1 = 11

Now we're getting somewhere! So the final question is- what two one-digit numbers give you a two digit number when SUBTRACTED by a two digit number?

The short answer is, none. Even if we change base, this problem simply does not work!

 

Sorry.

 Oct 25, 2016

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