Minimize with operators!
1 0 □ 1 0 □ 1 0 □ 1 0 □ 1 0
Fill in the squares with exactly one of each of these mathematical operators: + , − , × , ÷ . You are also allowed to use one ( 1 ) parenthesis ( ) . You may choose to use it or not to use it, and if you do, you cannot use it as a multiplication operator. Example:
1 0 + ( 1 0 × 1 0 ) is acceptable; 1 0 ( + 1 0 × 1 0 ) is not acceptable.
Now, what is the smallest positive integer that you can get from this expression?
Note : order of operations (BODMAS) applied.
The answer is 1.
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1 solution
Nice! Any way we can get 0?
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With the given conditions? I don't think so
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We xan get 0, but its not a positive integer, thus it is not an answer.
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@Ashish Menon – A 0 with the given conditions? How????
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@Hung Woei Neoh – To obtain 0 one should obtain an expression of either the form 0*10 (0/10) or of the form a - a (or -a+a) which can be tricky to find as the quantity of 10s is odd.
A priori (in a large sense of the term "a priori") , none of this forms seem anyway to be possible to obtain with the "conditions" of the problem so to say indeed.
To verify or try to prove it is impossible consider the 2 possible cases. What anyway will be the done can be described as a study of what would make it possible or it other words , can be described as a study of the conditions of possibility taken in it's inner structure (but it sounds too technical) anyway.
Since the quantity is odd it is expected for the "weight" of the combinations of numbers , especially since they are equal , to be greater in one part than other.
The result of all possible operations between two 10s are 100 , 1 , 20 , 0. Since the operators used are + , - , / , * it and parenthesis , though them anyway not necessarily it doesn't seem possible to obtain the form 0*10 either.
To obtain that form firstly observe that parenthesis would be necessary because otherwise you would have a 0 at some point followed by other operator. For the use of the parenthesis therefore which is considered as such necessary you have to put them such that they eventually lead to the form 0*10 which is not possible unless you put the parenthesis between 4 10s for the same reason as upper. T
Then if , you have to put the parenthesis between 4 10s which would have as a result 0 the operators between the 4 10s should not include * as that operator will be used to obtain the form 0*10 and using just / , + and - which gives different results the expression will not be certainly 0. So indeed , under the conditions stated as a result of the structure of the problem and the restrictions of the used operators it doesn't seem to be possible to obtain equal quantities or 0 anyway so I'm also let to wonder how it can be possible anyway.
So considering the structure of the use of operators I do agree with you that it's impossible under the given conditions.
But if Shiva says it's possible I'd love to see that way anyway.
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@A A – Cool explanation(+1)
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@Ashish Menon – It's not too rigorous but thanks. You can try to explain it better and make it rigorous if you want anyway.
@Hung Woei Neoh – Oops, sorry, I used 2 set of parentheses
Yes , there is another method to obtain it. Can you find it anyway ?
(10×10+10)÷10-10 (100+10)÷10-10 110÷10-10 11-10=1
( 1 0 − 1 0 + 1 0 − 1 0 ) × 1 0
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You didn't use ÷
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Ah, yes. My bad. Also, does anyone feel these comments are structured in a weird manner?
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@A Former Brilliant Member – You are not the first person to comment about it
( 1 0 − 1 0 ) × 1 0 + 1 0 ÷ 1 0
This expression will yield 1 as an answer:
( 1 0 − 1 0 ) × 1 0 + 1 0 ÷ 1 0 = 0 × 1 0 + 1 = 0 + 1 = 1
As 1 is the smallest positive integer available, and we are able to obtain this value, this is the answer we're looking for.
I haven't thought about it, but is there any other method to obtain 1 ?