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You are given that the numbers 1 , 2 , 3 , 4 , 5 , 6 are to be filled in the square boxes as shown above (without repetition) such that it represents a 2 digit integer subtracting two 2-digit integers.
Of all 6 ! = 7 2 0 possible arrangements, find the largest negative resultant number.
Details and Assumptions
As an explicit example, we can have 1 6 − 3 4 − 5 2 = − 7 0 and 1 2 − 5 3 − 6 4 = − 1 0 5 as a resultant number. We will choose the formal answer of − 7 0 because − 1 0 5 < − 7 0 .
See Part 1 .
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I got 12-64-53=-105... I thought you were looking for a large negative... As in most negative. This isn't so clear
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Yes. I thought exactly the same thing!
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me too. we were in the same school of thought.
How would you make it clearer?
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Maybe say it should be a negative number of smallest magnitude, or that by larger you mean closer to 0
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@James Cardus – The question is clear. It may be written as thus in order to maybe trick some people, but it is written correctly. The largest negative number is -1 (I at first put in -105, but that is why we have three tries at problems)
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@Chris Hambacher – I guess I just associate the term 'largest' with magnitude. I'm not great with semantics so you're probably right ;)
You could say 'smallest negative integer for which ∣ x ∣ is minimized;'
Yeah! exactly....The question was vague.
Thanks. I've updated the problem statement for clarity. Those who previously answered -105 will be marked correct.
The same happened to me.
The largest negative is − 1 , so I tried and found that 4 6 − 1 2 − 3 5 = − 1
Clearly, 65-43-21 = 1. Swapping out that 5 and 3's position, we have 63-45-21=-3. Therefore, the answer must be either -1, -2, or -3. You have 3 guesses and can try all 3 answers.
64-53-12=-1
and a nice trap for largest -ve integer
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4 6 − 1 2 − 3 5 = − 1