Eight 8 's
8 □ 8 □ 8 □ 8 □ 8 □ 8 □ 8 □ 8 = 1 0 0 0
What is the minimum number of operators that can be filled in the boxes to make the equation above true?
Note: Boxes can be left blank to denote concetanation of adjacent digits
The answer is 4.
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2 solutions
If one parenthesis is considered as 1 operation, there will be 4 operations. I'm pretty that's not the case so parentheses aren't allowed I Guess.
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A parenthetical is would be an operation if used like 8(8). Otherwise it is to denote order or sequence.
AFAIK parentheses are (grouping) operators in programming languages only, not in algebra, where they only specify the order of operators. But I'm not quite sure. Your solution wouldn't fit the pattern in either case, as )/ are two symbols and we can assume in the original problem that a box refers to maximum one symbol.
Why this is the minimum? Why you can't do it with less operators? I miss the proof.
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I think it would be good to firstly consider what makes the number of operators be minimized to have some basic elementary starting point of that proof.
As such observe that that number is minimized whenever you concatenate 2 eights anyway which will be the first crucial observation in the attempted proof.
The second observation to be done is that , if the operators used are just the basic arithmetic signs + , - , * , / , then any such expression can be seen as reducible at the cumulative sign + that meaning any such expression can be though as a cumulative of the same quantity (that being 8) for a number of times. This 2 observations , once well established and made , can help in fromualting a proof that indeed the found result is minimal or not as they speak once about the way the operators are mnimal and second the synthesize the behavour of operations.
Observe that for the ending result of any sequence of operators for a string of 8s to be possibly the number 1000 you need your last digit to be 0. The cumulative behaveour of number eight is that the final resulting number gets 0 in it's last digit when 8 is multiplied by 5 or , anyway , by a multiple of 5 so the idea of the answer there was exactly to make it reducible to 5 additions involving 8s.
Let's just be lazy and take for granted the result that the minimum is at least that of 5 operators used anyway and see if it can be done in less by the observations.
Then the number of numbers used will be any of 2 , 4 , 3 and it's clear that the sequence can't be made just from addition between those numbers from what was stated earlier so the numbers used should be 2 8s , 4 8s or 3s such that the resulting number of the operators used is 1000. Besides this add the condition that the quantity the nubmers should have should be large enough to give 1000 anyway.
The case of 2 numbers is immediately dismissed and I hope you agree with that anyway.
The case in which there are 3 numbers and 2 operators implies either one of the 3 numbers having at least 4 8s which will give you a number greater than 1000. This means that in the operators used you have to minimize the number by using subtraction and of course division such that you get 1000 where , naming the quantity being used for minimizing the number "subtracting quantity" that quantity used decreases as the nubmer from which it is subtarcted becomes bigger so for cases of numbers of 5 digits of 8 or 6 digits of 8 you have no possibility. That is because the quantity being the biggest increases at a faster rate anyway.
Finally the case of 4 numbers and 3 operators means that you have a maximal quantity of a nubmer of 5 digits which will not work , 4 digits in which case the maximal quantity of the remaining 8 eights will be 88*88 and decrease at a rate which doesn't work , 3 digits which also doesn't work or 8 digits in which case you have the representation of all 8s being of the form 88 88 88 88 which of course doesn't work as there is the same quantity and is not stabilized.
So , althought this is not a synthetic proof and there can be one that anyway so to say does better this reasoning can conclude that it's not possible anyway.
Anyway , that is a rushed proof but I hope you agree with what is said
Well , anyway.
(8888-888)/8=1000
What about (8888-888)/8
(8 8 8-8 (8/(8+8))-8) ((8+8)/8)
You can use 2 operators too , like 8888 - 888 / 8 anyway but I may miss something and please tell if I am wrong somewhere.
Also , maybe you should emphasize that the minimum number is the minimum quantity of operators not the minimum number of different operators to use. anyway.
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That's a good idea, but it doesn't obey the order of operations , unless you add parenthesis.
Yes , I realized that after some time of posting this anyway. The idea of using addition anyway may be said that is very good as well , as underlying it and coming up with it would be an insight into how 8 behaves under operations which is , to speak anyway further , generally done by anyone who wants to arrive at least semi-rationally at a solution.
concetanation is not a word. lol
Can we please specify in the question that parenthesis are not allowed? Otherwise it can be done with 2 operators: (8888 - 888) / 8 = 1000