Four Fours Challenge: Get 44 with ONLY 4s!

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The challenge is to use exactly five fours to achieve the number 44. One proposed solution is 44/4 + 44, although it raises questions about its rigor. Participants express a mix of humor and frustration regarding the challenge's complexity. There is a suggestion to avoid excessive editing of posts. The discussion highlights the playful nature of mathematical challenges while seeking valid solutions.
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Homework Statement


use five fours ONLY to get 44

Homework Equations





The Attempt at a Solution

 
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44/4 + 44. Kind of a silly question? :P

[Edit] If someone can come up with something rigorous(uniqueness and existence of such numbers) good on them, I just guessed and checked.
 
Last edited:
It used to say "get 55..." Nice edit!
 
44 mod 4 + 44

Stop editing your post lol
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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