Solve Einstein's Math Problem: 1-9=100

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The math problem requires using the numbers 1 through 9 in consecutive order to create an expression that equals 100 without using parentheses. Initial attempts included various combinations, but the user was unable to find a valid solution. Several solutions were proposed, including 1 + 23 * 4 + 5 - 6 + 7 - 8 + 9 = 100 and 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 * 9 = 100. The discussion highlights the challenge of adhering to the consecutive order requirement while achieving the target sum. Ultimately, the solutions provided successfully meet the problem's criteria.
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There's a problem on my math homework that I really don't get. I couldn't find it anywhere else on the internet, so I came here. I have to use the numbers
1, 2 ,3, 4, 5, 6, 7, 8, and 9 and I can multiply, divide, add, and subtract, but I can't use parenthesis
to make an expression that will equal 100. I played around for a while and didn't get anything. Plus, I can't use the expression
123-4-5-6-7+8-9=100, because that's the example they gave in the book.
So, could someone help me out here? Thanks.
:smile:
 
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23*5 - 9 - 8 - 7 - 1 + 6 + 4
That's also one.
 
Thanks! :smile:
 
Whoops! I forgot to say that the numbers had to be in consecutive order. Crap. Well, thanks anyway.
 
1+23*4+5-6+7-8+9=100:smile:
1+2+3+4+5+6+7+8*9=100:smile:
 
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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
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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