Integer solutions for equations A * [(B + C)(D - E) - F(G*H) ] / J = 1

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The discussion centers on solving the equation A * [(B + C)(D - E) - F(G*H)] / J = 10, with constraints that each variable is a unique, single-digit, nonzero number and specific relationships between the variables (C - B = 1 and H - G = 3). Initial attempts suggest that there are numerous solutions, with one approach yielding 720 distinct combinations when certain values are assigned. However, the uniqueness condition complicates the problem, as it requires all numbers to be non-repeating and non-zero. A calculation reveals 26 valid combinations for (B, C, G, H), leading to a total of 1560 special cases to analyze. Participants are encouraged to explore these combinations independently rather than seeking quick answers.
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Could anyone please answer this


You are confronted with the following formula:

A * [(B + C)(D - E) - F(G*H) ] / J = 10

Knowing that each variable is a unique, single-digit, nonzero number, and that C - B = 1, and H - G = 3, what is the number ABCDEFGHJ, where each letter is a digit? For example, if A = 4, B = 2, and C = 7, ABC would equal 427.

Thanks
 
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You have 3 equations and 9 unknowns, so most likely, you have lots of solutions.

Let us set A=J, H=3, G=0, whereby we must have (B+C)(D-E)=10

Setting now B=2, C must be 3, and D-E=2.

These particular choices determine 720 distinct solutions to your problem.

Or should your numbers be non-repeating? (Your uniqueness condition?)

Okay, I forgot that none of your numbers should be zero, either.

Then your problem is difficult, in the sense of being tedious:
If I have counted correctly, there exist 26 different quadruples (B,C,G,H) so that C-B=1, H-G=3, and all 4 numbers are unique, single digit numbers.

You will solve your problem on a trial&error basis restricting yourself to these 26 special cases.
 
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Aw come on, I just want the answer in a hurry!

It is Christmas, afer all.
 
Well, since there will exist for each such quadruple 10 combinations of D and E (since D-E must be positive), and the three remaining numbers can be placed in the A, J F positions, you are free to analyze the 26*10*6=1560 special cases occurring.

But don't count upon us to do it for you.
 
Ben Fido, please read and follow the rules for these forums.
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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