3
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$Given$:

$3$ = { A } + { B } + { C }

Where A, B, C represent 3 unique Pan Digital Expressions.

Each contains all the digits 1 to 9 occurs only once.

Expressions contain only plus, division signs . Left and Right brackets are present. No concatenations allowed.

Solve for A,B,C.

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closed as too broad by gabbo1092, hexomino, Bass, Omega Krypton, athin Jun 4 at 5:50

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ So we are only allowed to use + and $\div$? Can we use parenthesis like $A=(1+2) \div 3$? Or implied parenthesis by $A=\frac{1+2}{3}$? $\endgroup$ – Trenin May 31 at 15:00
  • $\begingroup$ Can we concatenate for example use number $123$? $\endgroup$ – Weather Vane May 31 at 15:02
  • $\begingroup$ Is the usual BEDMAS order of operations assumed? $\endgroup$ – Trenin May 31 at 15:04
  • $\begingroup$ @Trennin..sure..that’s what I meant left and right brackets. @weathervane..no concatenations allowed. All 3 expressions contain single digits 1 to 9, occurring only once....linked as needed by plus and division signs $\endgroup$ – Uvc May 31 at 15:05
  • $\begingroup$ As an example, one of the terms might look like...9/(1+6) $\endgroup$ – Uvc May 31 at 15:06
4
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OK - here is a real solution this time without concatenation.

I have a couple sub expressions.

$x=\frac{2+3}{5}=1$
$y=\frac{9+8+7}{6} \div 4=1$

Now I can construct $A,B,C$.

$$A= 1 \div \frac{\frac{2+3}{5}}{\frac{9+8+7}{6} \div 4}$$
$$B= \frac{2+3}{5} \div \frac{1}{\frac{9+8+7}{6} \div 4}$$
$$C= \frac{2+3}{5} \div \frac{\frac{9+8+7}{6} \div 4}{1}$$

Not sure if this is in the spirit of the puzzle since I am using the fact that $1 \div 1 = 1$ and getting a set of pan digital sub-expressions which equal to 1.

EDIT: I've added some other possibilities which don't bear as much resemblance to eachother.

$$A= 1 \div \frac{\frac{2+3}{5}}{\frac{9+8+7}{6} \div 4}$$
$$B= \frac{5+\frac{9+8+7}{6} \div 4}{1+2+3}$$
$$C=\frac{9+8+7+1}{5} \div \frac{\frac{6}{3}}{4 \div 2}$$

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  • $\begingroup$ I'd go for it, and call this a valid solution. Congrats! $\endgroup$ – Brandon_J May 31 at 15:57
  • $\begingroup$ As per puzzle, they all should be unique..you don’t have to use 1/1 trick..if nobody else comes close after few days, I will accept your answer...A, B, C as they stand now are just rearrangements $\endgroup$ – Uvc May 31 at 16:01
  • $\begingroup$ @Uvc Would you consider $\frac{5+\frac{9+8+7}{6} \div 4}{1+2+3}$ a different solution? $\endgroup$ – Trenin May 31 at 16:05
  • $\begingroup$ @Uvc Or how about $\frac{9+8+7+1}{5} \div \frac{\frac{6}{3}}{4 \div 2}$? $\endgroup$ – Trenin May 31 at 16:09
  • $\begingroup$ Sure..sufficiently different $\endgroup$ – Uvc May 31 at 16:10
4
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With pencil and paper:

A. $ \frac{3}{9} + \frac{1}{6} + \frac{ (4+8) / (5+7) }{2} = \frac{1}{3} + \frac{1}{6} + \frac{1}{2} = 1 $

B. $ \frac{1 + 2 + 3 + 4 + 5}{6 + 7 + 8 + 9} = \frac{15}{30} = \frac{1}{2} $

C. $ \frac{3 + 7 + 8 + 9}{1 + 2 + 4 + 5 + 6} = \frac{27}{18} = \frac{3}{2} $

Sum: $ 1 + \frac{1}{2} + \frac{3}{2} = 3$

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  • $\begingroup$ Very good one.... $\endgroup$ – Uvc May 31 at 16:40
  • $\begingroup$ I was looking for three $1$s but got the last two fairly quickly when I looked for them. From $\sum_1^9 n = 45$. $\endgroup$ – Weather Vane May 31 at 16:54
  • $\begingroup$ For fun..this is the way to do $\endgroup$ – Uvc Jun 1 at 13:26
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Here is one solution:

EDIT: Uses concatenations, so not a valid answer - sorry.

$$A=\frac{\frac{7852}{1963}}{4}$$
$$B=\frac{\frac{6952}{1738}}{4}$$
$$C=\frac{\frac{5796}{483}}{12}$$

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  • $\begingroup$ why the downvote? +1! $\endgroup$ – Omega Krypton May 31 at 15:20
  • $\begingroup$ Quite clear - no concatentions. $\endgroup$ – Weather Vane May 31 at 15:20
  • $\begingroup$ @WeatherVane ARG - Didn't see that.... Back to the drawing board... $\endgroup$ – Trenin May 31 at 15:21
2
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After wondering why people weren't doing very simple ones. I reread the rules realizing I misread the part where it says "Each contains all the digits 1 to 9 occurs only once." after correcting my self, I came up with this:

$$A = \frac{2+3+4+5+8}{6+7+9} \div 1 = \frac{22}{22} \div 1 = \frac{1}{1} = 1$$

$$B = \frac{\frac{1+8}{9}+7+6}{5+4+3+2} = \frac{\frac{9}{9}+13}{14} = \frac{1+13}{14} = \frac{14}{14} = 1$$

$$C = \frac{\frac{1+2}{3}+ \frac{4+5}{9}}{\frac{6+8}{7}} = \frac{\frac{3}{3} + \frac{9}{9}}{\frac{14}{7}} = \frac{1 + 1}{2} = \frac{2}{2} = 1$$

Sum: 1 + 1 + 1 = 3

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  • $\begingroup$ Welcome to Puzzling.SE, and thanks for the great answer! $\endgroup$ – Brandon_J Jun 1 at 19:04
  • $\begingroup$ Great effort!...once you realize that fractional route is better for this particular problem, solution becomes much neater..after you finish..you can look at my answer. $\endgroup$ – Uvc Jun 1 at 19:12
1
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Intended answer consists of 3 expressions with 3 fractions each:

enter image description here

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