3 as a Sum of 3 Pan Digital ExpressionsA bad and boring riddle, and an easy alphameticAlphametic twinsThe impossible digital sumFive Powers of Fives Produce Unique Pandigital Number…Solve for X..Tell me YFind the values of U, V, C based on the given relationship…useful for upcoming puzzlesPalindromic Pan digital Special Square wants you to reveal its RootUVC wants to give you a Helping Hand in Solving these Unique Set of Pan digital Fractional-Decimal RelationsStandard Sudoku Specialized with just 2 PrimesRep Digit Palindrome expressed as a Unique set of Pan Digital RelationsExpress 2222 as a sum of 3 powers in 2 Different Pan Digital Expressions
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3 as a Sum of 3 Pan Digital Expressions
A bad and boring riddle, and an easy alphameticAlphametic twinsThe impossible digital sumFive Powers of Fives Produce Unique Pandigital Number…Solve for X..Tell me YFind the values of U, V, C based on the given relationship…useful for upcoming puzzlesPalindromic Pan digital Special Square wants you to reveal its RootUVC wants to give you a Helping Hand in Solving these Unique Set of Pan digital Fractional-Decimal RelationsStandard Sudoku Specialized with just 2 PrimesRep Digit Palindrome expressed as a Unique set of Pan Digital RelationsExpress 2222 as a sum of 3 powers in 2 Different Pan Digital Expressions
$begingroup$
$Given$:
$3$ = 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.
mathematics no-computers
asked 9 hours ago
UvcUvc
1,159119
$endgroup$
|
show 8 more comments
$begingroup$
$Given$:
$3$ = 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.
mathematics no-computers
asked 9 hours ago
UvcUvc
1,159119
$endgroup$
$begingroup$
So we are only allowed to use + and $div$? Can we use parenthesis like $A=(1+2) div 3$? Or implied parenthesis by $A=frac1+23$?
$endgroup$
– Trenin
9 hours ago
$begingroup$
Can we concatenate for example use number $123$?
$endgroup$
– Weather Vane
9 hours ago
$begingroup$
Is the usual BEDMAS order of operations assumed?
$endgroup$
– Trenin
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
As an example, one of the terms might look like...9/(1+6)
$endgroup$
– Uvc
9 hours ago
|
show 8 more comments
$begingroup$
$Given$:
$3$ = 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.
mathematics no-computers
asked 9 hours ago
UvcUvc
1,159119
$endgroup$
$Given$:
$3$ = 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.
mathematics no-computers
mathematics no-computers
asked 9 hours ago
UvcUvc
1,159119
asked 9 hours ago
UvcUvc
1,159119
edited 8 hours ago
Uvc
asked 9 hours ago
UvcUvc
1,159119
asked 9 hours ago
UvcUvc
1,159119
asked 9 hours ago
UvcUvc
1,159119
1,159119
$begingroup$
So we are only allowed to use + and $div$? Can we use parenthesis like $A=(1+2) div 3$? Or implied parenthesis by $A=frac1+23$?
$endgroup$
– Trenin
9 hours ago
$begingroup$
Can we concatenate for example use number $123$?
$endgroup$
– Weather Vane
9 hours ago
$begingroup$
Is the usual BEDMAS order of operations assumed?
$endgroup$
– Trenin
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
As an example, one of the terms might look like...9/(1+6)
$endgroup$
– Uvc
9 hours ago
|
show 8 more comments
$begingroup$
So we are only allowed to use + and $div$? Can we use parenthesis like $A=(1+2) div 3$? Or implied parenthesis by $A=frac1+23$?
$endgroup$
– Trenin
9 hours ago
$begingroup$
Can we concatenate for example use number $123$?
$endgroup$
– Weather Vane
9 hours ago
$begingroup$
Is the usual BEDMAS order of operations assumed?
$endgroup$
– Trenin
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
As an example, one of the terms might look like...9/(1+6)
$endgroup$
– Uvc
9 hours ago
$begingroup$
So we are only allowed to use + and $div$? Can we use parenthesis like $A=(1+2) div 3$? Or implied parenthesis by $A=frac1+23$?
$endgroup$
– Trenin
9 hours ago
$begingroup$
So we are only allowed to use + and $div$? Can we use parenthesis like $A=(1+2) div 3$? Or implied parenthesis by $A=frac1+23$?
$endgroup$
– Trenin
9 hours ago
$begingroup$
Can we concatenate for example use number $123$?
$endgroup$
– Weather Vane
9 hours ago
$begingroup$
Can we concatenate for example use number $123$?
$endgroup$
– Weather Vane
9 hours ago
$begingroup$
Is the usual BEDMAS order of operations assumed?
$endgroup$
– Trenin
9 hours ago
$begingroup$
Is the usual BEDMAS order of operations assumed?
$endgroup$
– Trenin
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
As an example, one of the terms might look like...9/(1+6)
$endgroup$
– Uvc
9 hours ago
$begingroup$
As an example, one of the terms might look like...9/(1+6)
$endgroup$
– Uvc
9 hours ago
|
show 8 more comments
4 Answers
4
active
oldest
votes
$begingroup$
OK - here is a real solution this time without concatenation.
I have a couple sub expressions.
$x=frac2+35=1$
$y=frac9+8+76 div 4=1$
Now I can construct $A,B,C$.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac2+35 div frac1frac9+8+76 div 4$$
$$C= frac2+35 div fracfrac9+8+76 div 41$$
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.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac5+frac9+8+76 div 41+2+3$$
$$C=frac9+8+7+15 div fracfrac634 div 2$$
answered 8 hours ago
TreninTrenin
8,0031645
$endgroup$
$begingroup$
I'd go for it, and call this a valid solution. Congrats!
$endgroup$
– Brandon_J
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@Uvc Would you consider $frac5+frac9+8+76 div 41+2+3$ a different solution?
$endgroup$
– Trenin
8 hours ago
$begingroup$
@Uvc Or how about $frac9+8+7+15 div fracfrac634 div 2$?
$endgroup$
– Trenin
8 hours ago
$begingroup$
Sure..sufficiently different
$endgroup$
– Uvc
8 hours ago
add a comment |
$begingroup$
With pencil and paper:
A. $ frac39 + frac16 + frac (4+8) / (5+7) 2 = frac13 + frac16 + frac12 = 1 $
B. $ frac1 + 2 + 3 + 4 + 56 + 7 + 8 + 9 = frac1530 = frac12 $
C. $ frac3 + 7 + 8 + 91 + 2 + 4 + 5 + 6 = frac2718 = frac32 $
Sum: $ 1 + frac12 + frac32 = 3$
answered 8 hours ago
Weather VaneWeather Vane
3,5781118
$endgroup$
$begingroup$
Very good one....
$endgroup$
– Uvc
7 hours ago
$begingroup$
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
7 hours ago
add a comment |
$begingroup$
Here is one solution:
EDIT: Uses concatenations, so not a valid answer - sorry.
$$A=fracfrac785219634$$
$$B=fracfrac695217384$$
$$C=fracfrac579648312$$
answered 8 hours ago
TreninTrenin
8,0031645
$endgroup$
$begingroup$
why the downvote? +1!
$endgroup$
– Omega Krypton
8 hours ago
$begingroup$
Quite clear - no concatentions.
$endgroup$
– Weather Vane
8 hours ago
$begingroup$
@WeatherVane ARG - Didn't see that.... Back to the drawing board...
$endgroup$
– Trenin
8 hours ago
add a comment |
$begingroup$
Answer consists of 3 expressions with 3 fractions each
![ 3 as sum of 3 pandigital
]1
answered 5 hours ago
UvcUvc
1,159119
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
OK - here is a real solution this time without concatenation.
I have a couple sub expressions.
$x=frac2+35=1$
$y=frac9+8+76 div 4=1$
Now I can construct $A,B,C$.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac2+35 div frac1frac9+8+76 div 4$$
$$C= frac2+35 div fracfrac9+8+76 div 41$$
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.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac5+frac9+8+76 div 41+2+3$$
$$C=frac9+8+7+15 div fracfrac634 div 2$$
answered 8 hours ago
TreninTrenin
8,0031645
$endgroup$
$begingroup$
I'd go for it, and call this a valid solution. Congrats!
$endgroup$
– Brandon_J
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@Uvc Would you consider $frac5+frac9+8+76 div 41+2+3$ a different solution?
$endgroup$
– Trenin
8 hours ago
$begingroup$
@Uvc Or how about $frac9+8+7+15 div fracfrac634 div 2$?
$endgroup$
– Trenin
8 hours ago
$begingroup$
Sure..sufficiently different
$endgroup$
– Uvc
8 hours ago
add a comment |
$begingroup$
OK - here is a real solution this time without concatenation.
I have a couple sub expressions.
$x=frac2+35=1$
$y=frac9+8+76 div 4=1$
Now I can construct $A,B,C$.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac2+35 div frac1frac9+8+76 div 4$$
$$C= frac2+35 div fracfrac9+8+76 div 41$$
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.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac5+frac9+8+76 div 41+2+3$$
$$C=frac9+8+7+15 div fracfrac634 div 2$$
answered 8 hours ago
TreninTrenin
8,0031645
$endgroup$
$begingroup$
I'd go for it, and call this a valid solution. Congrats!
$endgroup$
– Brandon_J
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@Uvc Would you consider $frac5+frac9+8+76 div 41+2+3$ a different solution?
$endgroup$
– Trenin
8 hours ago
$begingroup$
@Uvc Or how about $frac9+8+7+15 div fracfrac634 div 2$?
$endgroup$
– Trenin
8 hours ago
$begingroup$
Sure..sufficiently different
$endgroup$
– Uvc
8 hours ago
add a comment |
$begingroup$
OK - here is a real solution this time without concatenation.
I have a couple sub expressions.
$x=frac2+35=1$
$y=frac9+8+76 div 4=1$
Now I can construct $A,B,C$.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac2+35 div frac1frac9+8+76 div 4$$
$$C= frac2+35 div fracfrac9+8+76 div 41$$
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.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac5+frac9+8+76 div 41+2+3$$
$$C=frac9+8+7+15 div fracfrac634 div 2$$
answered 8 hours ago
TreninTrenin
8,0031645
$endgroup$
OK - here is a real solution this time without concatenation.
I have a couple sub expressions.
$x=frac2+35=1$
$y=frac9+8+76 div 4=1$
Now I can construct $A,B,C$.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac2+35 div frac1frac9+8+76 div 4$$
$$C= frac2+35 div fracfrac9+8+76 div 41$$
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.
$$A= 1 div fracfrac2+35frac9+8+76 div 4$$
$$B= frac5+frac9+8+76 div 41+2+3$$
$$C=frac9+8+7+15 div fracfrac634 div 2$$
answered 8 hours ago
TreninTrenin
8,0031645
edited 8 hours ago
answered 8 hours ago
TreninTrenin
8,0031645
answered 8 hours ago
TreninTrenin
8,0031645
answered 8 hours ago
TreninTrenin
8,0031645
8,0031645
$begingroup$
I'd go for it, and call this a valid solution. Congrats!
$endgroup$
– Brandon_J
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@Uvc Would you consider $frac5+frac9+8+76 div 41+2+3$ a different solution?
$endgroup$
– Trenin
8 hours ago
$begingroup$
@Uvc Or how about $frac9+8+7+15 div fracfrac634 div 2$?
$endgroup$
– Trenin
8 hours ago
$begingroup$
Sure..sufficiently different
$endgroup$
– Uvc
8 hours ago
add a comment |
$begingroup$
I'd go for it, and call this a valid solution. Congrats!
$endgroup$
– Brandon_J
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@Uvc Would you consider $frac5+frac9+8+76 div 41+2+3$ a different solution?
$endgroup$
– Trenin
8 hours ago
$begingroup$
@Uvc Or how about $frac9+8+7+15 div fracfrac634 div 2$?
$endgroup$
– Trenin
8 hours ago
$begingroup$
Sure..sufficiently different
$endgroup$
– Uvc
8 hours ago
$begingroup$
I'd go for it, and call this a valid solution. Congrats!
$endgroup$
– Brandon_J
8 hours ago
$begingroup$
I'd go for it, and call this a valid solution. Congrats!
$endgroup$
– Brandon_J
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
@Uvc Would you consider $frac5+frac9+8+76 div 41+2+3$ a different solution?
$endgroup$
– Trenin
8 hours ago
$begingroup$
@Uvc Would you consider $frac5+frac9+8+76 div 41+2+3$ a different solution?
$endgroup$
– Trenin
8 hours ago
$begingroup$
@Uvc Or how about $frac9+8+7+15 div fracfrac634 div 2$?
$endgroup$
– Trenin
8 hours ago
$begingroup$
@Uvc Or how about $frac9+8+7+15 div fracfrac634 div 2$?
$endgroup$
– Trenin
8 hours ago
$begingroup$
Sure..sufficiently different
$endgroup$
– Uvc
8 hours ago
$begingroup$
Sure..sufficiently different
$endgroup$
– Uvc
8 hours ago
add a comment |
$begingroup$
With pencil and paper:
A. $ frac39 + frac16 + frac (4+8) / (5+7) 2 = frac13 + frac16 + frac12 = 1 $
B. $ frac1 + 2 + 3 + 4 + 56 + 7 + 8 + 9 = frac1530 = frac12 $
C. $ frac3 + 7 + 8 + 91 + 2 + 4 + 5 + 6 = frac2718 = frac32 $
Sum: $ 1 + frac12 + frac32 = 3$
answered 8 hours ago
Weather VaneWeather Vane
3,5781118
$endgroup$
$begingroup$
Very good one....
$endgroup$
– Uvc
7 hours ago
$begingroup$
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
7 hours ago
add a comment |
$begingroup$
With pencil and paper:
A. $ frac39 + frac16 + frac (4+8) / (5+7) 2 = frac13 + frac16 + frac12 = 1 $
B. $ frac1 + 2 + 3 + 4 + 56 + 7 + 8 + 9 = frac1530 = frac12 $
C. $ frac3 + 7 + 8 + 91 + 2 + 4 + 5 + 6 = frac2718 = frac32 $
Sum: $ 1 + frac12 + frac32 = 3$
answered 8 hours ago
Weather VaneWeather Vane
3,5781118
$endgroup$
$begingroup$
Very good one....
$endgroup$
– Uvc
7 hours ago
$begingroup$
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
7 hours ago
add a comment |
$begingroup$
With pencil and paper:
A. $ frac39 + frac16 + frac (4+8) / (5+7) 2 = frac13 + frac16 + frac12 = 1 $
B. $ frac1 + 2 + 3 + 4 + 56 + 7 + 8 + 9 = frac1530 = frac12 $
C. $ frac3 + 7 + 8 + 91 + 2 + 4 + 5 + 6 = frac2718 = frac32 $
Sum: $ 1 + frac12 + frac32 = 3$
answered 8 hours ago
Weather VaneWeather Vane
3,5781118
$endgroup$
With pencil and paper:
A. $ frac39 + frac16 + frac (4+8) / (5+7) 2 = frac13 + frac16 + frac12 = 1 $
B. $ frac1 + 2 + 3 + 4 + 56 + 7 + 8 + 9 = frac1530 = frac12 $
C. $ frac3 + 7 + 8 + 91 + 2 + 4 + 5 + 6 = frac2718 = frac32 $
Sum: $ 1 + frac12 + frac32 = 3$
answered 8 hours ago
Weather VaneWeather Vane
3,5781118
edited 7 hours ago
answered 8 hours ago
Weather VaneWeather Vane
3,5781118
answered 8 hours ago
Weather VaneWeather Vane
3,5781118
answered 8 hours ago
Weather VaneWeather Vane
3,5781118
3,5781118
$begingroup$
Very good one....
$endgroup$
– Uvc
7 hours ago
$begingroup$
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
7 hours ago
add a comment |
$begingroup$
Very good one....
$endgroup$
– Uvc
7 hours ago
$begingroup$
I was looking for three $1$s but got the last two fairly quickly when I looked for them. From $sum_1^9 n = 45$.
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– Weather Vane
7 hours ago
$begingroup$
Very good one....
$endgroup$
– Uvc
7 hours ago
$begingroup$
Very good one....
$endgroup$
– Uvc
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
add a comment |
$begingroup$
Here is one solution:
EDIT: Uses concatenations, so not a valid answer - sorry.
$$A=fracfrac785219634$$
$$B=fracfrac695217384$$
$$C=fracfrac579648312$$
answered 8 hours ago
TreninTrenin
8,0031645
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$begingroup$
why the downvote? +1!
$endgroup$
– Omega Krypton
8 hours ago
$begingroup$
Quite clear - no concatentions.
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– Weather Vane
8 hours ago
$begingroup$
@WeatherVane ARG - Didn't see that.... Back to the drawing board...
$endgroup$
– Trenin
8 hours ago
add a comment |
$begingroup$
Here is one solution:
EDIT: Uses concatenations, so not a valid answer - sorry.
$$A=fracfrac785219634$$
$$B=fracfrac695217384$$
$$C=fracfrac579648312$$
answered 8 hours ago
TreninTrenin
8,0031645
$endgroup$
$begingroup$
why the downvote? +1!
$endgroup$
– Omega Krypton
8 hours ago
$begingroup$
Quite clear - no concatentions.
$endgroup$
– Weather Vane
8 hours ago
$begingroup$
@WeatherVane ARG - Didn't see that.... Back to the drawing board...
$endgroup$
– Trenin
8 hours ago
add a comment |
$begingroup$
Here is one solution:
EDIT: Uses concatenations, so not a valid answer - sorry.
$$A=fracfrac785219634$$
$$B=fracfrac695217384$$
$$C=fracfrac579648312$$
answered 8 hours ago
TreninTrenin
8,0031645
$endgroup$
Here is one solution:
EDIT: Uses concatenations, so not a valid answer - sorry.
$$A=fracfrac785219634$$
$$B=fracfrac695217384$$
$$C=fracfrac579648312$$
answered 8 hours ago
TreninTrenin
8,0031645
edited 8 hours ago
answered 8 hours ago
TreninTrenin
8,0031645
answered 8 hours ago
TreninTrenin
8,0031645
answered 8 hours ago
TreninTrenin
8,0031645
8,0031645
$begingroup$
why the downvote? +1!
$endgroup$
– Omega Krypton
8 hours ago
$begingroup$
Quite clear - no concatentions.
$endgroup$
– Weather Vane
8 hours ago
$begingroup$
@WeatherVane ARG - Didn't see that.... Back to the drawing board...
$endgroup$
– Trenin
8 hours ago
add a comment |
$begingroup$
why the downvote? +1!
$endgroup$
– Omega Krypton
8 hours ago
$begingroup$
Quite clear - no concatentions.
$endgroup$
– Weather Vane
8 hours ago
$begingroup$
@WeatherVane ARG - Didn't see that.... Back to the drawing board...
$endgroup$
– Trenin
8 hours ago
$begingroup$
why the downvote? +1!
$endgroup$
– Omega Krypton
8 hours ago
$begingroup$
why the downvote? +1!
$endgroup$
– Omega Krypton
8 hours ago
$begingroup$
Quite clear - no concatentions.
$endgroup$
– Weather Vane
8 hours ago
$begingroup$
Quite clear - no concatentions.
$endgroup$
– Weather Vane
8 hours ago
$begingroup$
@WeatherVane ARG - Didn't see that.... Back to the drawing board...
$endgroup$
– Trenin
8 hours ago
$begingroup$
@WeatherVane ARG - Didn't see that.... Back to the drawing board...
$endgroup$
– Trenin
8 hours ago
add a comment |
$begingroup$
Answer consists of 3 expressions with 3 fractions each
![ 3 as sum of 3 pandigital
]1
answered 5 hours ago
UvcUvc
1,159119
$endgroup$
add a comment |
$begingroup$
Answer consists of 3 expressions with 3 fractions each
![ 3 as sum of 3 pandigital
]1
answered 5 hours ago
UvcUvc
1,159119
$endgroup$
add a comment |
$begingroup$
Answer consists of 3 expressions with 3 fractions each
![ 3 as sum of 3 pandigital
]1
answered 5 hours ago
UvcUvc
1,159119
$endgroup$
Answer consists of 3 expressions with 3 fractions each
![ 3 as sum of 3 pandigital
]1
answered 5 hours ago
UvcUvc
1,159119
answered 5 hours ago
UvcUvc
1,159119
answered 5 hours ago
UvcUvc
1,159119
answered 5 hours ago
UvcUvc
1,159119
1,159119
add a comment |
add a comment |
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$begingroup$
So we are only allowed to use + and $div$? Can we use parenthesis like $A=(1+2) div 3$? Or implied parenthesis by $A=frac1+23$?
$endgroup$
– Trenin
9 hours ago
$begingroup$
Can we concatenate for example use number $123$?
$endgroup$
– Weather Vane
9 hours ago
$begingroup$
Is the usual BEDMAS order of operations assumed?
$endgroup$
– Trenin
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
As an example, one of the terms might look like...9/(1+6)
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– Uvc
9 hours ago