MHB Find the total numbers to make A=21

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To achieve a total of A=21 using the numbers 1 through 9 with "+" and "-" operations, participants discuss various combinations and calculations. The consensus is that certain combinations can yield the desired total, while others are ruled out. Specifically, it is demonstrated that achieving A=12 is impossible due to the limitations of the number range and operations. The thread emphasizes the importance of strategic placement of "+" and "-" to reach the target. Ultimately, the challenge illustrates the mathematical exploration of combinations to achieve specific sums.
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$A=$$\square$1 $\square$2 $\square$3$\square$4 $\square$5$\square$6 $\square$7 $\square$8 $\square$9
randomly fill in each blank with eather $"+"$ or $"-"$ ,
(1) prove $A$ can not be $12$
(2) find the total numbers to make $A=21$
 
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Albert said:
$A=$$\square$1 $\square$2 $\square$3$\square$4 $\square$5$\square$6 $\square$7 $\square$8 $\square$9
randomly fill in each blank with eather $"+"$ or $"-"$ ,
(1) prove $A$ can not be $12$
(2) find the total numbers to make $A=21$

If there is + before each number we have sum of numbers = 45
when we convert a + to a - say before n we subtract the value by 2n, hence the number shall remain odd
so $A$ cannot be 12.
we need to make the result 21 so subtract 24 so we need to choose numbers whose sum is 12 and change plus to -
the numbers are (9,3), (9,2,1), (8,4), (8,3,1), (7,5) , (7,4,1), (7,3,2), (6,5,1), (6,4,2),(6,3,2,1), (5,4,3) (5,4,2,1) that is 12 combinations.
 

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