0 + 0 − 0 0 × 0 ÷ 0 = = 0 0
Put one of the integers 1 , 2 , … , 9 into each of the boxes, such that eight of these numbers are used once (and one number is not used at all) and both equations are true.
What is the sum of all possible values of the missing number?
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Another Scenario is (2 x 3)/6 = 1 ; 4 + 9 - 8 = 5, so we're also left with 7
And another way is to inverse position of 6 and 1 in your solution Joshua... I think this problem might be wrong...
Other solution... 1 × 6 / 3 = 2 and 7 + 5 − 8 = 4 . Where is the 9?...
1*6:2=3 4+8-5=7 9 is left out
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The only two possible scenarios are: 1. (9 x 2) / 3=6 , 8+4-7=5 which 1 is left out and 2. (8 x1) / 2 = 4, 9+3-5=7 which 6 is left out.