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I hadn't thought of that. That way the solution could be something else. Let's give it a try!
ā£ 1 = a
ā£ 2 = b
ā£ 3 = c
a = b + c + 5
b = ā(ab3)
c = min(a, b , 1)
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That's up to you to decide š
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1
,2
, and3
that label each panel count?