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Do the 1, 2, and 3 that label each panel count?
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 = āˆš(ab3) c = min(a, b , 1)
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That's up to you to decide šŸ˜œ
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