I solved it a different way. Take a toothpick from the 2nd group and place it over the equal sign to represent “not equal to”. 1 + 1 + 111 ≠ 1111
Colin
Move a toothpick over the equal sign to make ≠
cthellis
Kept trying to make it work in binary, but could only do that with "add a toothpick" not "move a toothpick" (Thought that would be perhaps a good spin on it. Or a round two.)
1 + 111 + 111 = 1111 1 + (7) + (7) = (15)
InfinitiveZ
1+11+11≠1111
Bret M. Green
I went with the roman numeral thing. Move 1 stick from the II stick and lay it across the IIII sticks. i + I + III = IIII
Angelique A
Solved it but with another way.. I think. If you take the horizontal stick on the second plus sign. Make a lopsided V with the first stick to make it a Roman 5. It would be V + 1 – 3 = 4. Don't know if that counts though.
Comments
8th grade is way harder than I remember.
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I solved it a different way.
Take a toothpick from the 2nd group and place it over the equal sign to represent “not equal to”.
1 + 1 + 111 ≠ 1111
Move a toothpick over the equal sign to make ≠
Kept trying to make it work in binary, but could only do that with "add a toothpick" not "move a toothpick" (Thought that would be perhaps a good spin on it. Or a round two.)
1 + 111 + 111 = 1111
1 + (7) + (7) = (15)
1+11+11≠1111
I went with the roman numeral thing. Move 1 stick from the II stick and lay it across the IIII sticks. i + I + III = IIII
Solved it but with another way.. I think. If you take the horizontal stick on the second plus sign. Make a lopsided V with the first stick to make it a Roman 5. It would be V + 1 – 3 = 4. Don't know if that counts though.