Solve the kenken using the bitwise operations indicated:
| 000⊕ 011 |
001 |
010 |
000 |
111⊕ 110 |
110⊕ 111 |
100 |
101 |
| 110∧ 110 |
111⊕ 000 |
111 |
101∨ 100 |
001 |
111⊕ 101 |
010 |
101⊕ 011 |
111 |
110 |
000 |
101 |
100 |
001∧ 011 |
001 |
010 |
| 011∨ 000 |
111 |
001∧ 011 |
001 |
010∧ 010 |
110 |
111MAJ 101 |
100 |
010 |
100∧ 100 |
011⊕ 101 |
110 |
001∨ 000 |
001 |
011 |
111⊕ 111 |
001 |
101 |
100∧ 100 |
111 |
011∨ 011 |
010 |
110 |
000 |
| 100 100 |
100⊕ 010 |
110 |
110⊕ 011 |
101 |
100∨ 000 |
111 |
111⊕ 001 |
| 110⊕ 101 |
011 |
011MAJ 001 |
010 |
111 |
100 |
000 |
110 |
The highlighted cells, in top to bottom, left to right reading order, give the bit string 000 110 111 101 110 001 000 100 110 100 010 010 111 101 110. Regrouping this bit string into 5-bit substrings (so that we can convert to numbers 1-26 and read as letters) yields the string 00011 01111 01110 00100 01001 10100 01001 01111 01110 or 3 15 14 4 9 20 9 15 14 in decimal or the final answer of CONDITION as letters.