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2 solutions
This is an interesting observation, given the examples in the problem, but note this definition isn't correct for the bitwise XOR in general.
For instance, we have 5 ⊕ 1 7 = 1 7 ⊕ 5 = 2 0 .
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In the original problem + sign was used instead of ⊕ . I was not solving it basing on XOR despite the title.
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@Chew-Seong Cheong Problem is fixed now. Sorry for inconvenience cause
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@Mohd Aasif – Not a problem. I have provided a solution.
- 1 5 1 0 ⊕ 1 4 1 0 = 1 1 1 1 2 ⊕ 1 1 1 0 2 = 0 0 0 1 2
- 0 0 0 1 2 means that 1 1 0 . So the final answer is 1.
1 5 1 0 ⊕ 1 4 1 0 = 1 1 1 2 ⊕ 1 1 0 2 = 1 2 = 1 1 0 = 1