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If a ⊕ B = n (n is a constant), then (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2
When a ♁ B = n (n is a constant), we can get (a + 1) ♁ B = n + 1, a ♁ (B + 1) = n-2.
Now we know that 1 ♁ 1 = 2, then 2008 ♁ 2008=____
If a ⊕ B = n (n is a constant), then (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2
When a ♁ B = n (n is a constant), we can get (a + 1) ♁ B = n + 1, a ♁ (B + 1) = n-2.
Now we know that 1 ♁ 1 = 2, then 2008 ♁ 2008=____
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