Deceptive looks
If a = 1 0 0 0 , b = 1 0 0 , c = 1 0 , and d = 1 , evaluate
( a + b + c − d ) + ( a + b − c + d ) + ( a − b + c + d ) + ( − a + b + c + d ) .
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2 solutions
a + b + c + d = 1 0 0 0 + 1 0 0 + 1 0 + 1 = 1 1 1 1 N o w , ( a + b + c − d ) + ( a + b − c + d ) + ( a − b + c + d ) + ( − a + b + c + d ) = a + b + c − d + a + b − c + d + a − b + c + d − a + b + c + d = a + b + c + d + a + b + c + d = 2 a + 2 b + 2 c + 2 d = 2 ( a + b + c + d ) = 2 ( 1 1 1 1 ) = 2 2 2 2
( a + b + c − d ) + ( a + b − c + d ) + ( a − b + c + d ) + ( − a + b + c + d )
= a + b + c − d + a + b − c + d + a − b + c + d − a + b + c + d
= 2 ( a + b + c + d )
= 2 ( 1 0 0 0 + 1 0 0 + 1 0 + 1 ) = 2 × 1 1 1 1 = 2 2 2 2