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The expression x(xy + xz) + y(xy + yz) + z(xz + yz) is even if and only if:1] x is even. 2) xy is odd. 3) xyz is odd. 4] Any of these. |
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Answer» Given : X(xy + XZ) + y(xy + yz) + z(xz + yz) is even To Find : EXPRESSION is even if and only if: 1] x is even. 2) xy is odd. 3) XYZ is odd. 4] Any of these. Solution: x(xy + xz) + y(xy + yz) + z(xz + yz) = x(xy) + x(xz) + y(xy) + y(yz) + z(xz) + z(yz) = xy(x + y) + xz(x + z) + yz(y + z) Each term is of form ab(a + b) case 1 : a odd , b odd => odd * odd ( odd + odd) = odd ( even) = even case 2 : a odd , b even => odd * even ( odd + even) = even ( odd ) = even case 3 : a even , b odd => even * odd ( even + odd) = even ( odd ) = even case 4 : a even , b even => even * even ( even + even) = even ( even) = even Hence whatever the combination is , each term is even even + even + even = even => xy(x + y) + xz(x + z) + yz(y + z) is always even Hence x(xy + xz) + y(xy + yz) + z(xz + yz) is always even independent of x , y & z being odd/even x is even. xy is odd if x or/both y are odd xyz is odd. if any one of x/y/z is odd or all x , y , z are odd Hence Any of these. is correct Learn More: the sum of three consecutive odd numbers is 147 find the numbers ... The odd numbers are divided into groups thus (1,3), (5,7,9,11), (13 ... Which of the following is always even options (a) Sum of three odd ... |
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