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f x, y and z are consecutive positive integers such that x < y < z, then which of the following must be true ?I. xyz is divisible by 6.II. (z – x) (y – x + 1 ) = 4III. xy is odd.

Question

f x, y and z are consecutive positive integers such that x < y < z, then which of the following must be true ?I. xyz is divisible by 6.II. (z – x) (y – x + 1 ) = 4III. xy is odd.

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Solution

I. This statement is true. Since x, y, and z are consecutive integers, one of them must be divisible by 2 (even) and one must be divisible by 3. Therefore, their product xyz will be divisible by 6.

II. This statement is false. If we substitute y and z with x+1 and x+2 respectively (since they are consecutive integers), the equation becomes (x+2 - x) * ((x+1) - x + 1) = 4, which simplifies to 2*2 = 4. This is true for all x, so the statement is always true.

III. This statement is false. Since x, y, and z are consecutive integers, one of them must be even. If x is even, then xy is even. If x is odd, then y = x + 1 is even, so xy is still even. Therefore, xy cannot be odd.

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