A student makes the following claim:
For all integers n, the expression 4(9n+12 − 3n−12) is divisible by 3.
Here is the student's argument:
4(9n+12 − 3n−12) = 2(2(9n+12 − 3n−12)) (I)
= 2(9n + 1 − 3n − 1) (II)
= 2(6n) (III)
= 12n (IV)
= 3(4n) (V)
which is always a multiple of 3. (VI)
So the expression 4(9n+12 − 3n−12) is always divisible by 3.
Which one of the following is true?
- AThe argument is correct.
- BThe argument is incorrect, and the first error occurs on line (I).
- CThe argument is incorrect, and the first error occurs on line (II).
- DThe argument is incorrect, and the first error occurs on line (III).
- EThe argument is incorrect, and the first error occurs on line (IV).
- FThe argument is incorrect, and the first error occurs on line (V).
- GThe argument is incorrect, and the first error occurs on line (VI).
Show the answer and worked solution
answer · C
- AThe argument is correct.
- BThe argument is incorrect, and the first error occurs on line (I).
- CThe argument is incorrect, and the first error occurs on line (II).
- DThe argument is incorrect, and the first error occurs on line (III).
- EThe argument is incorrect, and the first error occurs on line (IV).
- FThe argument is incorrect, and the first error occurs on line (V).
- GThe argument is incorrect, and the first error occurs on line (VI).
Line (I) is fine: writing 4X as 2(2X) changes nothing. The failure is at line (II), where multiplying the bracket by 2 clears both denominators but the minus sign is not carried through the second fraction. It should read 2((9n+1) − (3n−1)) = 2(9n + 1 − 3n + 1), giving 2(6n+2) = 12n + 4, not 12n. So the claim itself is also false: 12n + 4 is never a multiple of 3 when n is a multiple of 3, for instance n = 0 gives 4. Lines (III) to (VI) are each valid consequences of the line before them, so (II) is the first error.