Data Sufficiency

n is an integer such that n>= 0. For n > 0, the sequence t_n is defined as t_n = t_{n-1} + n. Given that t_0 = 3, is t_n even?

(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4


OA : B

@ Experts - How we can solve this in least possible time and is there any other way to tackle this type of problems rather than plugging in values & subsequently observing the pattern ?

RBBmba@2014 wrote:n is an integer such that n>= 0. For n > 0, the sequence t_n is defined as t_n = t_{n-1} + n. Given that t_0 = 3, is t_n even?

(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4

WRITE OUT a few terms and LOOK FOR A PATTERN.

tâ‚€ = 3.
t� = t₀ + 1 = 3 + 1 = 4.
t₂ = t� + 2 = 4 + 2 = 6.
t₃ = t₂ + 3 = 6 + 3 = 9.
t₄ = t₃ + 4 = 9 + 4 = 13.
tâ‚… = tâ‚„ + 5 = 13 + 5 = 18.
t₆ = t₅ + 6 = 18 + 6 = 24.
t₇ = t₆ + 7 = 24 + 7 = 31.
t₈ = t₇ + 8 = 31 + 8 = 39.

The resulting values after tâ‚€ -- 4, 6, 9, 13, 18, 24, 31, 39 -- imply the following pattern:
EVEN, EVEN, ODD, ODD, EVEN, EVEN, ODD, ODD...

Statement 1: n + 1 is divisible by 3
Statement 1 is satisfied by the options in red:
tâ‚€ = 3.
t� = t₀ + 1 = 3 + 1 = 4.
t₂ = t� + 2 = 4 + 2 = 6.
t₃ = t₂ + 3 = 6 + 3 = 9.
t₄ = t₃ + 4 = 9 + 4 = 13.
tâ‚… = tâ‚„ + 5 = 13 + 5 = 18.
t₆ = t₅ + 6 = 18 + 6 = 24.
t₇ = t₆ + 7 = 24 + 7 = 31.
t₈ = t₇ + 8 = 31 + 8 = 39.
Since t_n can be EVEN or ODD, INSUFFICIENT.

Statement 2: n - 1 is divisible by 4
Statement 2 is satisfied by the options in blue:
tâ‚€ = 3.
t� = t₀ + 1 = 3 + 1 = 4.
t₂ = t� + 2 = 4 + 2 = 6.
t₃ = t₂ + 3 = 6 + 3 = 9.
t₄ = t₃ + 4 = 9 + 4 = 13.
tâ‚… = tâ‚„ + 5 = 13 + 5 = 18.
t₆ = t₅ + 6 = 18 + 6 = 24.
t₇ = t₆ + 7 = 24 + 7 = 31.
t₈ = t₇ + 8 = 31 + 8 = 39.
t₉ = EVEN.
t�₀ = EVEN.
t�� = ODD.
t�₂ = EVEN.
t�₃ = EVEN.
In every case, t_n is EVEN.
SUFFICIENT.

The correct answer is B.

Thanks Mitch for your reply.

I did it in the same way but I landed up using 3+ minutes, so I thought whether there is any smarter way to solve this ?

RBBmba@2014 wrote:Thanks Mitch for your reply.

I did it in the same way but I landed up using 3+ minutes, so I thought whether there is any smarter way to solve this ?

tâ‚€ = 3.
t� = t₀ + 1 = 3 + 1 = 4.
t₂ = t� + 2 = 4 + 2 = 6.
t₃ = t₂ + 3 = 6 + 3 = 9.
t₄ = t₃ + 4 = 9 + 4 = 13.
tâ‚… = tâ‚„ + 5 = 13 + 5 = 18.
t₆ = t₅ + 6 = 18 + 6 = 24.
t₇ = t₆ + 7 = 24 + 7 = 31.
t₈ = t₇ + 8 = 31 + 8 = 39.

The sequence is composed of a repeating 4-term cycle:
EVEN, EVEN, ODD, ODD.
As indicated by the options in red, the cycle ENDS when n is a MULTIPLE OF 4.
The end of a cycle is always ODD.
The beginning of a cycle is always EVEN.

Statement 1: n + 1 is divisible by 3
Options for n+1 = 3, 6, 9, 12...
Options for n = 2, 5, 8, 11...

Case 1: n=5
Since tâ‚„ is the END of a cycle, tâ‚… is the BEGINNING of a new cycle, with the result that tâ‚… is EVEN.

Case 2: n=8
Since t₈ is the END of a cycle, t₈ is ODD.
INSUFFICIENT.

Statement 2: n - 1 is divisible by 4
Since t_(n-1) is the END of a cycle, t_n is the BEGINNING of a new cycle, with the result that t_n is EVEN.
SUFFICIENT.

The correct answer is B.