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# n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined

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Math Expert
Joined: 02 Sep 2009
Posts: 51184
n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined  [#permalink]

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11 Jun 2015, 03:46
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95% (hard)

Question Stats:

56% (02:57) correct 44% (03:11) wrong based on 335 sessions

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

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

Kudos for a correct solution.

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Re: n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined  [#permalink]

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11 Jun 2015, 04:33
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$$t_n$$=$$t_{n−1}$$+n
The above follows the below sequence
t0= Odd, t1=Even, t2=E,t3=O,t4=O,t5=E,t6=E,t7=O,t8=O,t9=E,t10=E. (E-Even, O-Odd)
So starting from t1 the sequence will be EEOOEEOOEEOOEEOO.......

(1) n + 1 is divisible by 3

n+1 being divisible by 3, n will be 2,5,8,11,14....
t2 = E, t5=E, t8=O
so we can't say whether tn is either odd or even.
Insufficient

(2) n - 1 is divisible by 4
n-1 being divisible by 4, n will be 5,9,13,17,21....
t5=E,t9=E,t13=E,t17=E ....

Since the sequence is cyclic (EEOO). So tn will be even for all the values of n, where n-1 is divisible by 4.
Sufficient

Ans: B
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Re: n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined  [#permalink]

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11 Jun 2015, 04:44
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Bunuel wrote:
n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined as $$tn = t_{n-1} + n$$. If $$t_0 = 3$$, is $$t_n$$ even?

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

Kudos for a correct solution.

$$tn = t_{n-1} + n$$

i.e. $$t1 = t_0 + 1 = 3+1 = 4$$
i.e. $$t2 = t_1 + 2 = 4+2 = 6$$
i.e. $$t3 = t_2 + 3 = 6+3 = 9$$
i.e. $$t4 = t_3 + 4 = 9+4 = 13$$
i.e. $$t5 = t_4 + 5 = 13+5 = 18$$
i.e. $$t6 = t_5 + 6 = 18+6 = 24$$
i.e. $$t7 = t_6 + 7 = 24+7 = 31$$
i.e. $$t8 = t_7 + 8 = 31+8 = 39$$
i.e. $$t9 = t_8 + 9 = 39+9 = 48$$
i.e. $$t{10} = t_9 + 10 = 48+10 = 58$$
i.e. $$t{11} = t_{10} + 11 = 58+11 = 69$$
and so on...

Eventually we can Observe that 2 terms are Even and the next two terms are odd and this cycle continues

Question : is $$t_n$$ even?

Statement 1: n + 1 is divisible by 3

i.e. n+1 can be 3 or 6 or 9 or 12 etc
i.e. n can be 2 or 5 or 8 or 11 etc

$$t_2$$ is Even
$$t_5$$ is Even
$$t_8$$ is Odd
$$t_{11}$$ is Odd.....i.e. INCONSISTENT answer

Hence, NOT SUFFICIENT

Statement 2: n - 1 is divisible by 4

i.e. n-1 can be 4 or 8 or 12 etc
i.e. n can be 5 or 9 or 13 etc

The terms in this pattern will be at the difference of 4 and the property of all terms at difference of 4 terms is alike

$$t_5$$ is Even
$$t_9$$ is Even
$$t_{13}$$ is EVEN.....i.e. CONSISTENT answer

Hence, SUFFICIENT

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Re: n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined  [#permalink]

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11 Jun 2015, 23:46
Bunuel wrote:
n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined as $$tn = t_{n-1} + n$$. If $$t_0 = 3$$, is $$t_n$$ even?

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

Kudos for a correct solution.

Listing the first couple of iterations:
$$n=0, t_0 = 3$$
$$n=1, t_1=1+1=2$$
$$n=2, t_2=2+2=4$$
$$n=3, t_3=4+3=7$$
$$n=4, t_4=7+4=11$$
$$n=5, t_5=11+5=16$$
$$n=6, t_6=16+6=22$$
$$n=7, t_7=22+7=29$$
So you can see that when $$n=even$$ the pattern is odd, even, odd, even, ..., or odd when n is a multiple of 4 and even otherwise, and when $$n=odd$$ the pattern is even, odd, even, odd, ..., or odd when n+1 is a multiple of 4, and even otherwise.

1: if $$n+1=3, n=2, t_2=4$$ but if $$n+1=9,n=8,t_8=37$$ not sufficient
2: n-1 is divisible by 4 means that n+1 is not a multiple of 4, and so $$t_n$$ will be even. sufficient. the answer is B.
Math Expert
Joined: 02 Sep 2009
Posts: 51184
Re: n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined  [#permalink]

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15 Jun 2015, 04:19
Bunuel wrote:
n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined as $$tn = t_{n-1} + n$$. If $$t_0 = 3$$, is $$t_n$$ even?

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

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Sequence problems are often best approached by charting out the first several terms of the given sequence. In this case, we need to keep track of n, tn, and whether tn is even or odd.
Attachment:

2015-06-15_1618.png [ 47.41 KiB | Viewed 3018 times ]

Notice that beginning with n = 1, a four-term repeating cycle of [even, even, odd, odd] emerges for tn. Thus, a statement will be sufficient only if it tells us how n relates to a multiple of 4 (i.e. n = a multiple of 4 ± known constant).

(1) INSUFFICIENT: This statement does not tell us how n relates to a multiple of 4. If n + 1 is a multiple of 3, then n + 1 could be 3, 6, 9, 12, 15, etc. This means that n could be 2, 5, 8, 11, 14, etc. From the chart, if n = 2 or n = 5, then tn is even. However, if n = 8 or n = 11, then tn is odd.

(2) SUFFICIENT: This statement tells us exactly how n relates to a multiple of 4. If n – 1 is a multiple of 4, then n – 1 could be 4, 8, 12, 16, 20, etc. and n could be 5, 9, 13, 17, 21, etc. From the chart (and the continuation of the four-term pattern), tn must be even.

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Re: n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined  [#permalink]

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13 Jul 2016, 05:40
Bunuel wrote:
n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined as $$tn = t_{n-1} + n$$. If $$t_0 = 3$$, is $$t_n$$ even?

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

Kudos for a correct solution.

t0= 3
t1= 3+1= 4 E
t2=4+2= 6 E
t3= 6+3=9 O
t4= 9+4= 13 O

We see that there is a sequence of EEOOEEOOEEOOEEOO

(1) n + 1 is divisible by 3
n can be 2, 5, 8 and hence it can be E or O
Not Sufficient.

(2) n - 1 is divisible by 4
n can be 5, 9, 13, 17 and it will always be E.
Sufficient.

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Re: n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined  [#permalink]

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17 Aug 2018, 19:17
1
my approach:
Tn=T0+n(n+1)/2
this is probably the quickest approach to tackle the problem.
Re: n is an integer such that n ≥ 0. For n > 0, the sequence tn is defined &nbs [#permalink] 17 Aug 2018, 19:17
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