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Question Stats:
59% (02:50) correct
41%
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n is an integer greater than or equal to 0. The sequence \(t_n\) for n > 0 is defined as \(t_n = t_{n-1} + n\). Given that \(t_0 = 3\), is tn even?
(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4
(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4
16 Mar 2011, 02:00
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See, t0 is odd.
Then we add an odd number (n=1), so t1 is even.
Then we add an even number(n=2), so t2 is even.
Then we add an odd number (n=3), so t3 is odd.
Then we add an even number(n=4), so t4 is odd.
Then we add an odd number(n=5), so t5 is even.
Then we add an even number(n=6), so t6 is even.
Then we add an odd number(n=7), so t7 is odd.
Then we add an even number(n=8), so t8 is odd.
Etc........
As you could see here the sequence is connected with divisibility by 4.
So (1) tells us about divisibility by 3 and it should be not sufficient. See countrexample: t2 is even (n+1 is 3), t8 is odd (n+1=9)
(2) alone should be sufficient since it tells us about the divisibility by 4, and we see that n which are divided by 4 with remainder of 1 or 2 is even. If the remainder is 0 or 3, it is odd. Here the reminder is 1 (n-1 is divided by 4), so the term should be even.
The answer is (B)
Then we add an odd number (n=1), so t1 is even.
Then we add an even number(n=2), so t2 is even.
Then we add an odd number (n=3), so t3 is odd.
Then we add an even number(n=4), so t4 is odd.
Then we add an odd number(n=5), so t5 is even.
Then we add an even number(n=6), so t6 is even.
Then we add an odd number(n=7), so t7 is odd.
Then we add an even number(n=8), so t8 is odd.
Etc........
As you could see here the sequence is connected with divisibility by 4.
So (1) tells us about divisibility by 3 and it should be not sufficient. See countrexample: t2 is even (n+1 is 3), t8 is odd (n+1=9)
(2) alone should be sufficient since it tells us about the divisibility by 4, and we see that n which are divided by 4 with remainder of 1 or 2 is even. If the remainder is 0 or 3, it is odd. Here the reminder is 1 (n-1 is divided by 4), so the term should be even.
The answer is (B)
16 Mar 2011, 04:57
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t0 = 3,
So t1 = t0 + n = 3 + n
t2 = 3 + n + n = 3 + 2n
t3 = 3 + 2n + n = 3 + 3n
So tn is even only if n is odd as odd + odd, if n is even then it will be odd + even = odd
From (1) we have n + 1 is divisible by 3 => n + 1 = 3k where k is an integer.
So n = 3k - 1 which can be values like 2, 5, 8, so (1) is insufficient.
From (2) we have n-1 = 4m where m is an integer
So n = 4m + 1, which is always odd
Hence (2) is sufficient, so the answer is B.
So t1 = t0 + n = 3 + n
t2 = 3 + n + n = 3 + 2n
t3 = 3 + 2n + n = 3 + 3n
So tn is even only if n is odd as odd + odd, if n is even then it will be odd + even = odd
From (1) we have n + 1 is divisible by 3 => n + 1 = 3k where k is an integer.
So n = 3k - 1 which can be values like 2, 5, 8, so (1) is insufficient.
From (2) we have n-1 = 4m where m is an integer
So n = 4m + 1, which is always odd
Hence (2) is sufficient, so the answer is B.
