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by mariah » Sun May 08, 2011 10:48 am
n is an integer greater than or equal to 0. The sequence tn for n > 0 is defined as
tn = tn-1 + n. Given that t0 = 3, is tn even?

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


(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

[spoiler]oa b [/spoiler]
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Source: — Data Sufficiency |
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by clock60 » Sun May 08, 2011 12:57 pm
hi mariah
my thinking, perhaps not optimal
t0=3
t1=4
t2=6
t3=9
t4=13
t5=18
t6=24
t7=31
t8=39... and so on
one thing to notice is that after t0 all terms have recurrent set
(even even) (odd odd) (even even) (odd odd) -i mean 2-even and immediately after 2-odd
we need to find tn-?
(1)n+1=3k. n=3k-1
k=1, n=2, t2=6-even
k=2, n=5, t5=18-even
k=3, n=8, t8=39-odd
as far as we have both even and odd tn,
i st insuff
(2)n-1=4k,
k=0, n=1, t1=4 even
k=1, n=5, t5=18-even
k=2, n=9, t9=40, even
i say that every 5th n will be even from recurrency, as i mentioned earlier
even even odd odd even even odd odd even even odd odd even even
for this reason st 2 suff
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by sourabh33 » Sun May 08, 2011 5:37 pm
IMO - B

T0 - 3
T1 - 3+n
T2 - 3+2n
T3 - 3+3n
Tn - 3+n.n

Now for Tn to be even n has to be odd

Statement 1

n+1 is divisible by 3
Testing numbers
Case 1 - n=2 ---> n+1 is divisible by 3
Case 2 - n=5 ---> n+1 is divisible by 3

Since n could be both even and odd, statement 1 is insufficient

Statement 2

if n - 1 is divisible by 4, the n will always be odd (we can test by picking nos.)

Therefore Statement 2 is sufficient
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by manpsingh87 » Sun May 08, 2011 11:11 pm
mariah wrote:n is an integer greater than or equal to 0. The sequence tn for n > 0 is defined as
tn = tn-1 + n. Given that t0 = 3, is tn even?

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


(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

[spoiler]oa b [/spoiler]
tn=tn-1+n;

1)
n+1 is divisible by 3;

adding 1 on both sides of the equation tn=tn-1+n we have;
(tn) +1= t(n-1)+n+1; as n+1 is divisible by 3; therefore n+1=3k;

(tn)- t(n-1)+1=3k;
i.e. (tn)-t(n-1) will result in a number which will be 1 less than a multiple of 3, i.e. 2,5,8 etc.

now consider values of tn and t(n-1);
t0=3;t1=4;t2=6;t3=9;t4=13;t5=18;t6=24;t7=31;t8=39;

now consider t2-t1=6-2; here t2 is even i.e. tn is even and tn-t(n-1)+1 is divisible by 3;
now consider t8-t7=39-31; here t8 is odd i.e. tn is odd and tn-t(n-1)+1 is divisible by 3;
as tn can be both even as well as odd hence 1 alone is not sufficient to answer the question.

2)tn=t(n-1)+n;
subtract 1 from both sides we have;
tn-1=t(n-1)+n-1;
tn-t(n-1)-1=n-1;
now since n-1 is divisible by 4; therefore n-1=4k;
tn-t(n-1)-1=4k;
right hand side is even; therefore left hand side must also be even; tn-t(n-1)-1; now since 1(which is odd) is subtracted from tn-t(n-1) which makes tn-t(n-1)-1 even; therefore tn-(tn-1) must be odd;
also tn-t(n-1) can be written as 4p+1; i.e. difference between them is of the form 5,9,13,,,;
t5-t4=18-13=5; t9-t8=48-39=9; (i.e. difference between the term is in a.p. with a common difference of 4) therefore next term will bet t13-t12=13; also if we observe the pattern of the terms of the series we will observe that two even terms are followed by two odd terms,, therefore pattern will be repeated after every 4 terms, therefore tn will always be an even.

hence answer should be B..!!!
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by kumadil2011 » Thu Nov 17, 2011 10:02 pm
Experts..

COULD YOU PLEASE HELP, HOW CAN WE SOLVE THIS USING ALGEBRA..AS NUMBER PICK IS DANGEROUS SOME TIMES IF WE DON'T PICK THE RIGHT ONES
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by neelgandham » Mon Nov 21, 2011 5:04 am
kumadil2011 wrote:Experts..
COULD YOU PLEASE HELP, HOW CAN WE SOLVE THIS USING ALGEBRA..AS NUMBER PICK IS DANGEROUS SOME TIMES IF WE DON'T PICK THE RIGHT ONES
Though not an expert, I will try to help you with this :-)
n is an integer greater than or equal to 0. The sequence tn for n > 0 is defined as
tn = tn-1 + n. Given that t0 = 3, is tn even?
(1) n + 1 is divisible by 3
(2) n - 1 is divisible by 4
t1 - t0 = 1 => t1 = to + 1
t2 - t1 = 2 => t2 = t1 + 2 = to + 1 + 2
t3 - t2 = 3 => t3 = t2 + 3 = t1 + 2 + 3 = to + 1 + 2 + 3

tn = (1+2+3+4+....n) + t0 = ((n)*(n+1)*0.5 )+3
(1) n + 1 is divisible by 3
Let n+1 = 3x
Then tn = ((n)*(n+1)*0.5 )+3 = (3x*(3x-1)*0.5) +3. In this case the value of 3x can be a multiple of 2 or 4, or the value of 3x-1 can be a multiple of 2 or 4.Let us do the IF analysis here

If 3x is a multiple of 2
tn = ((n)*(n+1)*0.5 )+3 = (3x*(3x-1)*0.5) +3 = odd number + 3 = even
If 3x is a multiple of 4
tn = ((n)*(n+1)*0.5 )+3 = (3x*(3x-1)*0.5) +3 = Even number + 3 = odd
You need not check for the values of (3x-1).
(2) n - 1 is divisible by 4
let n-1 =4x
Then tn = ((n)*(n+1)*0.5 )+3 = ((4x+1)*(4x+2)*0.5) +3 = ((4x+1)*(2x+1)) +3
(4x+1) = odd
(2x+1) = odd
(2x+1)*(4x+1) = odd
So tn = ((4x+1)*(2x+1)) +3 = odd + 3 = odd
Sufficient !

Hope that helps !
Anil Gandham
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by ollapodrida » Mon Nov 21, 2011 8:26 am
sourabh33 wrote:IMO - B

T0 - 3
T1 - 3+n
T2 - 3+2n
T3 - 3+3n
Tn - 3+n.n
This doesn't look right to me.
tn = tn-1 + n
t1 = t0+n = 3+1 = 4
t2 = t1+n = 4+2 = 6: This is not equal to 3+2(2)=7 ???
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by neelgandham » Mon Nov 21, 2011 8:28 am
ollapodrida wrote:
sourabh33 wrote: This doesn't look right to me.
tn = tn-1 + n
t1 = t0+n = 3+1 = 4
t2 = t1+n = 4+2 = 6: This is not equal to 3+2(2)=7 ???
It isn't !
tn = ((n)*(n+1)*0.5 )+3 is the correct notation IMO !
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